Download Math 10C Ch. 3 Lessons

Chapter 3: Factors and Products
3.1 Factors and Multiples of Whole numbers
Check out the picture of the belts on p. 134. In the belts, the patterns are 12 beads long
and 40 beads long. How many beads long must a belt be for it to be created using
either pattern?
Answer: We are looking for the number that both of them will divide into. The smallest
number that they both divide into is: 120.
Try the construct understanding on p. 134.
A. Get in groups of 2. List some powers of 2. Make another list of powers of 3.
Each of you pick a number from each list and multiply them to create a different
number. What are the factors of this number? What are some multiples of this
number?
Ex. Take 23 × 32 = 72 and 22 × 33 = 108
Answer:
Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Multiples of 72: 72, 144, 216, 288, etc.
Factors of 108: 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108
Multiples of 108: 108, 216, 324, etc.
B. Compare your number to your partner’s number. Which factors do the two
numbers have in common? Which factor is the greatest?
Answer: It looks like the greatest common factor in the list is 36. It looks like the least
common multiple is 216.
There has to be an easier way to find the greatest common factor and the least
common multiple. Before we discover this, we need to remember what a prime number
is. A prime number is any number that can only be multiplied by itself and 1. So,
when a factor of a number has exactly 2 divisors, 1 and itself, the factor is a prime
factor. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12. The prime factors of 12
are 2 and 3. To determine the prime factorization of 12, write 12 as a product of its
prime factors: 2 X 2 X 3 which should be written as: 22 × 3.
We often use a dot symbol for multiplication so that we don’t confuse the multiplication
symbol with the variable x. So it would look like:
22 ∙ 3
It is a really good idea to know all the prime numbers up to 30. Can you find those?
Answer: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Natural numbers greater than 1 that are not prime are composite.
Example #1:
Write the prime factorization of 3300.
Answer: In order to prime factorize a number, we need to make a factor tree. We will
divide 3300 by the smallest prime number to start. Please feel free to use your
calculator to help.
Ex. 3300
2
1650
2
825
5
165
5
3
33
11
Now, write it in prime factorization form: 22 ∙ 3 ∙ 52 ∙ 11.
There are a few different methods for finding the GCF. The first method is the Rainbow.
Check out p. 136 to see it. The second one is guess and check (also p. 136). The
easiest method is to use the factor trees to help us.
Example #2:
Determine the GCF (greatest common factor) of 138 and 198.
Answer: first step is to find the factor tree of each number.
138
2
198
69
3
2
23
99
3
33
3
11
To find the GCF, circle all the numbers that are common and then multiply together.
138
2
198
69
3
2
23
99
3
33
3
11
Since they have 2 and 3 in common, 2 X 3 = 6. Our GCF would be 6.
To generate multiples of a number, multiply the number by the natural numbers: that is,
1, 2, 3, 4, 5, etc. For example, some multiples of 26 are:
26 X 1 = 26, 26 X 2 = 52, 26 X 3 = 78, 26 X 4 = 104
Again, there are a few methods for finding the LCM (least common multiple). The first
method is just listing out multiples of each number (guess and check) until you find the
one that works. The second method is finding multiples of the highest number and
going backward from there. However, both of these methods involve guessing and
checking. The best method is to use our factor trees.
Example #3:
Determine the least common multiple of 18, 20 and 30.
Answer: Find the factor trees for all the numbers.
18
20
2
9
3
2
3
10
2
30
2
5
15
3
5
Write each number in its prime factorization form:
18: 2 ∙ 32
20: 22 ∙ 5
30: 2 ∙ 3 ∙ 5
To find the least common multiple, take the highest power of every number. Since we
have 2, 3 and 5, we need to take the highest power of each of those and multiply
together.
22 ∙ 32 ∙ 5 = 180
The least common multiple is 180.
Example #4
a) What is the side length of the smallest square that could be tiled with rectangles
that measure 16 cm by 40 cm?
Answer: Since the rectangles are going to go inside the square, they need to be
smaller than the square. That means we are going to need a multiple (the LCM). If it
helps, draw a picture.
16
cm
40
cm
To find the LCM, find a factor tree of both 16 and 40.
16
2
40
8
2
2
4
2
20
2
2
10
2
5
The prime factorization is:
16: 24
40: 23 ∙ 5
Take the highest power of each number: 24 ∙ 5 = 80.
Therefore, the side length of the smallest square is 80 cm.
b) What is the side length of the largest square that could be used to tile a rectangle
that measures 16 cm by 40 cm? Assume that the squares cannot be cut.
Answer:
16 cm
40 cm
This time we want a number to be smaller than our two numbers since the squares are
going inside. We want to find the GCF. Look at the factor trees.
16
2
40
8
2
2
4
2
20
2
2
10
2
5
Since they have three 2’s in common, the GCF is: 2 X 2 X 2 = 8.
So, the largest square has side length 8 cm.
Homework: p. 140 #3af, 5af, 6af, 8af, 9d, 10af, 11d, 12 – 14, 15a, 16ah, 17 – 21
3.2 Perfect Squares, Perfect Cubes and their roots
We will start today’s lesson by watching the fastest rubik’s cube solver on youtube.
Construct understanding:
In groups of 2 or 3, get integer tiles and linking blocks.
A. Use the tiles to model all the perfect squares from 1 to 100. How many can you
make?
B. Use the cubes to model all the perfect cubes from 1 to 100. How many can you
make?
Any whole number that can be represented as the area of a square with a whole
number side length is a perfect square. The side length of the square is the square
root of the area of the square.
Area = 25 square units
5 units
We write: √25 = 5
25 is a perfect square and 5 is its square root.
Any whole number that can be represented as the volume of a cube with a whole
number edge length is a perfect cube. The edge length of the cube is the cube root of
the volume of the cube.
6 units
Volume =
216 cubic units
3
We write: √216 = 6.
216 is a perfect cube and 6 is its cube root.
Example #1:
Determine the square root of 1296.
Answer: in order to discover the square root, we could just use the square root button
on our calculator. However, if we didn’t have the button available, we could make a
factor tree.
Ex.
1296
2
648
2
324
2
162
2
81
3
27
3
9
3
3
You will notice that there are four 2s and four 3s. We can split all our numbers evenly
into 2 groups:
Group 1: 2 X 2 X 3 X 3 = 36
Group 2: 2 X 2 X 3 X 3 = 36
Since 1296 divides into 2 equal groups, it is a perfect square with a square root of 36.
Example #2:
Determine the cube root of 1728.
Answer: we will start with our factor tree again.
1728
2
864
2
432
2
216
2
108
2
54
2
27
3
9
3
3
Notice that there are six 2s and three 3s. We can divide our numbers into 3 equal
groups.
Group 1: 2 X 2 X 3 = 12
Group 2: 2 X 2 X 3 = 12
Group 3: 2 X 2 X 3 = 12
Since it breaks into 3 equal groups, 1728 is a perfect cube and its cube root is 12. To
find a cube root using your graphing calculator, use Math #4 to find it.
In Example #1, √1296 is a radical with a radicand of 1296. The square root can be
2
written as √1296, with index 2, but the index on a square root is implied and often not
written.
radical
index
3
√1296
radicand
Example #3:
A cube has volume 4913 cubic inches. What is the surface area of the cube?
Answer: First, we need to remember some information about cubes. A cube is a
6-sided figure. Each side is the shape of a square. Since the area of a square is
𝐴 = 𝑠 2 , the surface area of the cube would be 6(area of a square) = 𝐴 = 6𝑠 2 . To find
the volume of a cube, we think of a rectangular prism. Its volume is lwh. Since a cube
has the same length for each side, the volume formula would be 𝑉 = 𝑠 3 .
Since we already have the volume, we need to go backwards to find the side length.
The opposite of cubing is cube rooting. Take the cube root of 4913 by using Math #4.
3
√4913 = 17
Now that we know the edge length, we can find the surface area by plugging the edge
length into it.
𝐴 = 6𝑠 2 = 6(17)2 = 1734 𝑖𝑛2
Homework: p. 146 #4 – 6 (first and last letter), 7a, 8a, 10, 12a, 13, 16
3.5 Simplifying polynomials
The second part of Chapter 3 is all about algebra. We are going to learn about
simplifying polynomials. Some of this should be a review from junior high.
What is a polynomial? Well, we need to go further back. What is a term? A term is a
number, a variable, or the product of numbers and variables. A polynomial is one or
more terms. There are specific names for one, two or three terms:
Monomial: one term
Binomial: two term
Trinomial: three terms
The degree of a polynomial is the highest exponent sum in a term. For example, in
𝑥 3 𝑦 6 − 2𝑥 2 𝑦 4 , the sum of the exponents in the first term is 9 and the sum of the
exponents in the second term is 6. Therefore, the degree of this polynomial is 9.
Simplifying
Simplifying is when we take brackets and we multiply them out. The main thing to
remember is that when you are multiplying, coefficients get multiplied and the
exponents on the variables get added.
There are three ways to multiply:
1. Algebra Tiles: do you remember these from junior high?
Let’s review:
 Small red square: represents 1 unit
 Long skinny green rectangle: represents x
 Large green square: represents 𝑥 2
 Turning the tile over to become white represents the negative values
Example:
The tile above represents: 𝑥 2 + 3𝑥 + 2.
How do we multiply algebra tiles? When you are given (2x + 1)(x – 2), first you set one
bracket vertically and one bracket horizontally. Then you fill in the inside area.
x
2x
-1
-
2
To find the final answer, add up the inside tiles. We have 2𝑥 2 tiles, 1 positive x, 4
negative xs, and 2 negative 1s. And any zero pairs cancel out….so one of the
positive xs cancels with the negative x. So our final answer is:
2𝑥 2 − 3𝑥 − 2
Algebra tiles only work for simpler equations, so we need other methods.
2. Rectangle method: Another method is to think in terms of a rectangle. Your
terms are going to be sides of a rectangle. We find the area of each individual
box. Let’s take the same example we just did.
2x
+1
x
-2
To find our answer, find the area of each small rectangle. Since that area = lw, do that
for each box.
Answer:
2x
+1
x
2𝑥 2
x
-2
−4𝑥
−2
2𝑥 2 + 𝑥 − 4𝑥 − 2 = 2𝑥 2 − 3𝑥 − 2
**note that we added x and -4x because they are like terms. For terms to be like, they
have to have the same variables and exponents. We also need to make sure our terms
are in descending numerical order. In other words, they go from highest down to
lowest.
Rectangle method is especially useful when you want to keep everything neatly
organized.
3. FOIL/Distributive method: this method is probably the most useful because it
works for every question. The FOIL method works when there are two binomials.
FOIL stands for: First, Outside, Inside Last. We will look at the distributive
method in a later example.
Ex. (x – 2)(2x + 1)
First:
Outside:
Inside:
Last:
(x)(2x) = 2𝑥 2
(x)(1) = 1x
(-2)(2x) = -4x
(-2)(1) = -2
To find the answer, collect like terms and put everything in order.
2𝑥 2 − 3𝑥 − 2
Homework: p. 167 #12ah, 13a, 16
3.7 Multiplying Polynomials
Now we are going to look at more complicated questions. We won’t use algebra tiles
anymore, but you can use either the rectangle method or the distributive method.
Example #1
Expand and simplify:
a) (2ℎ + 5)(ℎ2 + 3ℎ − 4)
Answer: to multiply out, either draw a rectangle and do the rectangle method or use the
distributive property. The distributive property runs on the principle that “everyone
comes to the party”. In other words, every term in the first bracket multiplies to every
term in the second bracket. Remember that coefficients get multiplied and exponents
get added.
= 2ℎ(ℎ2 ) + 2ℎ(3ℎ) + 2ℎ(−4) + 5(ℎ2 ) + 5(3ℎ) + 5(−4)
= 2ℎ3 + 6ℎ2 − 8ℎ + 5ℎ2 + 15ℎ − 20
= 2ℎ3 + 11ℎ2 + 7ℎ − 20
You are done when there are no more like terms and your terms are in order from
highest exponent to lowest.
b) (−3𝑓 2 + 3𝑓 − 2)(4𝑓 2 − 𝑓 − 6)
= −3𝑓 2 (4𝑓 2 ) − 3𝑓 2 (−𝑓) − 3𝑓 2 (−6) + 3𝑓(4𝑓 2 ) + 3𝑓(−𝑓) + 3𝑓(−6) − 2(4𝑓 2 )
−2(−𝑓) − 2(−6)
= −12𝑓 4 + 3𝑓 3 + 18𝑓 2 + 12𝑓 3 − 3𝑓 2 − 18𝑓 − 8𝑓 2 + 2𝑓 + 12
= −12𝑓 4 + 15𝑓 3 + 7𝑓 2 − 16𝑓 + 12
Example #2:
a) (2𝑟 + 5𝑡)2
Answer: when you have squared brackets, the easiest way to multiply out is to rewrite
out the bracket and FOIL.
(2𝑟 + 5𝑡)(2𝑟 + 5𝑡) = 4𝑟 2 + 10𝑟𝑡 + 10𝑟𝑡 + 25𝑡 2 = 4𝑟 2 + 20𝑟𝑡 + 25𝑡 2
b) (3𝑥 − 2𝑦)(4𝑥 − 3𝑦 + 5)
Answer: let’s do this one using the rectangle method.
4x
12𝑥 2
3x
−8𝑥𝑦
-2y
-3y
+5
−9𝑥𝑦
15𝑥
6𝑦 2
−10𝑦
Our answers need to go in alphabetical order and highest degree.
= 12𝑥 2 + 15𝑥 − 17𝑥𝑦 − 10𝑦 + 6𝑦 2
Example #3:
a) (2𝑐 − 3)(𝑐 + 5) + 3(𝑐 − 3)(−3𝑐 + 1)
Answer: since there are two sets of brackets we are going to do each bracket
separately. Do not multiply the 3 into the brackets yet.
= 2𝑐 2 + 10𝑐 − 3𝑐 − 15 + 3(−3𝑐 2 + 𝑐 + 9𝑐 − 3)
Now multiply the three in and collect like terms.
= 2𝑐 2 + 10𝑐 − 3𝑐 − 15 − 9𝑐 2 + 3𝑐 + 27𝑐 − 9
= −7𝑐 2 + 37𝑐 − 24
b) (3𝑥 + 𝑦 − 1)(2𝑥 − 4) − (3𝑥 + 2𝑦)2
Answer: (rewrite out the brackets first, then multiply)
= (3𝑥 + 𝑦 − 1)(2𝑥 − 4) − (3𝑥 + 2𝑦)(3𝑥 + 2𝑦)
= 6𝑥 2 − 12𝑥 + 2𝑥𝑦 − 4𝑦 − 2𝑥 + 4 − (9𝑥 2 + 6𝑥𝑦 + 6𝑥𝑦 + 4𝑦 2 )
Notice that we leave the subtraction sign in front of the brackets. Once you’ve
multiplied them out, remember that a subtraction sign means to change all the signs.
= 6𝑥 2 − 12𝑥 + 2𝑥𝑦 − 4𝑦 − 2𝑥 + 4 − 9𝑥 2 − 6𝑥𝑦 − 6𝑥𝑦 − 4𝑦 2
= −3𝑥 2 − 14𝑥 − 10𝑥𝑦 − 4𝑦 − 4𝑦 2 + 4
Homework: p. 186 #4 – 5(first and last letter), 8 – 10(first and last letter), 11, 13ab,
15af, 17a, 18a, 19a, 21a, 22a
3.3 Common Factors of a Polynomial
Now that we have learned to simplify (multiply out brackets), we are going to learn to go
backwards….put things back into brackets.
Let’s start with looking at algebra tiles. How many rectangles you can make out of the
following tiles?
We could create a variety of rectangles.
How does the fact that we can make rectangles help us factor?
Let’s take an easier example and check out factoring 4m + 12.The dimensions of each
rectangle represent the factors of the polynomial.
Length 4m + 12
Width 1
Answer: 1(4m + 12)
Length 2m + 6
Width 2
Answer: 2(2m + 6)
Length m + 3
Width 4
Answer: 4(m + 3)
When we write a polynomial as a product of factors, we factor the polynomial. The
diagrams above show that there are 3 ways to factor the expression 4m + 12. The first
two ways: 1(4m + 12) and 2(2m + 6) are incomplete because the second factor in each
case can be factored further. That is, neither 1 nor 2 is the greatest common factor of
4m and 12.
The third way: 4(m + 3) is complete. The greatest common factor of 4m and 12 is 4.
We say that 4m + 12 is factored fully when we write 4m + 12 = 4(m + 3); that is the
polynomial cannot be factored further.
Example #1
a) Factor 6n + 9
Answer: We can factor using algebra tiles or algebra itself. Let’s start with tiles. Can
we make 6n + 9 into a rectangle?
Notice that the width of the rectangle is 3 and the length of the rectangle is 2n + 3.
So the answer is: 3(2n + 3).
Instead of tiles, we can use algebra. To find the GCF (greatest common factor), we
need to find any numbers and letters that are common. In this case, the number in
common is 3. So to factor, we take out the 3 (which means we divide by 3).
6n + 9 = 3(2n + 3)
b) 6𝑐 + 4𝑐 2
Notice that the width is 2c and the length is 2c + 3
Answer: 2c(2c + 3)
If we were doing this using algebra, we would look to see what numbers and letters
were in common. Both terms can be divided by 2 and both terms have a “c” in it.
Answer: 2c(2c + 3)
When a polynomial has negative terms or 3 different terms, we cannot remove a
common factor by arranging the tiles in a rectangle. Instead, we can arrange the tiles
into equal groups.
Example #2
Factor the trinomial: 5 − 10𝑧 − 5𝑧 2
Answer: to solve using algebra tiles, put all the tiles into equal groups. Each group
would get a 1 tile, 2 negative z tiles and 1 negative 𝑧 2 tile.
Answer: 5(1 − 2𝑧 − 𝑧 2 ).
Using algebra tiles can be time consuming, so it is easier to use the GCF method. The
three terms have no variables in common, but they do have 5 in common. Factor out
the 5 by dividing each term by 5.
Answer: 5(1 − 2𝑧 − 𝑧 2 ).
Example #3
Factor the trinomial. −12𝑥 3 𝑦 − 20𝑥𝑦 2 − 16𝑥 2 𝑦 2
Answer: to factor, take out what is common. Each term can be divided by -4, each
term has an x and a y. Our GCF is -4xy.
−4𝑥𝑦(3𝑥 2 + 5𝑦 + 4𝑥𝑦)
Homework: p. 155 #4, 7ae, 8af, 9af, 10af, 11, 12, 14a, 16ace, 17, 18
3.5 Factoring Easy Trinomials 𝐱 𝟐 + 𝒃𝒙 + 𝒄
We are now going to look at factoring trinomials (three terms). We are going to start
with the trinomials that have an invisible 1 in front of the x-squared term.
Let’s look at the trinomial 𝑥 2 + 5𝑥 + 6. Could you make this into a rectangle?
The trick is to remember that the squares need to go diagonally across from each other.
Then we fill in the x tiles. Keep rearranging until it works.
The answer is going to be the dimensions of the rectangle.
(x + 2)(x + 3)
Now let’s examine the original trinomial: 𝑥 2 + 5𝑥 + 6
Our two factors involve the numbers 2 and 3. How do they relate to our original
trinomial? Notice that 2 and 3 multiply to the last term and add to the middle term. This
gives us an easier way to factor than to always use trinomials. Look for two numbers
that will multiply to the last term and add to the middle term.
Example:
a) 𝑥 2 − 2𝑥 − 8
Answer: Two numbers that multiply to -8 and add to -2 are -4 and 2. Therefore, our
answer is:
(𝑥 − 4)(𝑥 + 2)
b) 𝑧 2 − 12𝑧 + 35
Answer: Two numbers that multiply to 35 and add to -12 are -5 and -7. Therefore, our
answer is:
(𝑧 − 5)(𝑧 − 7)
Example
Factor: −24 − 5𝑑 + 𝑑2
Answer: First, put the terms in order. Then use the method we learned in the last
example.
𝑑 2 − 5𝑑 − 24
Answer: Two numbers that multiply to -24 and add to -5 are -8 and 3. Therefore, our
answer is:
(𝑑 + 3)(𝑑 − 8)
Sometimes we will combine different types of factoring. The next example combines
GCF and easy trinomial factoring
Example:
Factor: −4𝑡 2 − 16𝑡 + 128
Answer: Notice that each term has a GCF of -4 (note: if the squared term is negative,
factor out a negative GCF to make the question easier). So, first take out the GCF.
−4(𝑡 2 + 4𝑡 − 32)
(Remember to factor out means to divide out)
Now, it is an easy trinomial, so look for two numbers that multiply to -32 and add to 4.
The numbers are 8 and -4.
−4(𝑡 + 8)(𝑡 − 4)
Homework: p. 166 #4cd, 5d, 6cd, 8i, 10c, 11ah, 14ah, 15d, 17c, 19af, 20af, 21ef
3.6 Factoring Long Trinomials 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
Now we are going to look at trinomials that no longer have invisible 1 coefficients in
front of the squared term. We cannot factor them the same way. We will start with an
experiment, using algebra tiles to help us.
p. 168 Construct Understanding
2𝑥 2 + 15𝑥 + 7
5𝑥 2 + 4𝑥 + 4
2𝑥 2 + 5𝑥 + 2
2𝑥 2 + 9𝑥 + 10
6𝑥 2 + 7𝑥 + 2
5𝑥 2 + 11𝑥 + 2
For which of the above trinomials can the algebra tiles be arranged to form a rectangle?
Go in groups of 2 and try them out. Remember that the 1-tiles and the 𝑥 2 tiles must be
diagonally across from each other. There is only one trinomial that doesn’t work.
Using the algebra tiles is helpful, but sometimes time-consuming. So we have two other
methods that we can use to factor long trinomials. You may choose which method
makes more sense to you.
We will start with a method that we call lucky because if you are “lucky”, you will solve it
on the first try.
ex. 4ℎ2 + 20ℎ + 9
Lucky:
1. Break down the first and last terms. You will have to guess as to which
numbers you think will work. The first term could be 4 and 1 or 2 and 2. We
will try 2 and 2. The second term could be 9 and 1 or 3 and 3. We will try 3
and 3.
4ℎ2 + 20ℎ + 9
2h
2h
3
3
2. Put in the signs. Since all the terms are positive, we will use positive signs.
4ℎ2 + 20ℎ + 9
2h
2h
+
+
3
3
**If last term is negative you need two different signs. If last term is positive, you need
two identical signs.
3. Diagonal multiply to see if the middle term is correct.
4ℎ2 + 20ℎ + 9
2h
2h
+
+
3
3
(2h)(3) = 6h
(2h)(3) = 6h
Since this isn’t the middle term, we are incorrect. So we
6h + 6h = 12h
need to try again.
4ℎ2 + 20ℎ + 9
2h
2h
+
+
9
1
(2h)(9) = 18h
(2h)(1) = 2h
18h + 2h = 20h
since this is the middle term, this is the correct factoring.
Draw a horizontal line separating the factors. Whatever is above the line goes in
the first bracket, whatever is below the line goes in the second bracket.
4ℎ2 + 20ℎ + 9
2h
2h
Answer: (2h + 9)(2h + 1)
+
+
9
1
Let’s try the same question using a method called decomposition.
Decomposition:
ex. 4ℎ2 + 20ℎ + 9
1. Multiply the first and last numbers together. (4)(9) = 36
2. Find two numbers that multiply to the answer from step #1 and add to middle
term.
If you are not sure what the numbers are, just use your calculator to keep
dividing to find factors.
36/1 = 36
36/2 = 18
1 and 36 don’t add to 20
2 and 18 add to 20, so these are our 2 numbers
3. Split up the middle term with those two numbers. Remember that if there is a
letter with the middle term, there should be letters with those two numbers.
4ℎ2 + 20ℎ + 9
4ℎ2 + 18ℎ + 2ℎ + 9
4. Group into sets of two and take out the GCF of each group
4ℎ2 + 18ℎ + 2ℎ + 9
2ℎ(2ℎ + 9) + 1(2ℎ + 9)
5. We are now going to take out a bracket GCF. (2h + 9) becomes a single
bracket out front and whatever is left in front of each bracket, goes in the
second bracket.
(2ℎ + 9)(2ℎ + 1)
Let’s try another example, using both decomposition and lucky.
Ex. 6𝑘 2 − 11𝑘 − 35
Lucky:
6𝑘 2 − 11𝑘 − 35
3k
2k
- 7
+ 5
3k(5) = 15k
2k(-7) = -14k
15k – 14k = k
Since this is not the middle term, we need to try again.
6𝑘 2 − 11𝑘 − 35
3k
2k
- 5
+ 7
3k(7) = 21k
2k(-5) = -10k
21k – 10k =11k
signs.
Since this is at least the correct number, we just need to switch the
6𝑘 2 − 11𝑘 − 35
3k
2k
+ 5
- 7
Answer: (3k + 5)(2k – 7)
Decomposition:
ex. 6𝑘 2 − 11𝑘 − 35
1. Multiply the first and last numbers together. (6)(-35) = -210
2. Find two numbers that multiply to the answer from step #1 and add to middle
term.
If you are not sure what the numbers are, just use your calculator to keep
dividing to find factors. But always try dividing by 10 if the number ends in 0, just in
case that’s one of the factors. And in this case, it is!
-210/10 = -21 Your two factors are -21 and 10.
3. Split up the middle term with those two numbers. Remember that if there is a
letter with the middle term, there should be letters with those two numbers.
6𝑘 2 − 11𝑘 − 35
6𝑘 2 − 21𝑘 + 10𝑘 − 35
4. Group into sets of two and takeout the GCF of each group
6𝑘 2 − 21𝑘 + 10𝑘 − 35
3𝑘(2𝑘 − 7) + 5(2𝑘 − 7)
Answer: (3k + 5)(2k – 7)
Example:
a) 3𝑠 2 − 13𝑠 − 10
Use either lucky or decomposition to factor this. We will use lucky for example a) and
then decomposition for example b).
3𝑠 2 − 13𝑠 − 10
3s
1s
+ 2
- 5
3s(-5) + 2(1s) = -13s. Since middle term is correct, the answer is: (3s + 2)(s – 5)
b) 6𝑥 2 − 21𝑥 + 9
Notice that we actually have a GCF that we should take out first.
3(2𝑥 2 − 7𝑥 + 3)
What multiplies to 6 and adds to -7? The two numbers are -6 and -1.
2𝑥 2 − 6𝑥 − 1𝑥 + 3
2𝑥(𝑥 − 3) − 1(𝑥 − 3)
Answer:
3(𝑥 − 3)(2𝑥 − 1)
Homework p. 177 #5cd, 7cd, 12a, 13bh, 15aceg, 16, 18ef, 19g – j, 20a, 21a) i and vi
3.8 Factoring Special Polynomials
What patterns do you see in the trinomials and their factors?
4𝑥 2 + 20𝑥 + 25 = (2𝑥 + 5)2
9𝑥 2 − 12𝑥 + 4 = (3𝑥 − 2)2
The two trinomials above are what we call perfect square trinomials. They have the
following characteristics:
1. The first term is a perfect square.
2. The last term is a perfect square.
3. The middle term is found by taking the square root of the first term and
multiplying it to the square root of the last term and multiplying by 2.
If our trinomial fits all the characteristics, then we can factor it very quickly.
Example #1:
a) 4𝑥 2 + 12𝑥 + 9
Answer: the first term is a perfect square, the last term is a perfect square. The middle
term is the square root of the first term (2x) multiplied by the square root of the last term
(3) multiplied by 2. (2x)(3)(2) = 12x. Since this is the middle term, this is a perfect
square trinomial. It also means we can factor our trinomial quickly.
4𝑥 2 + 12𝑥 + 9 = (2𝑥 + 3)2
b) 4 − 20𝑥 + 25𝑥 2
Answer: the first term is a perfect square, the last term is a perfect square. The middle
term is the square root of the first term (2) multiplied by the square root of the last term
(5x) multiplied by 2. (2)(5x)(2) = 20x. Since this is the middle term, this is a perfect
square trinomial. It also means we can factor our trinomial quickly. Just make sure to
put a negative sign in the middle.
4 − 20𝑥 + 25𝑥 2 = (2 − 5𝑥)2
We can also factor trinomials that have multiple letters.
Example #2
a) 2𝑎2 − 7𝑎𝑏 + 3𝑏 2
Answer: since this a long trinomial, we can use lucky or decomposition. Use the
method of your choice. I will do decomposition.
What two numbers multiply to 2(3) = 6 and add to -7? The two numbers are -6 and -1.
2𝑎2 − 6𝑎𝑏 − 1𝑎𝑏 + 3𝑏 2
2𝑎(𝑎 − 3𝑏) − 𝑏(𝑎 − 3𝑏)
(𝑎 − 3𝑏)(2𝑎 − 𝑏)
b) 10𝑐 2 − 𝑐𝑑 − 2𝑑2
Answer: This time we will use lucky.
10𝑐 2 − 1𝑐𝑑 − 2𝑑 2
5c
2c
+ 2d
- 1d
5c(-1d) + 2c(2d) = -cd. Since middle term is correct, the answer is: (5c + 2d)(2c – d).
Another example of special quadratics is a difference of squares. A difference of
squares is a binomial of the form 𝑎2 − 𝑏 2 . It will have two terms and each term will be a
perfect square. It will also have a subtraction sign. To factor it, we need to take the
square root of each term and put them in two brackets. There will be one subtraction
and one addition sign.
𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏)
Notice that we need opposite signs because if we were to multiply out, we would need
the two middle terms (ab and –ab) to cancel:
(𝑎 − 𝑏)(𝑎 + 𝑏) = 𝑎2 + 𝑎𝑏 − 𝑎𝑏 − 𝑏 2 = 𝑎2 − 𝑏 2
See diagram at the top of p. 193 to illustrate a difference of squares pictorially.
Example #3
Factor each binomial:
a) 25 − 36𝑥 2
Answer: Both terms are perfect squares and there is a subtraction sign. Therefore, we
can factor (5 – 6x)(5 + 6x).
b) 5𝑥 4 − 80𝑦 4
Answer: In this question, we have to take out a GCF before we determine if it is a
difference of squares.
5(𝑥 4 − 16𝑦 4 )
Now it is a difference of squares, so we can factor:
5(𝑥 2 − 4𝑦 2 )(𝑥 2 + 4𝑦 2 )
However, notice that this is a sneaky question! We still have one more difference of
squares that we can factor. We can’t factor the second bracket because it is a sum of
squares and we don’t have a way to factor that. However, the first one can still be
factored, so our final answer is:
5(𝑥 − 2𝑦)(𝑥 + 2𝑦)(𝑥 2 + 4𝑦 2 )
Homework: p. 194 #4ag, 5 6ad, 7a i, vi, 8f, 10 - 13 (1st and last letter), 15i, 18, 20, 21ef