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Working with Polynomials:
A How-To Manual
Kent M. Willis
RaptorMath.com
Copyright © 2015 by Kent M. Willis. All rights reserved.
One of the most troublesome parts of algebra for many people is dealing with polynomials,
especially factoring them. It has so many rules, and “if this, then that” situations that a lot
of people are tempted to give up on it. Instead, you should take the time to work on these
skills until you feel fairly confident that you can handle one kind of factoring — then move
on to the next and practice until you conquer that one too.
Then comes the ultimate challenge: factoring any kind of polynomial. Don’t worry about
that one until you are ready for it — first you need to learn factoring one step at a time.
So, first we will cover the basics of how to add, subtract, multiply, divide, and evaluate
polynomials; these skills must be mastered before you begin factoring. When you are good
at them, master one kind of factoring at a time. Get plenty of practice — this stuff is not
something that you can merely read if you want to retain it.
Some places in this book have optional topics that appear in shaded boxes. Study these only
if you are interested in learning a little more. Often this is material on why something works
instead of merely how to use a procedure. Sometimes it is a topic that helps to prepare a
student for courses beyond Algebra, such as Calculus. You may always skip over the shaded
boxes without fear that you are missing out on something you need to learn for Algebra.
Sample of a Shaded Box
The material in a shaded box is optional.
1
Contents
1
Basics of Polynomials
4
1.1
Prime Factorization of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.1.1
Best Method for Prime Factorization . . . . . . . . . . . . . . . . . . . .
4
1.1.2
Using a Cheat Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Greatest Common Factor (GCF) . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.2.1
Finding the GCF of Natural Numbers . . . . . . . . . . . . . . . . . . .
6
1.2.2
Finding the GCF of Terms with Variables . . . . . . . . . . . . . . . . .
7
Evaluating Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.3.1
Plug-in Method of Evaluation . . . . . . . . . . . . . . . . . . . . . . . .
8
1.3.2
Synthetic evaluation: a nifty alternative . . . . . . . . . . . . . . . . . .
9
1.3.3
(Optional) Why synthetic evaluation is possible . . . . . . . . . . . . .
11
Adding and Subtracting Polynomials . . . . . . . . . . . . . . . . . . . . . . .
11
1.4.1
Adding Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.4.2
Subtracting Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
1.4.3
Working with more than one variable . . . . . . . . . . . . . . . . . . .
12
Multiplying Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
1.5.1
Multiplying One Term times One Term (One-by-One) . . . . . . . . . .
13
1.5.2
Multiplying One Term times Two or More Terms (One-by-More) . . .
14
1.5.3
Multiplying Two or More Terms times Two or More Terms (More-byMore) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
1.5.4
Procedure: Multiplying on the Grid . . . . . . . . . . . . . . . . . . . .
15
1.5.5
Procedure: Clarence the Clam . . . . . . . . . . . . . . . . . . . . . . . .
16
1.5.6
Multiplying More Things Together . . . . . . . . . . . . . . . . . . . . .
17
Special Products (Shortcuts) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.6.1
Squaring a binomial (Double-Stuff Shortcut) . . . . . . . . . . . . . . .
17
1.6.2
Multiplying a Sum and Difference (No-Stuff Shortcut) . . . . . . . . .
19
1.6.3
Finding Greater Powers of a Polynomial . . . . . . . . . . . . . . . . .
19
Dividing Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
1.7.1
Dividing one term by one term (one-by-one) . . . . . . . . . . . . . . .
21
1.7.2
Dividing two or more terms by one term (more-by-one) . . . . . . . .
21
1.7.3
Dividing two or more terms by two or more terms (more-by-more) . .
22
1.7.4
Alternate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
1.2
1.3
1.4
1.5
1.6
1.7
2
2
3
Introduction to Factoring
26
2.1
Earning a Black Belt in Factoring . . . . . . . . . . . . . . . . . . . . . . . . . .
26
2.2
How to know when you are done factoring . . . . . . . . . . . . . . . . . . . .
27
Fundamental Factoring Methods
28
3.1
White Belt: Factoring out the Greatest Common Factor (GCF) . . . . . . . . .
28
3.2
Yellow Belt: Factoring by Grouping . . . . . . . . . . . . . . . . . . . . . . . . .
29
3.2.1
Unfactorable polynomials . . . . . . . . . . . . . . . . . . . . . . . . . .
32
3.2.2
Some tips for oddball problems: . . . . . . . . . . . . . . . . . . . . . .
32
3.3
3.4
4
5
6
x2
+ bx + c . . . . . . . . . . . . . . . .
33
3.3.1
The Key Number Method . . . . . . . . . . . . . . . . . . . . . . . . . .
33
3.3.2
Special notes regarding negatives . . . . . . . . . . . . . . . . . . . . .
34
Orange Belt: Factoring Bare Trinomials
Purple Belt: Factoring General Trinomials
ax2
+ bx + c . . . . . . . . . . . . .
35
Special Pattern Factorizations
38
4.1
Green Belt: Factoring the Difference of Squares . . . . . . . . . . . . . . . . . .
38
4.2
Blue Belt: Factoring the Sum or Difference of Cubes . . . . . . . . . . . . . . .
39
4.3
Red Belt: Factoring Perfect Square Trinomials . . . . . . . . . . . . . . . . . . .
41
Mixed Factoring (Conclusion of Red Belt)
42
5.1
How to Choose the Correct Factoring Method . . . . . . . . . . . . . . . . . .
42
5.2
Things to Watch Out For . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
Brown Belt: Advanced Factoring
43
6.1
Basic Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
6.2
Advanced Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
6.3
Using Verified Roots to Factor Polynomials . . . . . . . . . . . . . . . . . . . .
48
6.3.1
Listing all possible rational roots . . . . . . . . . . . . . . . . . . . . . .
49
6.3.2
Using a Graphing Calculator to Narrow the Choices . . . . . . . . . . .
50
Analyzing graphs to find roots . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
6.4.1
Factors associated with roots . . . . . . . . . . . . . . . . . . . . . . . .
51
6.4.2
Number of “zero roots” (roots equal to zero) . . . . . . . . . . . . . . .
51
6.4.3
Using a graph to find non-zero real roots . . . . . . . . . . . . . . . . .
52
6.4.4
Adjusting for integer coefficients . . . . . . . . . . . . . . . . . . . . . .
53
6.4.5
Looking for multiplicity . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
6.4.6
Irrational and Irreal roots always come in pairs . . . . . . . . . . . . . .
53
6.4
A Appendix of Reference Materials
54
3
1
Basics of Polynomials
1.1
Prime Factorization of Integers
This is a process of breaking down a number into its prime factors. There are several ways
to do this. Textbooks usually show the tree method of factoring (which probably everyone
has seen at least once), but there are several shortcomings to this method:
• Suppose you are given a number that you can’t think of how to get there with multiplying two other numbers? Do you just give up and say it’s prime after 5 minutes, 20
minutes, or 4 weeks?
• The ends of the sticks with the prime factors are jagged and all over the place and not
in any order. It’s too easy to miss seeing one when you are collecting them for the final
answer.
• You have to sort your answer out at the end unless you are happy with something like
17 · 7 · 2 · 5 · 2 · 7 · 2. I would much rather see this: 23 · 5 · 72 · 17.
Fortunately, there is a method that fixes all these shortcomings.
1.1.1
Best Method for Prime Factorization
Keep a list of prime numbers at least up through 20 handy! 2, 3, 5, 7, 11, 13, 17, 19
1. We always start with 2 as the Number To Check (NTC) because it’s the smallest prime
number. We will repeat the following substeps, working our way up through the prime
numbers, small to large.
(a) See if the number is divisible by the NTC.
(b) If it is, do the division.
(c) See if the answer is also divisible by the NTC. Repeat until the answer is “no, it is
not divisible by the NTC.”
(d) Choose the next prime number from the list as the new NTC. Don’t skip to a larger
one or you won’t know when to quit. Repeat from step 1a for the new NTC.
2. Keep repeating for bigger and bigger NTCs until one of these things happens:
(a) the answer to a division problem is a prime number, or
(b) the answer to an attempted division is smaller than the NTC.
(c) If there were no successful divisions before the answer to an attempted division
was smaller than the NTC, then the number you were working with was prime.
3. The prime factors are the collection of numbers you divided by together with the prime
answer to the last division.
4
There is only one correct answer, and everyone should get the same one correct answer by
any valid method.
Example: Find the prime factorization of 60.
1. Try dividing by 2 repeatedly until it doesn’t work any more:
60 ÷ 2
30 ÷ 2
15
Notice the prime factors are collecting on the right side after the ÷ signs.
2. Now we’re done with the 2s, so try dividing by 3, since it is the next prime number:
60 ÷ 2
30 ÷ 2
15 ÷ 3
5
The answer to the last division was a prime number, so we are done. The answer is a
collection of the numbers on the right end of each line (including the lonely one on the
bottom line), all strung together with multiplication signs between them:
2 · 2 · 3 · 5 or 22 · 3 · 5
Always check your answer to see that each number in the list is really a prime number, and
that they multiply together to get back the number you started with.
Example Find the prime factorization of 24.
24 ÷ 2
12 ÷ 2
6÷2
3
Answer 2 · 2 · 2 · 3 or 23 · 3
Example Find the prime factorization of 35.
35 ÷ 2 doesn’t work
35 ÷ 3 doesn’t work
35 ÷ 5
7
Answer 5 · 7
Example Find the prime factorization of 194.
194 ÷ 2
97 ÷ 3 doesn’t work
97 ÷ 5 doesn’t work
97 ÷ 7 doesn’t work
97 ÷ 11 doesn’t work, but the answer to 97 ÷ 11 is less than 11.
So we stop. 97 must be prime.
Answer 2 · 97
5
1.1.2
Using a Cheat Sheet
Sometimes students are allowed to use a page of notes that includes the prime factorizations
of a multiptude of prime factorizations. Such a page is included in the appendix at the end
of the book.
1.2
Greatest Common Factor (GCF)
1.2.1
Finding the GCF of Natural Numbers
Procedure
1. Find the prime factors for each of the numbers (see page 4), organizing them into
lists.
2. Loop through the first number’s list of factors, checking to see if that factor is in
each and every list of factors. If it is not in every number’s list of factors, it is not
common, so skip it and continue on with the next factor in the first list.
3. If a factor is common to every list of factors, write down the factor in the common
factors list, joining factors together with multiplication dot, and put the exponent
to be the lowest exponent on any of the lists. Repeat step 2 with the next factor in
the first list.
4. If there is nothing in the common factors list, the answer is 1 (since 1 is a factor
of every number). If there is only one number in the common factors list, it is the
GCF. Otherwise, multiply all the common factors together.
Example Find the GCF for 12, 18, and 96
1. Each number gets its own list of prime factors:
12 = 22 · 3
18 = 2 · 32
96 = 25 · 3
CF =
2. Begin with the first factor of the first list (2 in this case). Since all three lists have a
2 to some power in them, then 2 is common.
3. Write down the lowest power of 2 that appears in any list. One way of thinking of
this is to ask, "Every term has at least how many factors of 2?" The list for 18 has
a plain 2 with no exponent. That is the lowest power of 2 in any of the three lists.
So write down a 2 in the common factors list.
Repeat these two steps for the next factor in the first list, namely 3. All three lists
have a 3 to some power, so it is common also. The lowest power of 3 that appears
in any list of factors is a plain 3 in the list for 12. So write a multiplication dot after
the 2 you found before and then write the 3 that you found in common.
The first list has no more factors, so we are done repeating steps 2 and 3. Our
common factor list looks like: 2 • 3
6
4. Multiplying the common factors together, we get the GCF: 6
Example: Find the GCF for 14 and 46
1. Each number gets its own list of prime factors:
14 = 2 • 7
46 = 2 • 23
CF =
2. Begin with the first factor of the first list (2 in this case). Since both lists have a 2 to
some power in them, then 2 is common.
3. Write down the lowest power of 2 that appears in any list. The list for 14 has a
plain 2 with no exponent. That is the lowest power of 2 in both lists. So write
down a 2 in the common factors list.
Repeat for the next factor in the first list. So check the 7. The list for 46 has no
factor of 7, so we skip 7.
The first list has no more factors, so we are done repeating steps 2 and 3. Our
common factor list looks like: 2
4. There is nothing to multiply in the common factors list, so the GCF is: 2
1.2.2
Finding the GCF of Terms with Variables
Procedure Just the same as for plain numbers, except it’s actually easier. We just say that the
variables are already in prime factored form. So find the lowest exponent for any letter
that is common to all the terms.
Example Find the GCF of 4x2 y and 3xyz
1. Each term gets its own list of prime factors:
4x2 y = 22 x2 y
3xyz = 3xyz
CF =
2. Begin with the first factor of the first list (2 in this case). Since the second list does
not have a 2, it is not common so we skip it.
Next factor is an x. Since both terms have an x, it is common.
3. The lowest exponent of x is the plain x with no exponent. So write it down as a
Common Factor.
The next factor in the first list is y. The second term also has a y, and the lowest
exponent is the plain y. It becomes the second factor in the Common Factors list.
The first list has no more factors, so we are done repeating steps 2 and 3. Our
common factor list looks like: x · y
4. Multiplying the common factors list together, we find the GCF is: xy
7
1.3
1.3.1
Evaluating Polynomials
Plug-in Method of Evaluation
When you are given a polynomial and a value for the variable in it, you can simply plug in
the value in place of each spot the letter appears. Then use the order of operations to simplify
the result, and you should end up with a plain number value.
Example Find the value of 2x3 − 3x2 + 8x − 6 if x = 2.
2x3 − 3x2 + 8x − 6
plugin
2 · 23 − 3 · 22 + 8 · 2 − 6
expos
2·8−3·4+8·2−6
mult
16 − 12 + 16 − 6
add
14
There are a few things to be careful about:
• The exponent on the x will transfer to the value of x, but not to the coefficient that is
multiplied times the x.
• Always take care of simplifying exponents before multiplying the coefficients.
• Be careful with signs!
• We can never solve a polynomial to find what x is equal to — we have to be told what
x is before we can do this evaluation.
We can also evaluate a polynomial with other letters and using negative values.
Watch out! Be sure to use parentheses around negative values that you plug in for the
variable.
Example Find the value of 2y3 − 3y2 + 8y − 6 if y = −1.
2y3 − 3y2 + 8y − 6
plugin
2(−1)3 − 3(−1)2 + 8(−1) − 6
expos
2(−1) − 3(1) + 8(−1) − 6
mult
−2 − 3 − 8 − 6
add
-19
Remember that the negative is also affected by the exponent. So, for each term in the polynomial, if the exponent is odd, you get a negative value; but if the exponent is even, the value
is positive.
8
1.3.2
Synthetic evaluation: a nifty alternative
Synthetic evaluation is a time-saving and mildly fun method that condenses evaluating a
polynomial down to a simple alternation between multiplying and adding. You won’t need
to find what (−2)5 is, or worry about exponents at all. It can be a little confusing at first, but
once you get used to it, you can zip through one of these evaluation problems in less than
half the time needed for the conventional standard method.
Here is the generic setup for a problem. We will evaluate things like ax3 + bx2 + cx + d when
x = n.
space for “carrying” numbers up
+
a
b
c
d
n
addition area above this line
multiplication row
The top rows (the “addition area” above the top horizontal line) are for numbers that are
added together in columns. The multiplication row is on the bottom. After adding the
numbers in the column above, the sum is written on the bottom row and then multiplied
by the number on the left edge (n as shown here), and the result of that is carried up to the
top of the next column. We rope off the right-most column with a second vertical line. The
answer will appear in the bottom right spot. Now we’ll step through the process in detail.
Some of the question marks in your head should start to disappear when you see the process
in action.
Procedure Synthetic Evaluation
For ease in explanation, we will be working an example as we go through the procedure. We will
evaluate 2x2 + 3x − 14 when x = 2.
1. Arrange the terms in order of descending powers (from highest power of x down to
lowest). Our example is already good to go.
2. Put in a placeholder for any “missing terms.” For example, x2 + x + 1 has no missing
terms, because from highest to lowest, the exponent on the x doesn’t skip anything.
But x3 + x + 1 has a missing term: there is no x2 term in it. Placeholders can be inserted
with a zero times the missing term. So x3 + x + 1 should be written as x3 + 0x2 + x + 1.
Our example has no missing terms.
3. Write the coefficents in a row with a little space between each one. Put a horizontal line
under the coefficients, and leave a blank row above the coefficient. Draw two vertical
lines, one to the left of all the coefficients and one singling out the rightmost coefficient.
The setup looks like this:
+
2
3
−14
9
4. Write the number you are plugging in for x in the bottom left corner. This is called the
“multiplication row.” Your setup should look like this:
+
2
3
−14
2
5. Now we begin a pattern. It goes like this: beginning on the left, add a column in the
addition area (above the horizontal line), put the answer in the same column in the
multiplication row (under the line), multiply by the number at the left of the multiplication row, and put the answer at the top of the next column (in the addition area). This
pattern repeats until all the columns are full. Here is what your work should look like
at each step of the way:
+
2
2
2
3
−14 Nothing to add to the 2, so it drops down
4
+
2
2
2
3
−14 Multiplied (2)(2), answer goes in the next column.
4
+
2
3
2
2
7
4
+
2
3
2
2
7
4
+
2
3
2
2
7
−14 Added 4+3
14
−14 Multiplied 2 times 7
14
−14 Added -14+14.
0
6. The bottom number in the last column is the answer to the evaluation. In this case, it is
0.
Example Find the value of 2y3 − 3y2 + 8y − 6 if y = −1.
Here is the setup:
+
2
−3 8
−6
−1
10
Here it is completely filled out:
−2
5
−13
−3
8
−6
−1 2 −5 13
−19
+
2
Answer: −19
1.3.3
(Optional) Why synthetic evaluation is possible
If we look at a polynomial such as 2x3 + 3x2 − 5x + 4, one of the problems we encounter with
evaluation is those exponents. We have to get the signs correct and not mess up anywhere
if we want the correct answer. But there is a way, using a little bit of factoring out a common factor (a method covered elsewhere), to avoid having any exponents at all. The only
operations will be addition and multiplication. Let’s take a look.
If we group all but the last term together, we can factor out a common x, and we have have:
x (2x2 + 3x − 5) + 4. For the sake of making things neater for later, let’s put the x after the
parentheses. It doesn’t change anything except make it easier to see a pattern. So we get
(2x2 + 3x − 5) x + 4. Now, inside the parentheses, we can do that same thing again, and we
get: ((2x + 3) x − 5) x + 4. If we had a higher order polynomial, we could keep repeating that
process as many times as needed until there are no more exponents higher than 1.
Now, if we are careful and thorough, we can see that there is a nice alternating pattern of
multiplication and addition. For any given input value (that gets plugged in for x), we
multiply 2 by the input, then add 3, then multiply by the input, subtract 5, multiply by
the input, then add 4. The result will be the same as taking 2 times the cube of the input
plus 3 times the square of the input minus 5 times the input plus 4 (as we would do by the
conventional method).
That is exactly what synthetic evaluation does for us, but we can use it without factoring and
doing so much writing. We condense it down to a very simple state and alternate between
multiplying and adding. Once you get used to it, you can zip through one of these evaluation
problems in less than half the time needed for the plug-in method.
1.4
1.4.1
Adding and Subtracting Polynomials
Adding Polynomials
There is only one thing we can do to add polynomials. That is to combine like terms. Anyone
who does more than that is trying to do the impossible, and will get a wrong answer.
Some people like to put one polynomial per line, lining up like terms under like terms above.
This is sometimes called “adding polynomials vertically.” Or you can do it horizontally, like
the examples below in “Subtracting Polynomials.”
Example Add 4x3 − 3x2 + 2x and 6x3 + 2x2 − 3x
11
4x3
−3x2
+2x
6x3
+2x2
−3x
10x3
− x2
−x
Answer: 10x3 − x2 − x
Example Add x2 − 2x + 5 and 4x2 − 2
Watch out for missing terms! This has a missing x term in the second polynomial. You can
either add in a 0x or just leave a gap. Either way gets the same answer.
x2
−2x +5
4x2
+0x −2 (Using a 0x placeholder)
5x2
−2x +3
x2
−2x +5
4x2
5x2
1.4.2
−2 (Leaving a gap instead)
−2x +3
Subtracting Polynomials
Instead of directly subtracting polynomials, we should distribute the minus sign and then
add. Why? It’s just about guaranteed that you’ll miss changing a sign if you are trying to
keep too many things in your head at once. So just distribute and then add. It works.
Note: You can work these in “vertical format” just as well. These will be worked horizontally
just as an example.
Example Subtract (14y3 − 6y2 + 2y) − (2y3 − 7y2 + 6)
(14y3 − 6y2 + 2y) − (2y3 − 7y2 + 6)
1.4.3
dist
14y3 − 6y2 + 2y − 2y3 + 7y2 − 6
like
12y3 + y2 + 2y − 6
Working with more than one variable
Watch out! Be sure that you only combine like terms. If they don’t match exactly on the
letters, don’t force the issue.
Example Simplify (5m3 n + 3m2 n2 − 4mn) − (7m3 n − m2 n2 + 6mn)
12
(5m3 n + 3m2 n2 − 4mn) − (7m3 n − m2 n2 + 6mn)
dist
5m3 n + 3m2 n2 − 4mn − 7m3 n + m2 n2 − 6mn
like
−2m3 n + 4m2 n2 − 10mn
Practice Time!
Add or subtract the polynomials as indicated:
1. 3x2 − 2x + 12 + x2 − 4
2. 4x3 − 3x − 3x2 − 5
3. 5x2 − 3x + 2 + x2 + 8x − 7 − 3x2 − 6x + 5
1.5
Multiplying Polynomials
One trick that we use a lot in Algebra is to break complicated problems down into a group
of simpler ones. This is certainly true for multiplying polynomials. We begin with the basic
building block of one term times one term. All the other kinds of multiplication are accomplished by doing a bunch of these basic ones and adding the results together. So learn well
how to do the simplest ones, because sometimes you may have 30 of them to do for one
complicated problem.
1.5.1
Multiplying One Term times One Term (One-by-One)
This is the basic building block for this entire section. All the other kinds of multiplication
are broken down to this level, so make sure you really get this.
1. Regroup the numbers and letters, putting similar things together.
2. Multiply them using exponent-combining techniques learned in previous sections.
3. If some element has nothing similar to it, just put it in as it is.
4. Normally, we write the answer with the number first (including a minus sign if needed),
then all the letters in alphabetical order.
5. We always get one single term for the answer of a one-by-one problem.
13
Example Multiply 3x4 · 4x
3x4 · 4x
regroup
3 · 4 · x4 · x
simplify
12x5
Example Multiply −3x4 · y2
−3x4 · y2
nothing to regroup
put it together
−3x4 · y2
−3x4 y2
Most people (after grasping the concept well) will skip writing down the steps in this process,
since they are easy enough to do in your head most of the time.
1.5.2
Multiplying One Term times Two or More Terms (One-by-More)
This turns out to be the usual kind of distributing. So the terms together in parentheses get
separated, and each one is multiplied by the term outside the parentheses. Each one of those
multiplications is just 1 term times 1 term, as described above.
• Treat the + and − signs as part of the term immediately after the sign, and then all the
terms are added together.
• We will get the same number of terms in the answer as we started with inside the
parentheses.
Example Find the product: 2x4 (3x2 + 2x − 5)
2x4 (3x2 + 2x − 5)
dist
2x4 · 3x2 + 2x4 · 2x + 2x4 (−5)
simp
6x6 + 4x5 − 10x4
Example Find the product: −8m3 (4m3 + 3m2 + 2m − 1)
−8m3 (4m3 + 3m2 + 2m − 1)
dist
−8m3 · 4m3 + (−8m3 )3m2 + (−8m3 )2m + (−8m3 )(−1)
simp
−32m6 − 24m5 − 16m4 + 8m3
14
1.5.3
Multiplying Two or More Terms times Two or More Terms (More-by-More)
This is the most complicated kind of polynomial multiplication and can be called “multiterm
distribution;” it has to be broken down into a bunch of one-by-one problems, each term from
the first parentheses times each term from the second parentheses. The answers for each of
these one-by-one may have some like terms, so we look to see if we can combine them. Then
all the terms are added together.
There are three systems that I know about for multiterm distribution: FOIL, Clarence the
Clam, and the Grid.
FOIL method: Many people have been taught to use the FOIL method for this, and it’s OK
as far as it goes. The main point of it is to remember all the combinations of what to multiply
together. But FOIL only works for two-by-two multiplication.
Clarence the Clam: Clarence the Clam works for two-by-two or more-by-more, so you can
skip the FOIL method and just have one method to learn. If you want to keep using FOIL, go
ahead, but you will need to use Clarence or another system of multiterm distribution when
you have more than 2 terms in a set of parentheses.
The Grid: The grid provides three advantages:
• it has a neat rectangular placement that helps make sure you don’t leave any term out,
• it helps you locate like terms that you can combine,
• and it also works in reverse to help you with factoring later on.
I recommend the grid for students who have never been good at multiplying polynomials
before.
The main point of all multiterm distribution methods is that each and every term from one
set of parentheses must be multiplied by each and every term in the other set of parentheses.
As long as you keep careful track and don’t skip any terms, you will be OK.
1.5.4
Procedure: Multiplying on the Grid
1. Write the first polynomial above the grid, leaving good spacing so the terms are separate with each one above a cell of the grid. Keep any minus signs with the term
following it.
2. Write the second polynomial on the left of the grid, with similar separation and spacing.
3. In each cell, write the product of the term at the top of its column times the term left of
its row.
4. Like terms will line up along diagonal lines — you can add them together to simplify.
15
Example Multiply 3x2 + 2x − 1 (2x − 7)
After writing both polynomials (steps 1 and 2), we get a grid like this:
3x2
−1
2x
2x
−7
Now multiply the term above by the term on the left by column and row and we get
(being careful with the signs):
3x2
2x
−1
2x
6x3
4x2
−2x
−7
−21x2
−14x
7
Notice that like terms are on the diagonals, and we can combine them:
Answer: 6x3 − 17x2 − 16x + 7
1.5.5
Procedure: Clarence the Clam
This is named for the shape of the arcs that help keep track of what is finished.
1. Rearrange the parentheses (if needed) to put the parentheses containing the fewest
terms first.
(a) Multiply the first term from the first parentheses times each term from the second
parentheses. This is like a one-by-n problem above.
(b) Repeat step 2 with the second term from the first parentheses times each term from
the second parentheses. Write the answer for each one-by-one problem below a
like term from step 2 (if there is a like term).
(c) Repeat again with the third term from the first parentheses and so on until you
run out of terms in the first parentheses.
(d) Combine all like terms.
Example Find the product: ( x2 + 3x + 5)( x − 4)
1. Rearrange to ( x − 4)( x2 + 3x + 5) to make things easier and neater.
2. Multiplying the x from the first parentheses times each of the terms in the second
parentheses, we get: x3 + 3x2 + 5x
3. Multiplying the −4 from the first parentheses times each of the terms in the second
parentheses, we get: −4x2 − 12x − 20 and when we line those terms up under like
terms from what we got on step 2, we have:
x3
+3x2
+5x
−4x2
−12x −20
4. We are out of terms in the first parentheses, so stop repeating.
5. Combining like terms, we get the answer: x3 − x2 − 7x − 20
16
1.5.6
Multiplying More Things Together
So far we only talked about multiplying two things together. But what if we stacked up
more, like 3( x + 1)( x − 4). In these cases, we can only multiply two at a time to get a partial
answer, which is enclosed in parentheses; then multiply the partial answer by the remaining
part(s). The best advice here is to save the simplest and easiest multiplication for last. Some
people refuse to believe me, and they end up with working something that is harder than it
had to be. So really and truly, get the hardest stuff done first, and save the easiest for last.
OK?
Example Simplify 3( x + 1)( x − 4)
We will save the multiplying by 3 for last, since it is the easiest!
3( x + 1)( x − 4)
clam
3( x2 − 4x + 1x − 4)
like
3( x2 − 3x − 4)
dist
3x2 − 9x − 12
Notice that the answer from multiplying the two sets of parentheses remains inside
parentheses so that the 3 is distributed. Always put these partial answers in parentheses. Then check to see if the parentheses are doing a job. If not, they can be deleted.
Practice Time!
Multiply the polynomials as indicated:
1. 3x2 − 2x + 12 x2 − 4
2. 4x3 − 3x
3x2 − 5
3. 3x (8x − 7) (−6x + 5)
1.6
1.6.1
Special Products (Shortcuts)
Squaring a binomial (Double-Stuff Shortcut)
One thing that comes along quite often is working with a polynomial times itself. If it happens to have two terms, we have a shortcut that makes your homework go quicker. Let’s
look at an example worked out the regular way from last section, and then dissect it a little
and come up with a general shortcut.
17
For example, let’s look at ( x + 4)2 . The first thing to do is to talk about what that means
and what it does not mean. Remember that an exponent of 2 is a shortcut way of writing
something multiplied by itself. So what is that something here? It is ( x + 4) taken as a unit.
Watch out! We absolutely CANNOT ever say that ( x + 4)2 means x2 + 42 . IT DOES NOT!
Sharing the power ONLY works with multiplication or division inside the parentheses. So
what is the correct way to write it out the long way? That would be ( x + 4)( x + 4). And if we
multiply that out using multiterm distribution, we get x2 + 4x + 4x + 16, and then combine
like terms, right?
But before we combine the like terms, let’s look at what we have. We do have an x2 and a 42 ,
like the warning above talked about. But we also have other stuff. We have 4x added in. Two
times, as a matter of fact. Where did they come from? From the first x times the last 4, and
the last x times the first 4. Since we’re talking about the first and last paren groups always
being the same, we can see that we will ALWAYS get a deal like that: multiply the two terms
together and then double it because it gets added in twice. That is why this Double-Stuff
shortcut works.
Procedure Double-Stuff Shortcut
1. Square the first term to get the first cookie.
2. Multiply the first and last terms together and then double it to get the double-stuff.
Be careful with the sign! If one of the terms is negative, this double-stuff term will
be negative and you would put − double-stuff. Otherwise, put + double-stuff.
3. Square the last term to get the last piece to get the second cookie. Put + (ALWAYS
plus) and this cookie at the end.
Example Simplify (y − 3)2
1. Squaring the first term (y) gives us y2 for the first cookie.
2. The double-stuff is 2(y)(−3) = −6y.
3. Squaring the last term (−3) gives us positive 9 for the second cookie.
Answer: y2 − 6y + 9
Example Simplify (10m + 7)2
1. The first term is 10m. So the first cookie is 100m2 .
2. The double-stuff is 2(10m)(7) = 140m.
3. The last term is 7. So the last cookie is 49.
Answer: 100m2 + 140m + 49
18
1.6.2
Multiplying a Sum and Difference (No-Stuff Shortcut)
This kind looks sort of like squaring a binomial except that one time we have a plus and one
time a minus, like ( x + 2)( x − 2). They may try to hide it in tricky ways, like changing the
order as in ( x + 2)(−2 + x ). But if can be put into the pattern of the same two terms, once
with a plus and once with a minus, then we have a good shortcut for it.
If we work ( x + 2)( x − 2) out the long way, we get x2 − 2x + 2x − 4, and the two middle
terms add to 0, so we end up with x2 − 4. It turns out that those middle terms will ALWAYS
drop out in this kind of multiplication. So we have another shortcut: the No-Stuff Shortcut.
Procedure No-Stuff Shortcut
1. Square the first term to get the first cookie.
2. Square the last term to get the last cookie.
3. Put a minus sign in between the cookies (there is no stuff!)
Example Simplify (3 + y)(3 − y)
1. The first term is 3. Don’t rearrange them when the sum/difference pattern is clear.
So the first cookie is 32 = 9.
2. The last term is y. We don’t worry about the sign, since it’s + in one case and − in
the other. So the last cookie is y2 .
3. Put together with a minus sign, we get the answer: 9 − y2
Example Simplify (10m + 7)(10m − 7)
1. The first term is 10m. So the first cookie is 100m2 .
2. The last term is 7. So the last cookie is 49.
3. Put together with a minus sign, we get the answer: 100m2 − 49
1.6.3
Finding Greater Powers of a Polynomial
There isn’t really much of a shortcut here, except to realize that we can break down higher
powers into groups of squares and use the double-stuff shortcut. After that, it gets messy
when we multiply the parital answers together.
Please note that there is a method for finding greater powers of a binomial, called “binomial
expansion.” However, it is fairly complicated for beginners and is usually considered worthwhile for quite high powers, such as ( x + y)6 or higher. By then, it is worth the time to learn
a new technique. For now, we will not worry about binomial expansion, and just carry on
with the procedures we are more familiar with.
Here is a glimpse of what binomial expansion looks like, but it is beyond the scope of this
book to explain what all this fancy stuff means.
19
Binomial Expansion
n
n n−k k
a b
( a + b) = ∑
k
k =0
n
where (nk) =
n!
k! (n − k )!
Example Simplify (3k − 2)4
Let’s split this into two pieces: (3k − 2)2 (3k − 2)2 .
Then we can use the double-stuff shortcut to change this to (9k2 − 12k + 4)(9k2 − 12k +
4).
Unfortunately, there is no shortcut for squaring a polynomial with more than two
terms, so we have to use normal multiterm distribution.
Answer: 81k4 − 216k3 + 216k2 − 96k + 16
Example Simplify (4x + 1)3
Let’s split this into two pieces: (4x + 1)(4x + 1)2
Then we can use the double-stuff shortcut to change this to (4x + 1)(16x2 + 8x + 1)
Unfortunately, there is no shortcut for this either, so we have to use normal multiterm
distribution.
Answer: 64x3 + 48x2 + 12x + 1
Practice Time!
Find the special products:
1. (5x − 2)2
2. (2x + 5y) (2x − 5y)
3. ( x + 5)3
1.7
Dividing Polynomials
There are three levels of polynomial division, going from pretty easy to pretty difficult. The
key to success with them is to get really good at the easy ones, and then see how the more
difficult ones use the easy kind. In effect, we break a complicated division problem down
into a series of easy ones!
20
1.7.1
Dividing one term by one term (one-by-one)
This is the simplest type of dividing with polynomials, and is a basic block for all the other
types. All we really do with this is to set up the division as a fraction and reduce it. You
should reduce it in steps, first the sign, then the numbers, then each of the letters in turn.
12x2
6x
Everything is positive so the answer will be positive.
Example Divide
Working with the plain numbers, we see that
12
6
reduces to 21 , or just plain 2 upstairs.
x
x2
reduces to , or just x upstairs.
x
1
Since the whole downstairs become a 1, just write the upstairs on the ground floor.
Working with the x terms, we see that
Answer: 2x
14
7x
The number parts reduce to 2 upstairs.
Example Divide
The letter parts don’t reduce, leaving x downstairs.
Answer:
1.7.2
2
x
Dividing two or more terms by one term (more-by-one)
This type is recognized by having only one term downstairs, with two or more terms upstairs. We break this into a series of one-by-one divisions by reversing what we do when we
add fractions. If we had to add the fractions 7x + 73 , since they have the same downstairs,
3
we would just add the tops. But x and 3 are not like terms, so our answer would be x+
7 .
3
Now let’s look at x+
7 as a division problem ( x + 3) ÷ 7. We can break it into two simpler
division problems: 7x + 73 , which is the reverse of what we did with adding the fractions. In
this illustration, it doesn’t help because neither of the fractions we broke it into reduce. Let’s
try an example that does do something.
Example Divide (12m6 + 18m5 + 30m4 ) ÷ 6m2
We break this into three fractions, one for each term upstairs. Notice the downstairs
12m6 18m5 30m4
part is repeated without change:
+
+
6m2
6m2
6m2
Then we reduce each fraction, independent of the others: 2m4 + 3m3 + 5m2
Sometimes we have subtractions to worry about. Just keep the sign with the term following
it and add all the fractions together. And sometimes we get messy fractional results. That is
OK. Just be sure that you don’t push everything upstairs just because it looks more finished
that way.
21
Example Divide
50m4 − 30m3 + 20m
10m3
Broken into three separate fractions, we have
50m4 −30m3
20m
+
+
3
3
10m
10m
10m3
Be careful! Note that the minus sign for the middle term followed along to the upstairs of the middle fraction.
2
Each of these is reduced for the final answer: 5m − 3 + 2
m
1.7.3
Dividing two or more terms by two or more terms (more-by-more)
This method is modeled after long division from elementary school, only it looks about ten
times worse with all the alphabet soup mixed in. There is no substitute for watching a problem in action, and I can’t everything you need into notes that don’t move and don’t talk. You
have to see it to believe it.
Procedure Long Division with Polynomials
1. Put the terms in order of descending powers and fill in any “missing” terms. Put the
dividend inside the division bracket and the divisor outside it.
2. Repeat this pattern:
(a) Divide the first term of the dividend by the first term of the divisor.
(b) Put the answer above the bracket.
(c) Multiply the divisor by the answer part you just wrote (not the whole thing every
time), and put that underneath the dividend, lining up like terms.
(d) Subtract by changing the signs of what you wrote below the divisor and then
adding the like terms. The highest order term will drop out if you are doing it
right. The remaining terms become the new dividend.
(e) Stop repeating if the power of x in the new dividend is lower than the power of x
in the divisor. At that point, the “new dividend” is called the remainder.
3. Write the answer as it appears above the bracket, and if there is a remainder, write it as
a fraction added on at the end. Remainder over divisor.
Watch out! Some important things that folks forget to do on these division problems:
• Always put polynomials in order of descending powers before dividing.
• Always check for “missing” terms. Fill them in with zero times the missing power
before you begin.
22
• Always add the remainder fraction, if the remainder is not zero. If the remainder is
negative, still put a plus sign and then the remainder fraction, with the minus sign
upstairs. The downstairs will always be the divisor you are dividing BY. The upstairs
is the remainder.
• Stop when the remainder has a lower power of x than the divisor you are dividing BY.
x2 + 5x + 6
x+3
First, we set up the problem. The number we are dividing up (the upstairs) is called
the dividend, and we put it inside a bracket. The number we are dividing by (the
downstairs) is called the divisor, and we put it outside the bracket. You should have a
setup like this now:
Example Divide
x+3
x2
+5x +6
Next, we pay attention to the first term of the dividend and the first term of the divisor.
Don’t worry about the rest of the terms. They will get worked out in their own good
time. We divide the first term inside the bracket by the first term outside the bracket,
x2
= x. This means the first part of the answer is x. Put it above the
and reduce:
x
bracket. You don’t need to worry about aligning it with anything. Just put it up on top
somewhere. You now have a setup like this:
x
x+3
x2
+5x +6
Next, we multiply x ( x + 3) and put that below the dividend. You then have:
x
x+3
x2
+5x +6
x2
+3x
Then, in order to subtract easily and accurately, we change the sign of each term that
we just wrote, then ADD the like terms. We get:
x
x+3
x2
− x2
+5x +6
−3x
2x
+6
We are ready to repeat the cycle, this time with
+2 up above the bracket. Now we have:
23
2x
= 2. Since that is positive, we put
x
x+3
x
+2
x2
+5x +6
− x2
−3x
2x
+6
Multiply the +2( x + 3) and put it below the dividend:
x+3
x
+2
x2
+5x +6
− x2
−3x
2x
+6
2x
+6
Change signs and then add:
x+3
x
+2
x2
+5x +6
− x2
−3x
2x
+6
−2x −6
0
The zero remainder says we can stop and the answer is above the bracket.
Answer: x + 2
The remaining examples will not be shown in step-by-step fashion, but you should be able
to duplicate them by following the same pattern.
Example Divide
2x+3
2x3 + 5x + x2 + 13
(Note the terms are out of order.)
2x + 3
x2
−x
+4
2x3
+ x2
+5x +13
2x3
+3x2
−2x2
+5x
−2x2
−3x
8x
+13
8x
+12
1
24
Answer: x2 − x + 4 +
Example Divide
x−2
1
2x + 3
x3 − 8
(Note the missing x2 and x terms.)
x−2
x2
+2x
+4
x3
+0x2
+0x −8
x3
−2x2
2x2
+0x
2x2
−4x
4x
−8
4x
−8
0
Answer: x2 + 2x + 4
Example Divide
together)
m2 + 3
2m5 + m4 + 6m3 − 3m2 − 18
(Note the strange alignment to keep like terms
m2 + 3
2m5
2m3
+ m2
−6
+ m4
+6m3
−3m2
2m5
+0m −18
+6m3
m4
m4
+0m3
−3m2
+3m2
−6m2
+0m −18
−6m2
−18
0
Answer: 2m3 + m2 − 6
3x3 + 7x2 + 7x + 8
(Sometimes we have fractions to deal with)
Example Divide
3x + 6
25
3x + 6
x2
+ 31 x
+ 35
3x3
+7x2
+7x
+8
3x3
+6x2
x2
+7x
x2
+2x
5x
+8
5x
+10
−2
Answer: x2 + 13 x + 35 +
1.7.4
−2
3x + 6
Alternate Method
There is another method that can be used to divide polynomials, but it is more abstract
and can be confusing to beginners. It is called “synthetic division,” and it is covered in the
advanced section near the end of the book. If you are anxious to try it, start with the basic
synthetic division on page 43. If you get good at synthetic division now, you will be ahead
of the game later on.
2
2.1
Introduction to Factoring
Earning a Black Belt in Factoring
Similar to the martial arts, you will earn “belts” of different colors as your skills in factoring
develop. Your instructor has a Black Belt in factoring, and will help you along your own
path. The different colors of the belts, from the most basic to most advanced are listed below. Generally, students only need to go through Red Belt until they take College Algebra.
College Algebra will require some more advanced methods detailed in the Brown Belt, and
teachers will need a Black Belt.
26
White
Factor out the GCF: Find factors that all the terms have in common, and pull
them out front.
Yellow
Factor by grouping: If four or more terms have no common factors,
sometimes we can group the terms by some commonality, find a common
factor for each little group, then find a common factor for the groups.
Orange
Key Number method: A step-by-step method for factoring trinomials like
x2 + 7x + 6 (ones with a bare x2 term).
Purple
General Trinomial method: A two step-by-step method for factoring
trinomials like 2x2 + 7x + 10 (with some number other than 1 times the x2
term).
Green
Difference of squares: A pattern-based method for factoring binomials like
x2 − 4.
Blue
Sum/difference of cubes: A pattern-based method for factoring binomials like
x3 + 8 or x3 − 8.
Red
Mixed factoring: Factoring a wide variety of the above types without being
told which method to use, as well as multi-step factorizations.
Brown
Advanced factoring: Limited-guess and check procedures for polynomials
not covered previously, such as x4 − 3x3 + −5x2 + 7x − 12.
Black
Mastery: Expertise gained through helping others learn how to factor.
When factoring, there is a definite order of steps to take. Taking things out of order can lead
to using the wrong method and generally leads to the wrong answer. Beginners usually feel
a little overwhelmed at first with all the different methods. But these methods are given
in a good order, and if you keep referring back to this section as you progress through the
different belts, it will all come together in the end. Sometimes the simple and easy stuff trips
people up more than the harder stuff, so always follow these steps and you should be OK.
2.2
How to know when you are done factoring
1. There are no addition or subtraction signs outside of parentheses.
2. Each and every factor inside parentheses cannot be factored any further.
3. The answer has been checked by multiplying out the answer to see if we get the original
problem back again.
27
3
Fundamental Factoring Methods
3.1
White Belt: Factoring out the Greatest Common Factor (GCF)
There are several ways to do this factoring. Usually people who have done this quite a few
times develop their own style. You begin to see common factors by inspection, and can pull
out more common factors later if you did not find all the common factors the first time. Just
be sure that when you are done, you really have found all the common factors.
For students who are not used to factoring, here is one sure-fire method for factoring out the
GCF.
Procedure: Factoring out the GCF
Example: Factor out the GCF:
1. Find the GCF for all of the terms
(see 1.2.2 on page 7).
2. Write the GCF first, followed by empty
parentheses for the “leftovers.”
3. Compute the leftovers term by term,
putting them inside the parentheses.
Here is how to do this, one term at a
time:
(a) Make a fraction with the original
term upstairs, and the GCF downstairs.
(b) Reduce the fraction. The downstairs should all cancel out.
(c) Put each term’s leftovers inside the
parentheses, making sure to keep
the plus or minus sign the original
term has.
Note: If the GCF is exactly identical to
one of the terms, the fraction will reduce to 1. Put the 1 in its place in the
leftovers.
4. Check the leftovers again to see if there
is another common factor.
5. Check your answer by distributing.
You should get the original problem
back.
Example: Factor out the GCF: 2x + 2
28
24x2 y4 z + 42x3 y2
1. Find the GCF of all numerical values, then find the GCF for each letter: 6x2 y2 .
2. Put the GCF first,with parentheses
after it for the “leftovers”, like this:
6x2 y2 (
).
3. The first term’s leftovers are:
24x2 y4 z
which reduces to 4y2 z and
6x2 y2
42x3 y2
the second term’s are: + 2 2
6x y
which reduces to +7x so the final
result is 6x2 y2 4y2 z + 7x . That
will be the final answer, as long as
we don’t find any trouble.
4. A quick check of the leftovers
inside the parentheses shows no
common factors for the leftovers.
That’s good.
5. Check: Distributing your answer
yields the original problem back.
That’s good, but don’t put the
problem down as the answer. Your
answer should have parentheses,
with no addition or subtraction
outside.
1. Find the GCF of all numerical values. GCF = 2. In this case, there is nothing to
additional for the GCF for each letter, since the second term has no letters at all.
2. Put the GCF first, with parentheses after it for the “leftovers”, like this: 2 (
)
2x
which reduces to x and the second term’s are:
3. The first term’s leftovers are:
2
+2
which reduces to +1 so the final result is 2 ( x + 1)
2
4. A quick check of the leftovers inside the parentheses shows no common factors
for the leftovers. That’s good.
5. Distributing your answer yields the original problem. Check!
3.2
Yellow Belt: Factoring by Grouping
Usually “factor by grouping” problems have exactly four terms, and we usually group the
terms in pairs. Later on, you might see some more peculiar “factor by grouping” problems,
and there are a few tips later on in this section for those.
There are two different approaches to factoring by grouping. The first one is the way that
almost everyone uses and you may have seen in a previous class. The second way uses a
grid, and it helps keep the plus and minus parts straight with very little effort. So, if you are
already used to factoring by grouping the usual way, feel free to keep using it. However, if
you have never had success at this kind of factoring, give the grid method a try. You might
like it!
Procedure Factoring by grouping (traditional method)
1. Pick the pairs by choosing the biggest ugliest part (BUP) and grouping it with the
term that resembles it the most. The choice is a judgment call, and it won’t stop
the show if you pick a different one from what I would.
2. Rearrange the terms if needed, and put parentheses around the pairs. There
should be a plus sign between the two sets of parentheses. If there is a minus
sign, hold on. You have a little extra work:
(a) Factor out a –1 from the second set of parentheses.
(b) Check that when you distribute the minus sign, everything would end up like
when you started.
3. Factor out the GCF (see page 28) for each set of parentheses. Then the leftovers
from the first set should be the same as the leftovers from the second set. If they
don’t match, try again, grouping the terms in a different way.
4. Treat the leftovers in parentheses as a single unit. It becomes the GCF for the two
sets. So factor it out, making a new set of parenthesized leftovers.
5. Check your answer by multiplying it out!
Example Factor by grouping: 2ab + 2b + 7a + 7
1. I would say the BUP is the first term, 2ab. The term that most resembles it is 2b.
29
2. There is no need to rearrange the terms this time. So putting the groups in parentheses, we get: (2ab + 2b) + (7a + 7). It’s good that there was a plus sign there
between the parentheses. No extra work!
3. Now to factor out the GCF for each group. Our example becomes: 2b ( a + 1) +
7 ( a + 1). And look! The stuff inside the parentheses matches! Yay!
4. Factoring out the ( a + 1) that the two parts have in common, we get: ( a + 1) (2b + 7).
Check your answer!
Example: Factor by grouping: 8x2 + 6x − 12xy − 9y
1. I would say the BUP is the first term, 8x2 . The term that most resembles it is 6x.
You might disagree and work it out another way. That’s OK.
2. There is no need to rearrange the terms this time. But if we
just merrily put the
2
groups in parentheses, we get something bad: 8x + 6x − (12xy − 9y), which
has a problem. What’s wrong? There is a minus sign there between the parentheses. That means extra work! Backing up, first we have to factor out –1 from the last
two terms: − (12xy + 9y). Then putting it together right, we have: 8x2 + 6x −
(12xy + 9y)
3. Now to factor out the GCF for each group. Our example becomes: 2x (4x + 3) −
3y (4x + 3). And look! The stuff inside the parentheses matches! Yay!
4. Factoring out the (4x + 3) that the two parts have in common, we get: (4x + 3) (2x − 3y).
5. Check your answer!
Procedure Factoring by grouping with the grid
1. Assuming you have a usual problem with four terms, make a 2-by-2 grid, and fill it in
with the four terms like this:
BUP (if it is positive)
term least like the BUP
or
BUP (if it is negative)
term least like the BUP
The remaining two terms go in any order into the last two cells, but the top left cell
must be positive for everything to work out correctly.
2. Factor out the GCF (see page 28) above each column and on the left of each row, making
the sign match the nearest cell.
3. Do a quick check to see if each cell is equal to the product of the GCF in its column and
the GCF on its row. This will catch over 90% of all errors. Do not continue if even one
of the four does not check!
4. Wrap the GCF parts in parentheses, and the answer is:
(the sum of the top terms)(the sum of the left terms)
30
5. Check your answer by multiplying it out!
The following examples are the same as we did using the usual method, so we should get
the same answer here.
Example Factor by grouping: 2ab + 2b + 7a + 7
1. I would say the BUP is the first term, 2ab. The term that least resembles it is 7.
Since the BUP is positive, it goes in the top left cell. The grid gets filled in like this:
GCFs
2ab
2b
7a
7
Note that the 7a and the 2b terms could trade places, and it would not change the
answer.
2. Now fill in the GCF for each column and each row:
GCFs
a
1
2b
2ab
2b
7
7a
7
3. The quick check gives a go, since 2ab = a · 2b, 2b = 1 · 2b, 7a = 7 · a, and 7 = 7 · 1.
4. Wrapping the top GCFs in parentheses (with a plus sign since they are positive)
and the side GCFs in like manner, we get: ( a + 1) (2b + 7).
5. Check your answer!
Example: Factor by grouping: 8x2 + 6x − 12xy − 9y
1. I would say the BUP is the first term, 8x2 . The term that least resembles it is −9y.
You might disagree and work it out another way. That’s OK. So we fill in the grid
like this:
GCFs
8x2
−12xy
6x
−9y
Again, the 6x and the −12xy could be switched, and you will still get the same
answer.
2. Filling in the GCF for each column and for each row, we get:
GCFs
2x
−3y
4x
8x2
−12y
3
6x
−9y
Note that the GCF for the right column is negative because the nearest cell was
negative.
31
3. The quick check gives a go: 8x2 = 4x · 2x, −12y = 4x (−3y), 6x = 3 · 2x, and
−9y = 3 (−3y).
4. Wrapping our GCFs in parentheses, we get: (2x − 3y) (4x + 3).
5. Check your answer!
3.2.1
Unfactorable polynomials
Sometimes there are polynomials that don’t factor out right. You can try grouping them
differently and see if it comes out that way. But some of them never can be factored by
grouping, and we would like to know how to spot them before we waste our time on them.
How to spot trouble before you fry your brain:
• Exactly one term is negative. With the sole exception of the second oddball problem
below, you can’t factor these by grouping. Because we put terms in pairs, you have to
have negatives in pairs also.
• None of the terms have any common factors. If you can’t match them up with another
term, it doesn’t do any good to group them. These can’t be factored.
3.2.2
Some tips for oddball problems:
• Usually six terms are grouped into two groups of three terms.
• Sometimes four terms are grouped three and one. Then the three are hopefully a perfect
trinomial square and the one lonely term is subtracted and is a perfect square. Then
you can factor those groups as a difference of squares. (You should not see any of these
until Red Belt or beyond.)
32
3.3
Orange Belt: Factoring Bare Trinomials x2 + bx + c
Sometimes instructors and textbooks only have one method for these factorizations, commonly called the “guess and check” method. While this works (with practice), it also has
drawbacks. If you find a polynomial that is unfactorable, you will always wonder if you
tried every guess that you could.
There is a clear, step-by-step method that will let you check every possibility quickly, leaving
no stone unturned. It is highly recommended, especially for beginners. It is called the “Key
Number” method. If you want help with the “guess and check” method, look in a textbook.
3.3.1
The Key Number Method
The Key Number method is used only for trinomials (three terms) with no number (or the
number 1) as the coefficient of the squared term. You can think of it as having a “bare x2
term.” This kind of factoring is quite common.
Examples of factoring problems that this method works for:
• x2 + 5x + 4
• x2 − 8x − 9
Note these have three terms, and the x2 term has no number in front of it. Examples of
factoring problems that this method will not work for:
Polynomial
Why the Key Number method won’t work
2x2 + 5x − 4
The x2 has a 2 in front of it.
x2 − 16
There are only two terms.
x2 − 4x + y2 − 4y
There are four terms.
33
Example: Factor 7x + x2 + 12
Procedure: Key Number Method
1. Make sure the terms are in descending
order of powers. Usually they are, but
some need to be rearranged.
2. When it is arranged in descending order of powers, the last term is the “key
number.”
3. Make a list of pairs of numbers that
multiply together to the key number.
Watch your signs; see the special notes
regarding negatives below.
4. Pick the pair that adds together to get
the middle term’s coefficient. If there is
no pair that works, the trinomial is unfactorable, or prime.
5. Set up two sets of parentheses with the
variable plus one of the pair in each.
6. Check your answer by multiplying it
out and combining like terms. You
should get the original trinomial back.
3.3.2
1. We need to rearrange the terms:
x2 + 7x + 12
2. The last term is the key number: 12
3. Starting with 1, then 2, 3, and so
on, list all the pairs of numbers
that multiply together to the key
number:
1
2
3
12
6
4
4. Pick 3 and 4, because 3 + 4 = 7, the
non-x part of the middle term.
5. Set up two sets of parentheses with
x plus one of the pair in each:
( x + 3) ( x + 4)
6. Check your answer!
Special notes regarding negatives
If the key number is positive, then both the numbers in the key pair have the same sign (+
or –). The sign of the middle term tells you what sign they both have.
Example A x2 + 5x + 4 has key number +, so key pair have same sign. middle term is +, so
both +
Example B x2 –5x + 4 has key number +, so key pair have same sign. middle term is −, so
both −
If the key number is negative, then one of the pair is positive and the other one is negative.
The sign of the middle term tells you the sign of the heavyweight.
Example C x2 − 3x − 4 has key number –, so key pair mixed signs. middle term is −, so
heavyweight −
Example D x2 + 3x − 4 has key number −, so key pair mixed signs. middle term is +, so
heavyweight +
34
Matching exercise
Match the following factored polynomials to Examples A through D above, based on the
signs of the numbers:
1. ( x − 4) ( x + 1)
2. ( x + 4) ( x + 1)
3. ( x − 4) ( x − 1)
4. ( x + 4) ( x − 1)
Example: Factor x2 − 7x + 12
1. Terms are in proper order.
2. The last term is the key number: 12
3. List all the pairs of numbers that multiply together to the key number. They must
match in sign because the key number is positive. And they must both be negative
because the middle term is negative.
−1
−2
−3
−12
−6
−4
4. Pick −3 and −4, because −3 + (−4) = −7, the non-x part of the middle term.
5. Set up two sets of parentheses with x plus one of the pair in each: ( x − 3)( x − 4)
6. Check your answer!
3.4
Purple Belt: Factoring General Trinomials ax2 + bx + c
There are several methods that I have seen to factor general trinomials, which are the kind
with a coefficient on the squared term, such as 3x2 + 5x − 12. These are different from the
trinomials that we used the Key Number method for; here we have trinomials (three terms)
with a number greater than 1 as the coefficient of the squared term.
The first method I learned is called the “guess and check” method. It has several drawbacks,
but it can be very quick if you are good at guessing the answer. It entails guessing the
answer and then using multiterm distribution to multiply out your guess, so you can check
to see if it multiplies back to the original problem. It does not, however, alert the student if
a polynomial is unfactorable. You are left feeling stupid if you can’t find the right answer to
guess. I give this method a big Mr. Yuck sitcker.
Later I learned a method called the “Texas Box” method and then the “A-C” method. They
both have merit, and are perfectly fine to use. But when I analyzed them in detail, I found
the two methods are actually identical except for the detail of how to finish up with factoring
by grouping. We already have two methods for factoring by grouping, so there is nothing
35
new to learn here about that. So I merged them together in this book, and I will call this one
thing the General Trinomial method.
Once this method has done its work of splitting the middle term into two, you can finish up
with factoring the four terms by grouping — either by the traditional method or by using the
grid. Although the grid looks a little scarier, students who use it tend to make fewer errors,
so I recommend it. You are always free to use the traditional method if you choose to.
Procedure: General Trinomial method
1. Make sure the terms are in descending order of powers. Usually they are, but
some need to be rearranged. Also, make sure the squared term is positive. If it is
not, factor out a –1 to force it positive. The method is difficult to use if the first term is
negative.
2. Take only the number part of the first term and multiply it by the last term to get
the “key number.” This is where the A-C method got its name — for Ax2 + Bx + C,
you multiply A and C together.
3. Make a list of pairs of numbers that multiply together to the key number. Watch
your signs just like in the Key Number method (Blue Belt).
4. Pick the key pair that add together to get the middle term’s coefficient (non-x part).
If there is no pair that works, the trinomial is unfactorable, or prime.
5. Take the middle term of the original trinomial, and tack the variable part of it onto
each one of the pair you just found. Replace the middle term by the two terms
(which add up to the original middle term).
6. Finish up by factoring by grouping (using either the grid or the traditional method).
Example Factor 6x2 − 17x + 10 using the General Trinomial method.
1. Terms are in proper order by powers.
2. Key number is 6 · 10 = 60.
3. The list of all pairs of numbers that multiply together to the key number. But the
signs must both be negative, because the key number is positive (meaning the
signs match) and the middle term is negative (meaning that both are negative).
Refer to the Blue Belt’s Special Notes Regarding Negatives on page 34.
-1
-2
-3
-4
-5
-6
-60
-30
-20
-15
-12
-10
4. The key pair that add together to get the middle term’s coefficient: –5 and –12,
because they add to –17.
5. The middle term as the variable x, so we take our pair and put the same variable on them: −5x and −12x. We then put them in place of the middle term. (If
either of the pair were positive, we would need to put a + sign in with it.) Our
polynomial is now: 6x2 − 5x − 12x + 10.
36
6. We finish up by factoring by grouping.
(a) If you use the traditional method, rearrange the terms to get the more similar parts together: 6x2 − 12x − 5x + 10. Then add parentheses, being careful to
change the signs for the second pair because of the negative sign: 6x2 − 12x −
(5x − 10).
This becomes 6x ( x − 2) − 5 ( x − 2) when we factor the GCF from each pair,
and so the final answer is ( x − 2) (6x − 5).
(b) If you use the grid, the terms can be arranged like this:
GCFs
6x2
−12x
−5x
10
Then find the GCFs for each column and each row:
GCFs
x
−2
6x2
6x
−12x
−5
−5x
10
So the final answer is ( x − 2) (6x − 5).
Note: You do not need to use both the grid and the traditional method for the finishing up
using grouping. Choose your favorite, and then excel with it.
Example Factor using the General Trimomial method: 20x2 + 3x − 2
Note: You could use guess-and-check if you like it.
1. The terms are in proper order.
2. The key number is (20) (−2) = −40.
3. The list of factor pairs is:
–1
40
–2
20
–4
10
–5
8
4. The key pair is −5 and 8, since they add to positive 3 (the middle term’s coefficient).
5. Replacing the middle term with the new pair, we get:
20x2 + 3x − 2 = 20x2 − 5x + 8x − 2
6. Factoring by grouping yields the final answer:
GCFs
4x
−1
5x
20x2
−5x
2
8x
−2
Answer: (5x + 2) (4x − 1)
37
4
Special Pattern Factorizations
4.1
Green Belt: Factoring the Difference of Squares
Use this table to help you determine perfect squares for factoring the difference of squares.
Square Root
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Square
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
Important: You can not factor the sum of squares. x2 + y2 is not factorable.1 Period. It is
prime. But the difference of squares, like x2 − y2 , is factorable. There is a pattern to it. First
we find the “square roots” (which we call A and B) and then we plug them into the pattern
(A + B)(A – B) as the answer.
Procedure: Factoring the difference of squares
1. Find the square root of the first term and call it A. Then find the square root of the
second term and call it B.
(a) To find the square root of a plain number, locate the number on the line labeled
“square” in the table above. The square root is immediately above the number
you found, in the line labeled “square root”. Or you can use the square root
function of your calculator.
(b) To find the square root of the letter parts (each letter individually), take half
the exponent, or if you prefer, divide the exponent by 2.
2. Plug the square roots into the pattern (A + B)(A – B).
3. Check your answer.
1 In an advanced method,
we find a way to do this using complex numbers. But there is still no way to factor
it using real numbers. So, unless you are working on advanced stuff in College Algebra, just remember that it
is not factorable.
38
Example: Factor x2 − 9y2
1. The first term’s square root is found by dividing the exponent by 2. This first
square root we call A:
A = x1 = x
The second term’s square root is found in two steps. First, locate 9 on the squares
line of the table. The square root of 9 is 3. Next, the square root of y2 is y (dividing
the exponent by 2). So the second square root we call B:
B = 3y
2. Plugging the square roots into the pattern (A + B)(A – B) we get:
( x + 3y)( x − 3y)
3. Check your answer!
Example: Factor 4x2 − 25y2 z4
1. The first term’s square root is found in two steps. First, locate 4 on the squares line
of the table. The square root is just above it. So the square root of 4 is 2. Next, we
look at the letter part, and divide the exponent by 2. This first square root we call
A:
A = 2x
The second term’s square root is found in three steps. First, locate 25 on the
squares line of the table. The square root of 25 is 5. Next, the square root of y2
is y, and lastly, the square root of z4 is z2 . So the second square root we call B:
B = 5yz2
2. Plug the square roots into the pattern (A + B)(A – B) to get:
(2x + 5yz2 )(2x − 5yz2 )
3. Check your answer!
4.2
Blue Belt: Factoring the Sum or Difference of Cubes
Use this table to help you determine perfect cubes for factoring the sum or difference of
cubes.
Cube Root 1 2 3
4
5
6
7
8
9
10
11
12
Cube
1
8
27
64
125
216
343
512
729
1000
1331
1728
This type of factoring requires using a pattern, kind of like the factoring the difference of
squares. But there are some differences. Namely, the pattern is different. And we use cube
roots instead of square roots. Also you can factor with either a + or – (sum or difference). So
we have two similar patterns: one for + and one for –.
Procedure: Factoring the sum or difference of cubes
1. Find the cube root of the first term and call it A. Then find the cube root of the
second term and call it B.
39
(a) To find the cube root of the number part, locate the number on one of the lines
labeled “cube” in the table above. The cube root is immediately above the
number you found, in the line labeled “cube root”.
(b) To find the cube root of the letter parts (each letter individually), take onethird the exponent, or if you prefer, divide the exponent by 3.
2. Make a list to refer to while plugging things into the pattern. The list should contain five things: A, B, A2 , AB, and B2 . To find A2 , multiply the A cube root times
itself. To get AB, multiply the two cube roots together. And to get B2 , multiply the
B cube root times itself.
3. Select the pattern that fits the original problem. Here are the two patterns. They
are very similar.
A3 − B3 = ( A − B) A2 + AB + B2 A3 + B3 = ( A + B) A2 − AB + B2
4. Plug the five items from your list into the correct pattern.
5. Check your answer.
Special Notes Regarding the Patterns
When selecting the correct pattern, there is some additional information that may help you
make the right choice, and maybe even memorize the patterns. Memorization helps your
speed greatly.
In the first set of parentheses, put A + B or A − B, whichever sign matches the original
problem. For x3 − y3 , the first parentheses contain x − y; for x3 + y3 , the first parentheses
contain x + y.
Remember: same sign.
The second set of parentheses contains 3 terms. The first is A2 . Square each number and
double the exponent on each letter. The second term is AB (the two cube roots multiplied
together), with the opposite sign as was used in first set of parentheses. And the last is B2
(always positive). So for x3 − y3 , the last set of parentheses would contain x2 + xy + y2 . For
x3 + y3 , the last set of parentheses would contain x2 − xy + y2 . The first and last terms are
ALWAYS plus, and the middle terms is the opposite of the sign used in the first parentheses.
Remember: positive, opposite sign, positive.
Example: Factor 27x3 + 64y6
1. Find A. The cube root of 27 is 3, and the cube root of x3 is x1 or just plain x. So
A = 3x
Find B. The cube root of 64 is 4, and the cube root of y6 is y2 . So B = 4y2
2. Find A2 : A2 = (3x )(3x ) = 9x2
Find AB: AB = (3x )(4y2 ) = 12xy2
Find B2 : B2 = (4y2 )(4y2 ) = 16y4
3. Choose the pattern with + like the original problem: A3 + B3 = ( A + B) A2 − AB + B2
4. Plug our As and Bs into the pattern, to get: (3x + 4y2 )(9x2 − 12xy2 + 16y4 )
5. Check your answer.
40
4.3
Red Belt: Factoring Perfect Square Trinomials
There is one more factoring method to cover that is optional for beginners. Without it, you
can always use either the Key Number method or the General Trinomial method, whichever
is applicable, and get the same result. The advantage to learning this method is that it is a
little faster and easier if you are good at recognizing and dealing with patterns. So if you
hate the Green and Blue Belts’ problems with patterns, you can skip this part at first. But be
sure to come back and learn it when you are more advanced — it can save you a ton of time.
This kind of factoring is very specialized, but if it applies to a given problem, it saves a lot
of time. Most of the procedure below has to be completed before you know whether the
method works or not, so you will want to do a quick check first to save time.
Put the terms in order of descending powers (for one variable) and then check whether the
first and last terms are both perfect squares. If so, this method is worth a shot.
Procedure: Perfect trinomial square shortcut
1. Rearrange the terms (if needed) into descending order of powers.
2. Find the square roots of the first and last terms (square roots, just as in Yellow Belt).
Call them A and B.
3. Multiply the two square roots together, and double the answer. Call this 2AB.
4. Check to see if this 2AB matches the middle term of the original problem. Or if
−2AB matches, that is good, too. Either way is OK. However, if it does not match,
this method will not work. Stop and try another way.
5. Select the pattern that matches the original problem. The patterns are:
A2 + 2AB + B2 = ( A + B)2
A2 − 2AB + B2 = ( A − B)2
6. Plug in your A and B. The answer is ( A + B)2 or ( A − B)2 , whichever matches the
pattern.
7. Check your answer.
Example: (This one works!) Factor: 4x2 − 12x + 9
1. Terms are in descending order of powers. Great!
2. The square roots A and B are: A = 2x, B = 3
3. Now find 2AB: (2)(2x )(3) = 12x
4. Is that equal to the middle term, or else the negative of the middle term? Yes, 12x
is the negative of −12x
5. Choose the pattern that matches: A2 − 2AB + B2 = ( A − B)2
6. Plug in your A and B. The answer is (2x − 3)2
7. Check your answer.
Example: (This one fails!) Factor: x2 + 3x + 9
41
1. Terms are in descending order of powers. Great!
2. The square roots A and B are: A = x, B = 3
3. Now find 2AB: (2)( x )(3) = 6x
4. Is that equal to the middle term, or else the negative of the middle term? No,
6x does not match 3x. Since it does not match, we can’t continue this method.
The next step would be to try the Key Number method, since the squared term
does not have a number part. The General Trinomial method would be used if the
squared term did have a number in front of it.
5
Mixed Factoring (Conclusion of Red Belt)
Now that you are good at all these different kinds of factoring, it is time to learn how to
factor a polynomial without being told which method to use. There are definite clues about
which method to use, and also several pitfalls to learn about so that you don’t fall into them.
First, study this section to learn what you can from a book, then practice determining which
method(s) to use for a random problem.
5.1
How to Choose the Correct Factoring Method
1. Always and always factor out all common factors first thing. (White Belt) If there aren’t
any common factors, that’s OK. But you’ve got to check. Some of the methods below
will not work if you don’t.
2. Count the terms. Then choose the correct method as follows:
• 4 or more terms: factor by grouping (Yellow Belt)
• 3 terms:
– perfect square trinomial (Red Belt, if possible)
– Key number (if it has a bare x2 term) (Orange Belt)
– General trinomial (otherwise, such as: 3x2 + 5x − 4) (Purple Belt)
• 2 terms:
– difference of squares (if exponents are multiples of 2) (Green Belt)
– difference of cubes or sum of cubes (if exponents are multiples of 3) (Blue Belt)
3. Check each factor to see if it can be factored further.
4. Check your answer. If you multiply the factors together, you should get the original
problem back again.
42
5.2
Things to Watch Out For
There are three things that people commonly forget to do, any of which can leave them with
the wrong answer:
• forgetting to first and always check for common factors,
• forgetting to see if a factor can be factored further, and
• forgetting to check the answer by multiplying it out.
Every factoring problem answer should be checked!
6
Brown Belt: Advanced Factoring
These advanced factoring techniques are generally not used until College Algebra or beyond.
6.1
Basic Synthetic Division
There is one special technique with polynomials that can help in advanced factoring that
requires our attention by the time you take College Algebra. In lower levels of Algebra, we
covered division with polynomials, but most students do not get any exposure to a nifty
time-saving technique called “synthetic division.” If you have seen it before, it’s still good to
review it and make sure that you are practiced up on it.
Another reason to look at it now is that most instructors and textbooks that deal with synthetic division say that there are many problems that it cannot be used on, because they are
too involved. These notes describe how to use synthetic division on any polynomial division
with a single variable. If there are two or more different variables, then it may be true that
you can’t use synthetic division. But that should be the only thing that must stop you if you
want to use the method.
You might want to review the section on synthetic evaluation (see page ??), because synthetic
division uses the same steps. As strange as it sounds, dividing a polynomial by x − 3 can be
accomplished by evaluating the polynomial at x = 3. All we need to do is take the non-x
part of the downstairs divisor, change the sign, and evaluate.
Procedure Synthetic Division
For ease in explanation, we will be working an example as we go through the procedure. We will
divide 2x2 + 3x − 14 by x − 2.
1. Arrange the terms in order of descending powers (from highest power of x down to
lowest). Our example is already good to go.
43
2. Put in a placeholder for any “missing terms.” For example, x2 + x + 1 has no missing
terms, because from highest to lowest, the exponent on the x doesn’t skip anything.
But x3 + x + 1 has a missing term: there is no x2 term in it. Placeholders can be inserted
with a zero times the missing term. So x3 + x + 1 should be written as x3 + 0x2 + x + 1.
Our example has no missing terms.
3. Write the coefficents in a row with a little space between each one. Leave space for
more numbers above them, and draw a line below them. Determine the degree of the
divisor. Draw two vertical lines as shown. You need to make the number of columns
to the left of the first vertical line match the number of columns to the right of the
second line, and both these should be equal to the degree of the divisor. Our example
has a divisor of degree 1, so we need one column on the left of the first vertical line and also one
column to the right of the second vertical line. The setup looks like this:
+
2
3
−14
÷
4. Look at the polynomial you are dividing by (the downstairs if written as a fraction).
The term with the highest power of x (or other variable) determines whether we need
the division row or not. If its leading coefficient is 1, omit the division row. But if it is
anything else, put ÷ and the coefficient in the row just below the horizontal line. Now
write the opposite of the coefficient of all the lower power terms on the bottom row.
If there are any missing terms in the divisor, put a 0 in their place. Our example has a
leading coefficient of 1, so we will omit the division row. Your setup should look like this:
+
2
3
−14
2
Note: when dividing by x − 2, you are actually evaluating at x = 2, and for x + 2, you
are evaluating at x = −2.
5. Now we begin a pattern. It goes like this: starting at the left after the first vertical
line, add all the numbers in the leftmost column of the addition area together, and
put the answer under the line (in the division row if there is one; otherwise in the
multiplication row), divide by the number in the division row (if there is one) and write
the answer below that in the multiplication row, multiply by each of the numbers in the
multiplication row, and put the answer in the next column in the addition area above
the line. If there are additional numbers to multiply by, put their products stepping
over to the right. This pattern repeats until all the non-remainder columns are full.
Here is what your work should look like at each step of the way:
Important: Do not multiply or divide in the remainder columns! Do ONLY the addition
for remainders!
44
+
2
2
2
3
−14 first column only has a 2, so we drop it down
4
+
2
2
2
3
−14 Multiplied (2)(2), answer goes into a higher row in the next column.
4
+
2
3
2
2
7
4
+
2
3
2
2
7
4
+
2
3
2
2
7
−14 Added 4+3
14
−14 Multiplied 2 times 7
14
−14 Added -14+14.
0
1. The bottom row represents the answer to the division. You can read it from right to
left, kind of backwards. Keep going with higher powers of x as needed.
4
14
+
2
3
−14
2
2
7
0
N/A
x’s
plain number
remainder
Answer: 2x + 7 (Since the remainder is zero, we can drop that part)
3x3 + 2x2 − 7x − 4
using synthetic division.
x+3
Here is the setup for the division:
Example Divide
+
3
2
-7
−4
−3
Now we fill in all the blanks, using the usual pattern of adding down, then multiplying
to begin the next column:
45
+
3
−9
21
−42
2
−7
−4
14
−46
−3 3 −7
This time we did not get a zero remainder, so we will express it as a fraction (over the
x + 3 that we are dividing by).
Answer: 3x2 − 7x + 14 +
6.2
−46
x+3
Advanced Synthetic Division
Any student who is already familiar with synthetic division will notice that the setup shown
herein has a different appearance from what many other books use. The reason for that is that
most other books wimp out on the tougher problems and say that they can’t be done with
synthetic division. I changed the appearance (but not the method behind the procedure) to
make it more fluid for all types of problems. You are free to use the more conventional setup
if you want, but it is less graceful and you will need to adapt it somewhat for the tougher
problems.
Here is the generic setup for a problem:
ax3 + bx2 + cx + d
.
mx2 + nx + p
addition area, number
a
+
÷
m
−n − p
b
c
d
of rows varies
division row (sometimes not needed)
multiplication row
The top rows, the “addition area” above the top horizontal line, are just like we had for
synthetic evaluation. Those numbers are added together in columns. Depending on the
problem, there may be more rows going up higher and higher, but they are all for addition,
no matter how high we go. The division row (marked with the division symbol ÷) is not
always needed — we only include it when the leading coefficient of the divisor is not 1
(m 6= 1). The multiplication row on the bottom, is just like for synthetic evaluation, but
sometimes we need more columns on the left side of the line. The number of columns on
the left side of the first vertical line will match the degree of the divisor (the polynomial we
are dividing by). The second vertical line marks the cutoff point for the remainder, and we
always need the same number of columns to the right of it as we have to the left of the first
vertical line. So there is always a symmetry between left and right.
Note: For the simplest division problems (which are very common), the setup we use
matches exactly what we used for synthetic evaluation (see page 9).
2x3 − 5x2 − 22x − 15
using synthetic division.
Example Divide
2x + 3
46
Note: This example requires a division row, since the leading coefficient downstairs is
not equal to 1. Here is the setup:
+
2
−5 −22
−15
÷2
−3
Now we fill in the blanks, by taking in order: add, divide, and multiply.
−3
+
2
12
−5 −22
15
−15
÷2 2 −8 −10
0
−3 1 −4
*
−5
Watch out! * Remember to never, ever divide or multiply in the remainder column(s).
Answer: x2 − 4x − 5
6x4 − 17x3 − 61x2 − 23x + 16
Example Divide
2x2 − 7x − 15
Note: This example requires not only a division row, since the leading coefficient downstairs is not equal to 1, but also two columns on the left and on the right. Here is the
setup:
+
6
−17 −61
−23 16
÷2
7
15
Now we fill in the blanks, by taking in order: add, divide, and multiply.
21
+
6
÷2
6
7
15
45
−17 −61
−23 16
3
Notice the 3 · 15 goes into the next column after the 3 · 7. The vertical position is not
important, so just stack it on top of the heap.
47
14
21
−17 −61
+
6
÷2
6
4
3
2
7
15
45
30
−23 16
This time the 2 · 7 and the 2 · 15 do not end up in the same row. That’s OK. Only the
column position is important.
21
14
-7
45
30
-15
−23
16
−17 −61
+
6
÷2
6
4
−2
0
1
3
2
−1
*
*
7
15
Now that we have completed the non-remainder part, all that is left is to do ONLY
the addition for the remainder. Also, note that the remainder part has the same kind
of power business going on as the non-remainder part. In this example, we have the
plain number part as the right-most column, and the x-term coefficient to the left of it.
Since it’s zero in this example, we don’t have any x-term in the top of the remainder
fraction.
Watch out! * Remember to never, ever divide or multiply in the remainder column(s).
1
Answer: 3x2 + 2x − 1 + 2
2x − 7x − 15
Practice Problems
Try synthetic division on these puppies:
1.
3x3 − 2x2 + 5x − 4
x+2
2.
y5 − 1
y−1
3.
2k3 + 5k − 2
2k + 1
6.3
Using Verified Roots to Factor Polynomials
This section is missing a lot of detail, and it is doubtful that any student would be able to
master all the techniques mentioned just from this material. However, it is hoped that it will
48
be sufficient for a review; it may help suggest enough that you might remember how to get
through some of the techniques. Remember to get as much help as you need from other
sources, including coming to class!
6.3.1
Listing all possible rational roots
There are a couple of things you should routinely do when confronted with a higher order
polynomial to factor. Always check to see if there are common factors to factor out. Then
arrange the terms in descending order of powers if they are not already. Then you are ready
to begin making the list. Here is how to make your list of all possible rational roots:
Procedure Listing all possible rational roots
1. Make a list of all the factors of the plain number term, just like what you do with
the key number method. These are the possible numerators of the roots.
2. Make the same kind of list of all the factors of the coefficient of the highest power
term (the first term, right?). These are the possible denominators of the roots.
3. Combine the numerators from step 1 with each possible denominator from step 2,
and list them twice: once positive and once negative. This makes the entire list of
all possible rational roots for the polynomial. Nothing else is possible.
Example Find all possible rational roots for x2 + 5x + 6.
1. All possible factors for 6 are: 1, 6, 2, 3. These will be the upstairs parts.
2. All possible factors for 1 are: 1. The only downstairs part is 1, so in effect, we don’t
have fractions for this example.
3. The final list will be the list from step 1, with plus and minus: 1, −1, 2, −2, 3,−3,
6, −6
Remember that this list is not a list of all the roots. It is only a list that must contain all the
rational roots. There may be irrational roots or complex roots, and they will not be in the list.
And, almost always, there will be some duds in the list that are not roots at all. The list can
even be all duds! All we know is that there are not going to be any rational roots that are not
in the list.
Example Find all possible rational roots for 2x3 + 15x2 − 32x + 7
1. All possible factors of 7: 1, 7
2. All possible factors of 2: 1, 2
3. The final list is: 1, 7, 21 , 72 , −1, −7, − 21 , − 72 .
49
6.3.2
Using a Graphing Calculator to Narrow the Choices
If you graph a polynomial function on a graphing calculator (aka “grakulator”), the real
roots will be obvious on the graph as points where the graph touches the x-axis. Use the
grakulator to calculate the “zeros” (which is what it calls the roots), but remember that it
only gives decimal approximations.
Figure out which of the possible rational roots would be approximately the same value as
p
what the grakulator says, and then use synthetic division to verify that the ± value is
q
indeed an exact root.
Example Find the real roots of 3x4 + 4x3 − x2 + 4x − 4
In the shortest cheater’s way of expressing it, the possible rational roots are: ±
we look at the graph, we see two places that the graph touches the x-axis:
1, 2, 4
. When
1, 3
We ignore the y values, which should be zero or very close to it. So the grakulator says that
we have two roots: one near –2 and the other near 0.67.
p
Which of the ± values are near those? –2 is obvious, but 0.67 is not. However, the only
q
2
1
way to get a positive fraction less than 1 is to choose or . The grakulator can be used to
3
3
1
find decimal equivalents of these fractions if you need help, and we find that ≈ 0.33 and
3
2
2
≈ 0.67. Therefore, is the guy to try.
3
3
2
Use synthetic evaluation to check that you get exactly zero for –2 and again for . That nasty
3
old 0.67 will NOT get you a zero remainder. There is more on this in the next section.
6.4
Analyzing graphs to find roots
Since the exponent of the leading term tells us exactly how many roots to account for, there
are many cases where we can find all the roots for a polynomial. We use a combination of
techniques: analyzing the graph, synthetic division, finding zero roots — and sometimes
people use Descartes’ Rule of Signs to figure out how many positive roots and how many
negative roots there are.
As you account for a root, you will be able to factor out the factor associated with that root.
And once you have accounted for all the roots except the last two, you can use the quadratic
50
formula or any other method for solving quadratic equations to find the last two roots pretty
easily.
We have several tools at our disposal to accomplish this task. Remember that every root we
find will make it easier to find the others. Just take it one step at a time, and start with the
easiest steps first.
Watch out! When you are finding the roots of a polynomial function, remember that some
roots are multiple roots. So if you know that there must be two real roots, but you can only
find one, remember the two roots can have the same value!
6.4.1
Factors associated with roots
Every root has a factor associated with it. It is of the form ( x − root). So if you find that 2 is
a root, then ( x − 2) is the factor that goes with it, and you can factor it out of the polynomial.
Watch your signs, especially with negative roots. If you find that −3 is a root, then ( x − (−3))
is its factor, but we will simplify the double negative and say ( x + 3) is the factor.
6.4.2
Number of “zero roots” (roots equal to zero)
The easiest roots to find are the ones that occur where x = 0. These are called “zero roots.” If
all the terms in the polynomial have a common factor of x to some power, factor the common
part out. Whatever power of x you can factor out of all the terms gives you a zero root, and
the exponent tells you the multiplicity of it.
Example Find the number of zero roots in P( x ) = 2x3 − 3x2 .
Since all terms have x to some power in them, we can factor out the smallest power of x
from all the terms. This is x2 . Then P( x ) = x2 (2x − 3). From this we see there is a zero
root with multiplicity 2. The remaining root can be found easily by other methods.
Example Find the number of zero roots in P( x ) = 2x3 − 3x.
Since all terms have x to some power in them, we can factor out the smallest power of
x from all the terms. This is x. Then P( x ) = x (2x2 − 3). From this we see there is a zero
root with no multiplicity. The remaining roots can be found easily by other methods.
Example Find the number of zero roots in P( x ) = 2x3 − 3.
Since there is a term that has no x, we can’t factor out any power of x from all the terms.
Therefore there is no zero root. The 3 roots must be found by other methods, but the
chore may not be easy.
Remember to factor out all the factors associated with zero roots. So, with the second example above, 2x3 − 3x would be factored out to x (2x2 − 3), and then you would set to work
finding the roots of the leftover part in parentheses. Each root you find makes it easier to
find the other roots.
51
6.4.3
Using a graph to find non-zero real roots
Your graphing calculator will show you approximately where the graph touches or crosses
the x-axis. These places are real roots, but remember that the graphing calculator provides
only approximate values. You will need to use synthetic division to verify them as actual exact
roots, and some of them will not be rational.
p
Therefore, use the ± list of possible rational roots, and see which of them are very close to
q
the decimal approximations that show up as real roots. Then use synthetic division to verify
the exact value and, as a nice bonus, factor out the associated factor.
Example Find a real root of f ( x ) = 3x3 + x2 + 12x + 4.
The first thing to look for is zero roots, but there are none. Next we graph the polynomial function and examine the graph for where it crosses the x-axis:
We only pay attention to the x value where the graph touches the x-axis. In this case,
p
that is −0.333. That is not an exact value, so we need to look for a ± possible root
q
that is approximately equal to the −0.333 that the grakulator came up with. Of all the
1, 2, 4
1
possibles: ±
, only − ≈ −0.333, so that is the most likely suspect of being the
1, 3
3
exact root. We need to verify this using synthetic division.
−1
0
−4
+
3
1
12
4
− 13
3
0
12
0
Since the remainder is 0, then −1/3 is the exact value of a real root. Note that −0.333
would not have come out with zero remainder, because it is only an approximate root,
not exact.
Note: Remember that the middle part of the bottom line will be extremely useful for finding
further roots. It can be interpreted as 3x2 + 0x + 12 or as 3x2 + 12 (ignoring the 0x term in
the middle). There is no need to go back to the original function to find the other roots. As
a matter of fact, do not go back to the original unless you like to work way too hard all the
time!
52
Once you have found a root and verified it using synthetic division (with 0 remainder), use the quotient part in the middle of the bottom line to find further roots,
and each successive one will become easier to find. If you ignore this, it will likely
become impossible to finish finding all the roots!
6.4.4
Adjusting for integer coefficients
In the last example, we found
a
root with a fractional value. If we wrote what we found in
1
factored form, it became x −
3x2 + 12 . This form has a couple of things we don’t like
3
to have if we can avoid it: the fraction in the first parentheses, and the common factor of 3
in the second parentheses. It is pretty common, as in this case, for these two hiccups to help
each other out.
If we factor out the common 3 from the second parentheses,
and multiply it instead to the
2
contents of the first parentheses, we get (3x − 1) x + 4 and now have integer coefficients
and no common factors within one set of parentheses. That is better and easier to work with.
6.4.5
Looking for multiplicity
The second thing that the graph can tell you about is multiplicity of roots. Look at the area
where the graph approaches the x-axis. If it flattens and goes horizontal at the axis, then it is
a multiple root.
If you see evidence of multiplicity, either even or odd, use synthetic division to factor out the
same root again.
6.4.6
Irrational and Irreal roots always come in pairs
The last stage of finding all the roots and factors of a polynomial function comes when you
have found all of them except the last two roots. If you followed my advice and worked your
way from the easiest roots to find (the zero roots), then the medium ones that grakulators and
synthetic division help you find, then the last two are going to be the hardest kind to find.
The good news is that you have excellent ways to find the last two roots. Most people use
the quadratic formula, but sometimes the square root property or factoring can work. Many
times, however, you will end up with two irrational roots or two complex roots. Remember
they always come in conjugate pairs.
53
A
Appendix of Reference Materials
This appendix contains materials for easy reference.
54
2=2
3=3
4 = 22
5=5
6=2•3
7=7
8 = 23
9 = 32
10 = 2 • 5
11 = 11
12 = 22 • 3
13 = 13
14 = 2 • 7
15 = 3 • 5
16 = 24
17 = 17
18 = 2 • 32
19 = 19
20 = 22 • 5
21 = 3 • 7
22 = 2 • 11
23 = 23
24 = 23 • 3
25 = 52
26 = 2 • 13
27 = 33
28 = 22 • 7
29 = 29
30 = 2 • 3 • 5
31 = 31
32 = 25
33 = 3 • 11
34 = 2 • 17
35 = 5 • 7
36 = 22 • 32
37 = 37
38 = 2 • 19
39 = 3 • 13
40 = 23 • 5
41 = 41
42 = 2 • 3 • 7
43 = 43
44 = 22 • 11
45 = 32 • 5
46 = 2 • 23
47 = 47
48 = 24 • 3
49 = 72
50 = 2 • 52
51 = 3 • 17
52 = 22 • 13
Prime Factorizations of Natural Numbers 2 through 409
53 = 53
104 = 23 • 13
155 = 5 • 31
3
54 = 2 • 3
105 = 3 • 5 • 7
156 = 22 • 3 • 13
55 = 5 • 11
106 = 2 • 53
157 = 157
56 = 23 • 7
107 = 107
158 = 2 • 79
2
3
57 = 3 • 19
108 = 2 • 3
159 = 3 • 53
58 = 2 • 29
109 = 109
160 = 25 • 5
59 = 59
110 = 2 • 5 • 11
161 = 7 • 23
60 = 22 • 3 • 5
111 = 3 • 37
162 = 2 • 34
4
61 = 61
112 = 2 • 7
163 = 163
62 = 2 • 31
113 = 113
164 = 22 • 41
2
63 = 3 • 7
114 = 2 • 3 • 19
165 = 3 • 5 • 11
6
64 = 2
115 = 5 • 23
166 = 2 • 83
65 = 5 • 13
116 = 22 • 29
167 = 167
2
66 = 2 • 3 • 11
117 = 3 • 13
168 = 23 • 3 • 7
67 = 67
118 = 2 • 59
169 = 132
2
68 = 2 • 17
119 = 7 • 17
170 = 2 • 5 • 17
69 = 3 • 23
120 = 23 • 3 • 5
171 = 32 • 19
2
70 = 2 • 5 • 7
121 = 11
172 = 22 • 43
71 = 71
122 = 2 • 61
173 = 173
72 = 23 • 32
123 = 3 • 41
174 = 2 • 3 • 29
2
73 = 73
124 = 2 • 31
175 = 52 • 7
3
74 = 2 • 37
125 = 5
176 = 24 • 11
75 = 3 • 52
126 = 2 • 32 • 7
177 = 3 • 59
2
76 = 2 • 19
127 = 127
178 = 2 • 89
7
77 = 7 • 11
128 = 2
179 = 179
78 = 2 • 3 • 13
129 = 3 • 43
180 = 22 • 32 • 5
79 = 79
130 = 2 • 5 • 13
181 = 181
80 = 24 • 5
131 = 131
182 = 2 • 7 • 13
81 = 34
132 = 22 • 3 • 11
183 = 3 • 61
82 = 2 • 41
133 = 7 • 19
184 = 23 • 23
83 = 83
134 = 2 • 67
185 = 5 • 37
2
3
84 = 2 • 3 • 7
135 = 3 • 5
186 = 2 • 3 • 31
85 = 5 • 17
136 = 23 • 17
187 = 11 • 17
86 = 2 • 43
137 = 137
188 = 22 • 47
87 = 3 • 29
138 = 2 • 3 • 23
189 = 33 • 7
3
88 = 2 • 11
139 = 139
190 = 2 • 5 • 19
89 = 89
140 = 22 • 5 • 7
191 = 191
2
90 = 2 • 3 • 5
141 = 3 • 47
192 = 26 • 3
91 = 7 • 13
142 = 2 • 71
193 = 193
92 = 22 • 23
143 = 11 • 13
194 = 2 • 97
4
2
93 = 3 • 31
144 = 2 • 3
195 = 3 • 5 • 13
94 = 2 • 47
145 = 5 • 29
196 = 22 • 72
95 = 5 • 19
146 = 2 • 73
197 = 197
96 = 25 • 3
147 = 3 • 72
198 = 2 • 32 • 11
97 = 97
148 = 22 • 37
199 = 199
2
98 = 2 • 7
149 = 149
200 = 23 • 52
2
2
99 = 3 • 11
150 = 2 • 3 • 5
201 = 3 • 67
100 = 22 • 52
151 = 151
202 = 2 • 101
101 = 101
152 = 23 • 19
203 = 7 • 29
2
102 = 2 • 3 • 17
153 = 3 • 17
204 = 22 • 3 • 17
103 = 103
154 = 2 • 7 • 11
205 = 5 • 41
206 = 2 • 103
207 = 32 • 23
208 = 24 • 13
209 = 11 • 19
210 = 2 • 3 • 5 • 7
211 = 211
212 = 22 • 53
213 = 3 • 71
214 = 2 • 107
215 = 5 • 43
216 = 23 • 33
217 = 7 • 31
218 = 2 • 109
219 = 3 • 73
220 = 22 • 5 • 11
221 = 13 • 17
222 = 2 • 3 • 37
223 = 223
224 = 25 • 7
225 = 32 • 52
226 = 2 • 113
227 = 227
228 = 22 • 3 • 19
229 = 229
230 = 2 • 5 • 23
231 = 3 • 7 • 11
232 = 23 • 29
233 = 233
234 = 2 • 32 • 13
235 = 5 • 47
236 = 22 • 59
237 = 3 • 79
238 = 2 • 7 • 17
239 = 239
240 = 24 • 3 • 5
241 = 241
242 = 2 • 112
243 = 35
244 = 22 • 61
245 = 5 • 72
246 = 2 • 3 • 41
247 = 13 • 19
248 = 23 • 31
249 = 3 • 83
250 = 2 • 53
251 = 251
252 = 22 • 32 • 7
253 = 11 • 23
254 = 2 • 127
255 = 3 • 5 • 17
256 = 28
257 = 257
258 = 2 • 3 • 43
259 = 7 • 37
260 = 22 • 5 • 13
261 = 32 • 29
262 = 2 • 131
263 = 263
264 = 23 • 3 • 11
265 = 5 • 53
266 = 2 • 7 • 19
267 = 3 • 89
268 = 22 • 67
269 = 269
270 = 2 • 33 • 5
271 = 271
272 = 24 • 17
273 = 3 • 7 • 13
274 = 2 • 137
275 = 52 • 11
276 = 22 • 3 • 23
277 = 277
278 = 2 • 139
279 = 32 • 31
280 = 23 • 5 • 7
281 = 281
282 = 2 • 3 • 47
283 = 283
284 = 22 • 71
285 = 3 • 5 • 19
286 = 2 • 11 • 13
287 = 7 • 41
288 = 25 • 32
289 = 172
290 = 2 • 5 • 29
291 = 3 • 97
292 = 22 • 73
293 = 293
294 = 2 • 3 • 72
295 = 5 • 59
296 = 23 • 37
297 = 33 • 11
298 = 2 • 149
299 = 13 • 23
300 = 22 • 3 • 52
301 = 7 • 43
302 = 2 • 151
303 = 3 • 101
304 = 24 • 19
305 = 5 • 61
306 = 2 • 32 • 17
307 = 307
308 = 22 • 7 • 11
309 = 3 • 103
310 = 2 • 5 • 31
311 = 311
312 = 23 • 3 • 13
313 = 313
314 = 2 • 157
315 = 32 • 5 • 7
316 = 22 • 79
317 = 317
318 = 2 • 3 • 53
319 = 11 • 29
320 = 26 • 5
321 = 3 • 107
322 = 2 • 7 • 23
323 = 17 • 19
324 = 22 • 34
325 = 52 • 13
326 = 2 • 163
327 = 3 • 109
328 = 23 • 41
329 = 7 • 47
330 = 2 • 3 • 5 • 11
331 = 331
332 = 22 • 83
333 = 32 • 37
334 = 2 • 167
335 = 5 • 67
336 = 24 • 3 • 7
337 = 337
338 = 2 • 132
339 = 3 • 113
340 = 22 • 5 • 17
341 = 11 • 31
342 = 2 • 32 • 19
343 = 73
344 = 23 • 43
345 = 3 • 5 • 23
346 = 2 • 173
347 = 347
348 = 22 • 3 • 29
349 = 349
350 = 2 • 52 • 7
351 = 33 • 13
352 = 25 • 11
353 = 353
354 = 2 • 3 • 59
355 = 5 • 71
356 = 22 • 89
357 = 3 • 7 • 17
358 = 2 • 179
359 = 359
360 = 23 • 32 • 5
361 = 192
362 = 2 • 181
363 = 3 • 112
364 = 22 • 7 • 13
365 = 5 • 73
366 = 2 • 3 • 61
367 = 367
368 = 24 • 23
369 = 32 • 41
370 = 2 • 5 • 37
371 = 7 • 53
372 = 22 • 3 • 31
373 = 373
374 = 2 • 11 • 17
375 = 3 • 53
376 = 23 • 47
377 = 13 • 29
378 = 2 • 33 • 7
379 = 379
380 = 22 • 5 • 19
381 = 3 • 127
382 = 2 • 191
383 = 383
384 = 27 • 3
385 = 5 • 7 • 11
386 = 2 • 193
387 = 32 • 43
388 = 22 • 97
389 = 389
390 = 2 • 3 • 5 • 13
391 = 17 • 23
392 = 23 • 72
393 = 3 • 131
394 = 2 • 197
395 = 5 • 79
396 = 22 • 32 • 11
397 = 397
398 = 2 • 199
399 = 3 • 7 • 19
400 = 24 • 52
401 = 401
402 = 2 • 3 • 67
403 = 13 • 31
404 = 22 • 101
405 = 34 • 5
406 = 2 • 7 • 29
407 = 11 • 37
408 = 23 • 3 • 17
409 = 409