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Unit 7
Factoring Polynomials
Introduction
In Unit 2 we learned how to solve linear equations. Linear equations have a
single variable whose exponent is one. Below is an example of a linear equation.
3(x + 2) – x + 4 = 4x – 22
In this equation the only variable we use is ‘x’ so this is the single variable from our
definition above. Each instance of ‘x’ could actually have the understood exponent of
one written with it, if we needed an exponent. If we added the understood exponent
of one, our example equation would be:
3(x1 + 2) – x1 + 4 = 4 x1 – 22
The above equation has one variable ‘x’ and in each case there is an exponent of
one. These equations are very useful. The slope-intercept form of a linear equation
is y = m x + b. We saw many ways in Unit 4 in which this type of equation can model
situations that have rates of change in Unit 4. A linear equation can model a company’s
sale, a person’s pay, the altitude of a balloon, the relationship or a person’s height
to foot size, etc.
Linear equations are not the only type of equation that we can use to model
situations. In this unit we are going to learn about a type of equation called
quadratics. These equations have a squared term. An example of a quadratic equation
is given below.
x2 + 5x = -6
We will need to learn how to solve this type of equation. One of the primary techniques
that can be used to solve a quadratic equation is called factoring. Factoring is a term we
should know. Factoring means we should find numbers that are multiplied together. In this
unit we are going to find expressions that can be multiplied together.
Unit 7
Factor
Vocabulary and Concepts
Used as a verb, it means to find the values of monomials or binomials
that multiply to produce a given polynomial.
Key Concepts for Exponents and Polynomials
Raising a Monomial to a Power
Types of pairs of factors
To raise a monomial to a power, we multiply the
power times ALL the exponents in the monomial.
Pairs of factors can be positive, negative, or mixed.
Standard Form for a Polynomial Standard form for a polynomial has a first term
that is positive, and all of the terms are arranged
in descending order of exponents.
Zero Product Theorem
If two numbers have a product of zero then one, or both,
of the values must be zero. If ab = 0 then a = 0 or b = 0.
Trial and Error Factoring Used when the highest ordered term has a coefficient other
than one. We must try to use the factors of the first
and last terms to produce the middle term.
Difference of Perfect Squares A type of binomial factored into sum and difference
binomials. Two perfect squares are separated by a
minus sign. Ex: a2 – b2 = (a + b)(a – b)
Unit 7 Section 1
Objective

The student will raise a monomial to a power.
A number or variable raised to a second power means to multiply two of the variables
or number times each other. A number or variable raised to a third power means to
write down the factor three times and multiply them together. A number or variable
raised to a fourth power means to write down the factor four times and multiply them together.
A monomial raised to the ‘n’th power means to write down the monomial
‘n’ times and multiply them together. The examples below show us how to raise a
monomial to a power.
Example A
Example B
(5x3)2
(4x5y)3
(-2a3)4
means
means
means
(5x3)(5x3)
Example C
(4x5y)(4x5y)(4x5y)
(-2a3)(-2a3)(-2a3)(-2a3)
and results in
and results in
and results in
25x6
64x15y3
16a12
Writing out the monomials the required number of times and multiplying will produce
correct answers every time. However, when the power becomes larger and larger,
writing out the monomial six or seven times would be inefficient. We need to examine
these examples closely so we can draw some conclusions. First, we should try to find a
pattern in the way the exponents work. The table below may help us understand the
process better.
Example
Example
Example
Example
A
B
B
C
Exponent
inside the
parentheses
Power
outside the
parentheses
Resulting
Exponent
3
5
1
3
2
3
3
4
6
15
3
12
( the understood exponent
of ‘y’ )
If we ask ourselves, “What arithmetic do we use with the exponent from the monomial and the
power to get the resulting exponent?”, the answer appears to be to multiply! And this gives us
our rule for “raising a monomial to a power”.
To raise a monomial to a power, we multiply the power times ALL the exponents
in the monomial.
This rule applies to ALL exponents in the monomial. This includes the understood
exponents of both the variables and the coefficients. If we examine the coefficients
in examples A through C on the previous page we can see this clearly.
Example A
Example B
Example C
Coefficient
51
41
(-2)1
Power
2
3
4
Result
52 = 25
43 = 64
(-2)4 = 16
Our approach to raising a monomial to a power should be to put in ALL the understood
exponents, then multiply all the exponents by the power, and simplify the coefficient.
The examples below show us how to use the process.
Example A Simplify (3a2b)4
(3a2b)4
(31a2b1)4
34a8b4
81a8b4
Put in the understood exponents.
Multiply the power ‘4’ times each of the monomial’s exponents.
Simplify the coefficient.
Example B Simplify (-2x3yz -2)5
(-2x3yz -2)5
(-21x3y1z -2)5
(-2)5x15y5z -10
-32x15y5z -10
Put in the understood exponents.
Multiply the power ‘5’ times each of the monomial’s exponents.
Simplify the coefficient.
In this example we needed to be careful of the negative coefficient and the
negative exponent.
Example C Simplify (-3n2m5)4
(-3n2m5)4
(-31n2m5)4
(-3)4n8m20
81n8m20
Put in the understood exponent.
Multiply the power ‘4’ times each of the monomial’s exponents.
Simplify the coefficient.
In this example we had a negative coefficient, but because it was raised to an
even power the simplified coefficient is a positive 81.
Example D
Simplify (6x4yz -3 )-2
(6x4yz -3 )-2
(61x4y1z -3 )-2
6-2x-8y-2z 6
1 -8 -2 6
x y z
36
Put in the understood exponents.
Multiply the power ‘-2’ times each of the monomial’s exponents.
Simplify the coefficient.
The negative power must be used carefully. In particular, we must remember
a negative exponent means the value or variable belongs to the denominator of
a fraction. That is why we have a fractional coefficient.
Example E
Simplify (3a5bc)2(-2ab2)3
This example shows us using monomials raised to a power, and multiplication.
To simplify this expression we must use the Order of Operations and use the
powers first, and then do the multiplication.
(3a5bc)2(-2ab2)3
(9a10b2c2)(-8a3b6)
-72 a13b5c8
We raise each monomial to their powers first.
Now we multiply the coefficients, and add the exponents.
VIDEO LINK: Youtube: Power to a Power
Exercises Unit 7 Section 1
Fill in the blanks.
1. If an expression has a monomial that is raised to a power then we should put in
the ___________________ exponents, __________________ the power times
all the exponents and simplify the ____________________.
Simplify the following
2. (a2)4
3. (x7y3)2
4. (3x2)4
5. (-4a6)2
6. (-x2y)3
7. (0.4x4)3
8. (-5a2)4
9. (-3x2y5)3
10. (.2x5)2
1
11. ( a5)3
2
1
12. (- xy)2
3
13. (1.5y9)2
14. (2xy3)6
15. (a-3)4
16. (w-2)-5
17. (7xy-4)-2
Simplify, then multiply the resulting monomials. Show your work.
18. (ab2)3(a3b)4
19. (xy3)2(x2)4
20. (2a4b3)3(3a5b)2
21. (-ab)3(-a2b)6
22. (x2)3(x3)4(x5)2
23. (4ab-2)2(.5ab3)2
24. Which expression is the completely simplified form of (-2ab2c -1d-4)3
a.
 8a 3b 6
c 3 d  4
b.
 8a 3b 6
c3d 4
 6a 3 b 6
c3d 4
c.
d.
6a 3 b 6
c 3 d  4
25. Which expression is the completely simplified form of (x2y -3)-2
a.
y 6
x 4
x4
y5
b.
26. Given the expression
c.
x4
y6
d.
y6
x4
(-5w3x-1yz-4)-2
a. Simplify the expression completely.
b. Explain how the coefficient will be raised to the "negative two" power.
c. Explain how you raised w3 to the negative two power and why it
belongs in either the numerator or denominator of your answer.
Read the problems carefully to find the two data points described in the problem, and
then write an equation to model the situation in slope-intercept form.
27. A hot air balloon has lifted off the ground. After 4 minutes, it was 220 feet high.
The balloon is rising at a steady rate. Write an equation to model how the altitude
is increasing. (Hint: What was the balloon’s altitude when it first started, and what
was its time when it started? )
(
,
) (
,
)
m = __________
equation __________________
28. A concert at a large arena just finished. After 5 minutes there were 11800 people
still in the arena. After 11 minutes there were 5200 people left in the arena.
a. Write an equation to model how fast people are leaving.
(
,
) (
,
)
m = __________
equation __________________
b. How many people were at the concert?
Explain how you found your answer!
c. How many minutes will it take to empty the arena?
Explain how you found your answer!
Unit 7 Section 2
Objective

The student will factor trinomials of the form x2 + bx + c where
b > 0 and c > 0.
Solving problems often means reversing operations. If we buy 5 tickets to a show
at $9 each we can use multiplication to find the total. However, if we know we bought
5 tickets for $45 then to find the price of each ticket we must divide. After we learned
how to solve problems with multiplication we then learned how to solve problems with
the reverse, with division.
We can solve area problems like the one below with binomial multiplication.
Given the rectangle to the right with the length and width
listed as expressions, find the expression for the area.
(2x + 5)(x – 1)
2x2 – 2x + 5x - 5
2x2 + 3x – 5
2x + 5
x-1
If there are problems we can solve with binomial multiplication there will also be problems
we can solve by reversing the process of multiplication.
In reversing the process of binomial multiplication there is a key type of questions we
need to use. A sample of that question follows.
“What two numbers add up to 11 and multiply to 24?”
Finding the answer to this question involves being able to list the pairs of factors for 24.
Below is a list of these pairs.
24
1, 24
2, 12
3, 8 This is the answer since 3 + 8 = 11
4, 6
In listing the pairs of factors we started with the number 1 and worked our way up to
4 in order. By listing the pairs in order we make sure we don’t skip any pairs.
This question and our ability to find the pairs of factors is critical because in the process
of binomial multiplication we will end up multiplying two numbers AND adding those same
two numbers. The examples that follow demonstrate this arithmetic.
Example A
Example B
(a + 4)(a + 9)
+ 9a + 4a + 36
a2 + 13a + 36
(x + 7)(x + 2)
+ 2x + 7x + 14
x2 + 9x + 14
a2
x2
In example A the two
numbers in our binomials
are 4 and 9. When we
examine the product we
can see that the constant
or last term is 36 (4 ∙ 9).
The middle term has a
coefficient of 13 (4 + 9).
Our answer has the two
numbers we multiplied,
and then added.
In example B the two
numbers in our binomials
are 7 and 2. When we
examine the product we
can see that the constant
or last term is 14 (2 ∙ 7).
The middle term has a
coefficient of 9 (2 + 7).
Our answer has the two
numbers we multiplied,
and then added.
Example C
(z + 8)(z + 3)
+ 3z + 8z + 24
z2 + 11z + 24
z2
In example C the two
numbers in our binomials
are 8 and 3. When we
examine the product we
can see that the constant
or last term is 24 (8 ∙ 3).
The middle term has a
coefficient of 11 (3 + 8).
Our answer has the two
numbers we multiplied,
and then added.
To reverse the binomial multiplication we must find the two numbers that multiply
to the last term, and add to become the coefficient of the middle term. Below are some
examples that show us how to use our question from the previous page to find the
binomial factors when we are given the trinomial product.
Example A
Factor x2 + 11x + 30
First we must find “two numbers that multiply to 30 and add up to 11”.
To do this we will list all the pairs of factors of 30.
30
1, 30
2, 15
3, 10
5, 6
This is the pair of factors that adds up to 11.
These are the values that belong in our binomials so our answer is
(x + 5)(x + 6)
You can check your answer by using the FOIL process and multiplying.
Example B
Factor a2 + 6a + 9
First we must find “two numbers that multiply to 9 and add up to 6”.
To do this we will list all the pairs of factors of 9.
9
1, 9
3, 3 This is the pair of factors that adds up to 6.
These are the values that belong in our binomials so our answer is
(a + 3)(a + 3), which in this case is best written as (a + 3)2
Example C
Factor z2 + 43x + 42
First we must find “two numbers that multiply to 42 and add up to 43”.
To do this we will list all the pairs of factors of 42.
42
1, 42 This is the pair of factors that adds up to 43.
2, 21
3, 14
6, 7
These are the values that belong in our binomials so our answer is
(z + 1)(z + 42)
When we are listing our pairs of factors, if we can see that a pair adds
up to the numbers we need then it may not be necessary to write out
all the pairs. For instance, in the above example the very first pair we
listed was correct so we would not necessarily have to list the rest.
When we list our binomial answers the order of the binomials is not critical since
multiplication is commutative. For instance:
(x + 7)(x + 11) = (x + 11)(x + 7)
The binomials can be written in either order and are correct.
VIDEO LINK: Youtub: Factoring a Simple Trinomail
Exercises Unit 7 Section 2
1. Find a pair of factors that multiply to 20 and add up to 9.
2. Find a pair of factors that multiply to 18 and add up to 19.
3. Find a pair of factors that multiply to 16 and add up to 8.
4. Find a pair of factors that multiply to 21 and add up to 10.
5. Find a pair of factors that multiply to 36 and add up to 20.
6. Find a pair of factors that multiply to 15 and add up to 16.
7. Find a pair of factors that multiply to 24 and add up to 10.
8. Find a pair of factors that multiply to 28 and add up to 16.
9. Find a pair of factors that multiply to 6 and add up to 5.
10. Find a pair of factors that multiply to 90 and add up to 23.
Factor the following trinomials.
(Reverse the FOIL method for multiplying and list your answer as a binomial product. )
11.
x2 + 8x + 15
12. a2 + 12a + 35
13. x2 + 20x + 36
14.
a2 + 14a + 40
15. x2 + 11x + 28
16. a2 + 9a + 18
17.
x2 + 7x + 12
18. a2 + 13a + 42
19. x2 + 15x + 26
20.
x2 + 6x + 5
21. x2 + 6x + 8
22. a2 + 14a + 40
23.
y2 + 27y + 50
24. z2 + 18z + 32
25. x2 + 25x + 100
26.
y2 + 8y +16
27. t2 + 14t + 45
28. w2 + 10w + 25
29.
v2 + 2v + 1
30. b2 + 14b + 33
31. u2 + 26u + 105
32. Mike factored the trinomial x2 + 11x + 30 into the binomial
product (x + 2)(x + 15).
a. Prove Mike is wrong algebraically and show your work.
b. Factor the trinomial correctly and explain how you did the factoring.
c. Prove your answer is correct algebraically and show your work.
33. The trinomial a2 + 14a + 48 is equivalent to
a. (a + 2)(a + 24)
b. (a + 4)(a + 12)
c. (a + 8)(a + 6)
c.
(a + 48)(a + 1)
Perform Common Factoring for the following ( Factor out the GCF and reverse the
Distributive Property. )
34. 5x - 20y
35. 18a2 - 9a
36. 7ax - 14ay
37. 3x2 + 21x - 12
38. 8x3 + 6x2 + 14x
39. az3 - az2 + az
Simplify
40. (3xy3)2
41. (2a3)4
42. (w3)-5
43. (11x3y4z)2
Simplify then multiply the resulting monomials. Show your work.
44. (a2b3)4(ab)4
45. (-2ab3)3(4a3b)2
Unit 7 Section 3
Objective

The student will factor trinomials of the form x2 + bx + c where
b < 0 and c > 0.
This section is very closely related to Section 2. The only difference is the sign
of the coefficient of the middle term. In this objective b < 0 which means b is
negative. A sample of this type of quadratic is given below.
x2 – 14x + 40
Factoring this type of trinomial is no different from the previous section. We must still
ask ourselves the question:
“What two numbers add up to -14 and multiply to 40 and add up to -14?”
We will still need to list all the pairs of factors and choose the correct pair of values.
If we want a positive number when we multiply, the signs can either both be positive or both
be negative. In order to add up to a negative number, however, the two factors will
both need to be negative. The list of pairs of factors of 40 are given below.
40
-1,
-2,
-4,
-5,
-40
-20
-10 This pair of factors adds up to -14.
-8
Now that we have our pair of factors we can write the binomials that are the answer.
(x – 4)(x – 10)
Example A
Factor x2 - 15x + 56
First we must find “two numbers that multiply to 56 and add up to -15”.
To do this we will list all the pairs of factors of 56.
56
-1, -56
-2, -28
-4, -14
-7, -8 This pair of factors that adds up to -15.
These are the values that belong in our binomials so our answer is
(x – 7)(x – 8)
We must be able to decide what kind of pair of factors we need to factor a trinomial.
We have two patterns for the signs of the terms that tell us which kind of factors we
need.
Positive Pair
Negative Pair
x2 + bx + c
x2 – bx + c
When both the middle term and the last term are positive, then both the factors in
the parentheses are positive. If the middle term is negative and the last term is
positive, then the factors in the parentheses are negative.
VIDEO LINK: Youtube: Simple trinomials with a negative pair of factors
Exercises Unit 7 Section 3
Examine each of the trinomials below and decide what kind of pair of values we need
to use in order to factor the expressions. List your answer as either positive or
negative.
1.
x2 + 8x + 15
2. a2 – 12a + 35
3. x2 – 20x + 36
4.
a2 + 14a + 40
5. x2 + 11x + 28
6. a2 – 9a + 18
Find the asked for pairs of factors. List your answers with signs as needed.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Find a pair of factors that multiply to 12 and add up to -7.
Find a pair of factors that multiply to 21 and add up to 10.
Find a pair of factors that multiply to 16 and add up to -17.
Find a pair of factors that multiply to 36 and add up to 15.
Find a pair of factors that multiply to 54 and add up to 29.
Find a pair of factors that multiply to 70 and add up to -17.
Find a pair of factors that multiply to 2 and add up to -3.
Find a pair of factors that multiply to 48 and add up to 19.
Find a pair of factors that multiply to 100 and add up to -20.
Find a pair of factors that multiply to 44 and add up to 15.
List the kind of pair of values required to factor the trinomials, and then factor the
trinomials.
17.
x2 + 12x + 20
18. a2 – 22a + 40
19. x2 – 20x + 51
20.
a2 + 11a + 18
21. x2 – 16x + 60
22. a2 + 13a + 40
23.
x2 – 7x + 10
24. a2 + 17a + 30
25. x2 – 12x + 36
26.
x2 + 5x + 6
27. x2 – 14x + 48
28. a2 – 18a + 80
29.
y2 + 18y + 32
30. z2 – 30z + 29
31. x2 + 25x + 66
32.
y2 + 9y + 20
33. t2 – 8t + 7
34. w2 + 14w + 49
35.
v2 - 4v + 3
36. b2 + 4b + 4
37. u2 + 28u + 75
38. Jean factored the trinomial x2 - 15x + 36 into the binomial
product (x - 18)(x - 2).
a. Prove Jean's answer is incorrect algebraically and show your work.
b. Factor the trinomial correctly and explain how you did the factoring.
c. Prove your answer is correct algebraically and show your work.
39. The trinomial a2 - 17 a + 72 is equivalent to
a. (a - 9)(a - 8)
b. (a - 36)(a - 2)
c. (a - 4)(a - 18)
c.
(a - 12)(a - 6)
Perform Common Factoring ( Factor out the GCF and reverse the distributive property. )
40. 7ax – 21ay
41. 15a3 – 21a2
42. 2ax - 18ax2
43. 5x2 + 20x - 105
44. -2x3 + 6x2 - 20x
45. a2z - 9az + 20z
46. Find the slope of each line by counting from point to point and list the y-intercept.
a.
b.
c.
Simplify
47. (a5)3
48. (5x2)3
3
50. ( a5b)3
4
49. (-2a2)4
51. The slope of the line passing through the two points (2, 5), (-1, 7) is
a. -
3
2
b. -
2
3
c.
3
2
d.
2
3
Unit 7 Section 4
Objective

The student will factor trinomials of the form x2 + bx + c where c < 0.
In this section we are again factoring trinomials. The inequality c < 0 means the last
term or constant in the trinomial is negative. This will not change our process but it
does require we find factors that can multiply to a negative number. In order for two
numbers to multiply to a negative answer there must be one positive number and one
negative number. For instance, the factors for -10 could be -2 and 5, one negative
number and one positive number. The example below shows us how to find the
factors for a negative constant.
Given
x2 + 3x – 28
factor the trinomial
To begin the process we must still ask ourselves the question:
What numbers multiply to -28 and add up to 3?
So we must list all the pairs of factors for -28.
-28
1, -28
-1, 28
2, -14
-2, 14
4, -7
-4, 7 This pair of factors adds up to 3.
So out answer is (x – 4)(x + 7)
In this example we had to list each pair of factors with alternating signs. We must be
careful here because reversing the signs will not give us the correct answer. We can
use the FOIL process for checking our answer to show this.
(x – 4)(x + 7)
+ 7x – 4x – 28
x2 + 3x – 28
x2
correct sign/answer
(x + 4)(x – 7)
– 7x + 4x – 28
x2 – 3x – 28
x2
incorrect sign
Only one pair of factors with the appropriate signs will give us the correct answer.
When we factor trinomials now we must first determine the kind of pair of factors we
want to use. There are three kinds of pairs of values to choose from. Both factors can
be positive. The two factors could both be negative. The two factors could have
one positive and one negative value, which we will call mixed. If we look at the
last two terms in the trinomial we can choose the correct kind of pair. The diagram
below shows us the best way to make this choice.
Start
Is the last term of the trinomial
negative?
YES
Mixed Pair of Factors
NO
Is the middle term of the trinomial
positive?
YES
Positive Pair of Factors
NO
Negative Pair of Factors
The examples below show us how to use the diagram.
Example A
Given
a2 + 6a – 16
list the type of pair of factors we want to use.
First we look at the last term. Since the last term is NEGATIVE (-16)
we will use a mixed pair of factors.
Example B
Given
x2 + 11x + 30
list the type of pair of factors we want to use.
First we look at the last term. Since the last term is POSITIVE (+30)
we go to the second question. The middle term is POSITIVE (+11)
so we will use a positive pair of factors.
Example C
Given
y2 – 14y + 40
list the type of pair of factors we want to use.
First we look at the last term. Since the last term is POSITIVE (+40)
we go to the second question. The middle term is NEGATIVE (-14)
so we will use a negative pair of factors.
We check or prove that a pair of binomials is the correct factoring of a trinomial
by using the FOIL process for binomials multiplication. Below are examples of the
type of problem we will be asked to do.
Example A
Prove: 10x2 + 13x – 3 can be factored into (2x + 3)(5x – 1)
(2x + 3)(5x – 1)
10x2 -2x + 15x – 3
10x2 + 13x - 3
Example B
Prove: 6a2 – 29ab + 28b2 can be factored into (3a – 4b)(2a – 7b)
(3a – 4b)(2a – 7b)
– 21ab – 8ab + 28b2
6a2 – 29ab + 28b2
6a2
VIDEO LINK: Youtube: Factoring Trinomials with c < 1
Exercises Unit 7 Section 4
Examine each of the trinomials below and decide what kind of pair of values we need
to use in order to factor the expressions. List your answer as positive, negative,
or mixed.
1.
x2 – 8x + 15
2. a2 – 2a – 35
3. y2 + 13x + 42
4.
a2 – 14a + 40
5. x2 + 11x – 26
6. a2 – a – 56
Find the asked for pairs of factors. List your answers with signs as needed.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Find a pair of factors that multiply to 36 and add up to -15.
Find a pair of factors that multiply to -22 and add up to -9.
Find a pair of factors that multiply to -3 and add up to -4.
Find a pair of factors that multiply to 4 and add up to 5.
Find a pair of factors that multiply to 60 and add up to -17.
Find a pair of factors that multiply to 14 and add up to -9.
Find a pair of factors that multiply to -6 and add up to 1.
Find a pair of factors that multiply to 48 and add up to -26.
Find a pair of factors that multiply to 121 and add up to 22.
Find a pair of factors that multiply to -42 and add up to 11.
List the kind of pair of values required to factor the trinomials and then factor the
trinomials.
17.
x2 + 12x – 28
18. a2 – 3a – 40
19. x2 – x – 20
20.
a2 + 10ab + 9b2
21. x2 – 16x + 60
22. a2 + 13az – 30z2
23.
x2 + 7x + 10
24. a2 – 18a + 32
25. x2 + 9x – 22
26.
x2 + 5x – 6
27. x2 – 15xy + 50y2
28. a2 + 18a + 65
29.
y2 + yz – 2z2
30. z2 – 14z + 13
31. x2 + 16x + 64
32.
y2 – 8y + 16
33. t2 – 8t - 33
34. w2 – 18w + 81
35.
v2 – 4v – 5
36. b2 – 6b + 9
37. u2 + 27u + 110
38. Harry factored the trinomial x2 - 9x - 36 into the binomial
product (x - 6)(x + 6).
a. Prove Harry's answer is incorrect algebraically and show your work.
b. Factor the trinomial correctly and explain how you did the factoring.
c. Prove your answer is correct algebraically and show your work.
39. The trinomial a2 - 3a - 40 is equivalent to
a. (a + 4)(a - 10)
b. (a - 5)(a + 8)
c. (a + 2)(a - 20)
c.
(a - 8)(a + 5)
Perform Common Factoring for the following. ( Factor out the GCF and reverse the Distributive
Property. )
40. 9bz – 24az
41. 12x3 – 20x
42. 22ax – 2a
43. 3x2 + 6x – 72
44. -4x3 – 10x2 - 22x
45. a2z - 5az + 15z
Prove the listed trinomial will factor into the given binomials.
46. Prove: 5y2 – 28y – 12 can be factored into (5y + 2)(y – 6)
47. Prove: 7z2 – 27z – 4 can be factored into (z – 4)(7z + 1)
48. Prove: 4a2 + 12ab + 9b2 can be factored into (2a + 3b)2
49. Prove: 8x2 – 22x + 5 can be factored into (4x – 1)(2x – 5)
Unit 7 Section 5
Objective

The student will factor trinomials of the form ax2 + bx + c.
In this section we our trinomials have a new coefficient. We can still use many of
the same ideas from the previous sections. We will be using a method called
“trial and error” or "guess and check" to do these problems, however. The key to
this method is knowing how to find the middle term from a binomial product. The
middle term is the sum of the OUTER terms and the INNER terms products. The
example below illustrates this.
Outer Terms (2x)(7) = 14x
(2x + 5)(3x + 7)
Inner Terms (5)(3x) = 15x
The Sum 29x
Using FOIL we get 6x2 + 14x + 15x + 35
6x2 + 29x + 35
The ‘29x’ can be found using the full FOIL process, or we can find it just using “OI”.
The "trial and error" or "guess and check" method depends on our ability to find the
middle term quickly and confirm that we have found the correct binomials. The trial
and error process can be described by the algorithm or set of steps given below.
Step
Step
Step
Step
Step
1.
2.
3.
4.
5.
Decide what kind of pair of factors we need.
Factor the first and last terms.
Set up binomials using the factors from step 2.
Use the sum of the Outer and Inner terms’ products - check correctness.
Repeat Steps 3 and 4 until we find the right combination.
The examples that follow show us how to use the algorithm.
Example A
Factor
2x2 + 11x + 5
Step 1. We need a POSITIVE pair of factors
2x2 + 11x + 5
1, 2
(2x + 5)(x + 1)
1, 5 Step 2. Factor the first and last terms.
(2x + 1)(x + 5)
Step 3. Setting up the binomials
To set up the binomials we have to use the ‘1’ and ‘2’ as coefficients of ‘x2’
because those were the factors of the coefficient of ‘x2’. The ‘1’ and ‘5’ were
factors of the constant at the end of the trinomial, so we use them at the
end of the binomials. We have two sets of binomials because there are two ways
to combine the ‘1’ and ‘5’. The ‘1’ can be in the first binomial, or the second.
Example A (continued)
Step 4. Using the Outer and Inner products.
(2x + 5)(x + 1)
(2x)(1) = 2x
(5)(x) = 5x
7x
(2x + 1)(x + 5)
(2x)(5) = 10x
(1)(x) = 1x
11x
The original problem was 2x2 + 11x + 5 so the right hand set of binomials
must be correct because it will give us the correct middle term.
Our answer is (2x + 1)(x + 5)
Example B
3x2 – 13x + 14
Factor
Step 1. We need a NEGATIVE pair of factors.
3x2 – 13x + 14
1, 3
-1, -14
-2, -7
Step 2. Factor the first and last terms.
Step 3. Setting up the binomials
(x – 1)(3x – 14)
(x – 14)(3x – 1)
(x – 2)(3x – 7)
(x – 7)(3x – 2)
To set up the binomials we have to use the ‘1’ and ‘3’ as coefficients of ‘x2’
because those were the factors of the coefficient of ‘x2’. The ‘-1’ and ‘-14’
and the ‘-2’ and ‘-7’ were factors of the constant at the end of the trinomial so
we use them at the end of the binomials. We have four sets of binomials
because there are four ways to combine the ‘-1’with ‘-14’ and the ‘-2’ with ‘-7’.
Step 4. Using the Outer and Inner products.
(x – 1)(3x – 14)
(x – 14)(3x – 1)
(x)(-14) = -14x
(-1)(3x) = -3x
-17x
(x)(-1) = -1x
(-14)(3x) = -42x
-43x
(x – 2)(3x – 7)
(x)(-7) = -7x
(-2)(3x) = -6x
-13x
(x – 7)(3x – 2)
(x)(-2) = -2x
(-7)(3x) = -21x
-23x
The combination that gives us the middle term ‘-13x’ is the third set of binomials
and our answer is (x – 2)(3x – 7).
Example C
Factor
7x2 + 19x – 6
Step 1.We need a MIXED pair of factors.
7x2 + 19x – 6
1, 7
1, -6
-1, 6
2, -3
-2, 3
Step 2. Factor the first and last terms.
Step 3. Setting up the binomials
There are actually eight combinations of the factors but we will set up four
of them and test them to see if one of them is correct and change it as needed.
(x + 1)(7x – 6)
(x – 6)(7x + 1)
(x – 2)(7x + 3)
(x – 3)(7x + 2)
Step 4. Using the Outer and Inner products.
(x + 1)(7x – 6)
(x – 6)(7x + 1)
(x)(-6) = -6x
(1)(7x) = 7x
1x
(x)(1) = 1x
(-6)(7x) = -42x
-41x
(x – 2)(7x + 3)
(x)(3) = 3x
(-2)(7x) = -14x
-11x
(x – 3)(7x + 2)
(x)(2) = 2x
(-3)(7x) = -21x
-19x
None of these four pairs gives us the positive ‘19x’ we need. However, we
did get an answer of ‘-19x’. If we reverse the signs in the binomials we can
get the middle term.
(x + 3)(7x – 2)
(x)(-2) = -2x
(3)(7x) = 21x
19x
This combination of factors gives us the positive ‘19x’ and so we know our
answer is (x + 3)(7x – 2).
Example D
Factor
4x2 – 13x + 3
Step 1. We need a NEGATIVE pair of factors.
4x2 – 13x + 3
1, 4
2, 2
-1,-3
Step 2. Factor the first and last terms.
Step 3. Setting up the binomials
The leading coefficient has two pairs of factors, so we need to set up pairs
of binomials that try both pairs of factors. Because one of the pairs is ‘2’ and ‘2’,
switching the factors won’t change the binomials; therefore, we only have three
combinations to check.
(x – 1)(4x – 3)
(x – 3)(4x – 1)
(2x – 3)(2x – 1)
Step 4. Using the Outer and Inner products.
(x – 1)(4x – 3)
(x)(-3) = -3x
(-1)(4x) = -4x
-7x
(x – 3)(4x – 1)
(x)(-1) = -1x
(-3)(4x) = -12x
-13x
(2x – 3)(2x – 1)
(2x)(-1) = -2x
(-3)(2x) = -6x
-8x
The combination that is our answer this time is (x – 3)(4x – 1)
VIDEO LINK: Youtube: Factoring with a non-zero leading coefficient
Exercises Unit 7 Section 5
Factor the following
1. 5x2 + 26x + 5
2. 7x2 – 10x + 3
3. 2x2 + 9x – 11
4. 3a2 + 17a + 10
5. 5z2 + 8z – 4
6. 6y2 – 25y + 14
7. 22x2 + x – 5
8. 15a2 + 19a + 6
9. 4z2 + 17z - 15
List the kind of pair of values needed and then factor the trinomials.
10.
x2 + 19x + 34
11. a2 – 21a + 38
12. x2 + 19x + 18
13. a2 – 2ab – 80b2
14. x2 – 12x + 36
15. a2 + 15az – 34z2
16.
x2 – 12x + 11
17. a2 + 19a + 70
18. x2 - 14x – 72
19.
x2 + 9x – 10
20. x2 – 17xy + 70y2
21. a2 + 8a + 16
Perform Common Factoring for the following.
22. 16by – 12bz
23. 18x – 22x4
24. 24a2x – 12a2
25. 6x3 + 12x2 – 24x
26. -3a3 – 15a2 – 27a
27. a3z - 5a2z + 15az
Unit 7 Section 6
Objective

The student will factor the difference of perfect squares.
There is one more type of factoring we need to master. This type of factoring is called
the difference of perfect squares. Below is a list of the perfect squares of the natural
numbers.
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, …
These are the perfect squares because each one of these numbers is calculated by
taking one of our counting numbers and squaring the number as shown below.
(1)2, (2)2, (3)2, (4)2, (5)2, (6)2, (7)2, (8)2, …
Variable terms can also be perfect squares. If a variable has an even exponent, then it
is a perfect square as demonstrated below.
a2,
x4 = (x2)2,
z6 = (z3)2,
y8 = (y4)2, …
The difference of perfect squares indicates that we are subtracting. Using these
representations, a sample of a difference of perfect squares would be:
x2 – 9
The ‘x2’ and the ‘9’ are both perfect squares and the minus sign between them is a
difference. Factoring a difference of perfect squares means we should be able to
find a pair of binomials whose product is the perfect squares. The type of binomials
that will give perfect squares is sum and difference binomials. Below are some
examples of multiplying sum and difference binomials.
Example A
Example B
(x + 9)(x – 9)
(2w – 5)(2w + 5)
x2 – 9x + 9x + 81
4w2 + 10w – 10w – 25
x2 – 81
4w2 – 25
Example C
(3a2 + 4b)(3a2 – 4b)
9a4 – 12a2b + 12a2b – 16b2
9a4 – 16b2
We should remember the short cut for multiplying the sum and difference binomials
is to use just the First and Last terms in the binomials. This is because the middle
terms always add up to zero and cancel out. The examples below show us how to
use the short cut.
Example A
Example B
Example C
(a – 12)(a + 12)
(2x + 10)(2x – 10)
(6y2 +5z)(6y2 – 5z)
a2 – 144
4x2 – 100
36y4 – 25z2
Factoring the difference of perfect squares means to reverse the short cut for
sum and difference multiplication. The examples below show us reversing this
process.
Example A
Example B
Example C
x2 – 49
25a2 – 4
16y6 – z2
(x + 7)(x – 7)
(5a – 2)(5a + 2)
(4y3 – z)(4y3 + z)
In Example A we see
‘x2’ which is a perfect
square and ‘49’ which
is also a perfect
square. We use ‘x’ and
‘7’ as factors in the
binomial. Since there
is a minus sign in the
original binomial, we
have to use a ‘+’ and a
‘-‘ in the binomial
factors.
In Example B we see
‘25a2’ which is a perfect
square and ‘4’ which is
also a perfect square.
Since it has the minus
sign between them, we
use ‘5a’ and ‘2’ in the
binomials. Because we
have a ‘-4’ we have to
have a ‘+’ and a ‘-‘ in
the binomials.
In Example C we see
‘16y6’ which is a perfect
square and ‘z2’ which is
also a perfect square.
Since it has the minus
sign between them, we
use ‘4y3’ and ‘z’ in the
binomials. Because we
have a ‘–z2’ we have to
have a ‘+’ and a ‘-‘ in
the binomials.
This method of factoring can only be used with the difference of perfect squares.
We will still need to be able to use trial and error, common factoring, or all the
techniques we learned for trinomials.
VIDEO LINK: Youtube: Factoring the difference of perfect squares
Exercises Unit 7 Section 6
Factor the following
1.
x2 – 64
2.
y2 – 36
3. a2 – 4b2
4. 81x2 – 16
5. 9x2 – 25y2
6. x2 – y2
7. a2b2 – 1
8.
49 – z2
1 2
11.
z – 100
4
9. 121x4 – 144
13. x2 + 11x – 12
14. 5a2 + 2a – 7
15. z2 + 20z + 64
16. x2 – x – 90
17. a2 – 49b2
18. 3x2 + 14x + 8
19.
a2 – 2ab – 8b2
20. x2 + 14xy + 40y2
21. a2 + 15az – 54z2
22.
4x2 + 16x – 9
23. 21a2 – 10a + 2
24. y4 – 144x4
25.
x2 + 19x + 88
26. 6x2 – 31xy + 18y2
27. a2 + 22a + 121
10. y6 – 169
12. w2 – 0.36
Factor the following
28.
x2 – 2x – 99
29. b2 – 24b + 144
30. a2 – 23a + 60
Perform Common Factoring for the following.
31. 24b2 – 8bc
32. 14x2 – 26x4
33. 4a2 – 12a3b
34. 5x3 + 10x2 – 20x
35. a3x2 – a2x + ax2
36. -2a4 – 14a3 – 22a2
37. Which of the following is the correct factoring of 14y2 - y - 3
a. (14y +1)(y - 3)
b. (7y - 3)(2y + 1)
c. (14y - 1)(y + 3)
d. (7y + 3)(2y - 1)
38. Which of the following is the correct factoring of 8w2 – 14w – 15
a. (8y - 15)(y + 1)
b. (4w - 5)(2w + 3)
c. (4w + 3)(2w - 5)
d. (4w + 5)(2w - 3)
39. Maria factored the trinomial 6x2 - 23x - 4 into the binomial
product (3x - 2)(2x + 2).
a. Prove Maria's answer is incorrect algebraically and show your work.
b. Factor the trinomial correctly and explain how you did the factoring.
c. Prove your answer is correct algebraically and show your work.
40. Find the equation of each line in slope-intercept form.
a.
b.
c.
1
41. Given the line y = - x + 2 find the equation of the perpendicular line that
3
passes through the point (2, -5).
Unit 7 Section 7
Objective

The student will factor polynomials using multiple techniques.
Factoring some polynomials requires the use of more than one technique. The key to
successfully factoring polynomials is recognizing when to use a given technique. Some
basic ideas for this task are:

Always look to perform common factoring FIRST.

If there are two terms and a minus sign in between, use the difference of perfect
squares.

If there are three terms, use our question, “What multiplies to the last term and
adds to the middle term”.

If there is a trinomial with a leading coefficient use “trial and error” factoring.

If you have factored a polynomial into quantities look to see if the quantities
can be factored.
The examples below show us how to combine the various types of factoring.
Example A
Factor
2x2 + 18x + 28
2(x2 + 9x + 14)
2(x + 7)(x + 2)
2x2 + 18x + 28
There is a common factor of ‘2’ so we do common factoring.
If we look into the parentheses we can factor the trinomial.
We are done since “x + 7” and “x + 2” will not factor.
Example B
Factor
3x3 – 6x2 – 45x
3x3 – 6x2 – 45x There is a common factor of ‘3x’ so we do common factoring.
3x(x2 – 2x – 15) If we look into the parentheses we can factor the trinomial.
3x(x – 5)(x + 3)
Example C
Factor
20ax3 – 5ax
20ax3 – 5ax
There is a common factor of ‘5ax’ so we do common factoring.
2
5ax(4x – 1)
In the parenthesis we see the difference of perfect squares.
5ax(2x + 1)(2x – 1)
Example D
Factor
60x3 – 28x2 – 8x
60x3 – 28x2 – 8x
There is a common factor of ‘4x’ so we do common factoring.
2
4x(15x – 7x – 2) In the parenthesis we need to use trial and error factoring.
4x(5x + 1)(3x – 2)
Example E
Factor
2x5 – 58x3 + 200x
2x5 – 58x3 + 200x There is a common factor of ‘2x’ so we do common factoring.
2x(x4 – 29x2 + 100) In the parenthesis we can factor the trinomial.
2x(x2 – 4)(x2 – 25) There is a difference of perfect squares in each parentheses.
2x(x + 2)(x – 2)(x + 5)(x – 5)
Example F
Factor
3x3 – 30x2 + 75x
3x(x2 – 10x + 25)
3x(x – 5)(x – 5)
3x(x – 5)2
3x3 – 30x2 + 75x
There is a common factor of ‘3x’ so we do common factoring.
In the parenthesis we need to use trial and error factoring.
The best way to write this is as a binomial square.
VIDEO LINK: Youtube: Factoring with multiple techniques
Exercises Unit 7 Section 7
Set A
Factor the following polynomials completely
1. 4x2 + 36x + 80
2. a3 + 12a2 – 28a
3. xz2 – 5xz + 6x
4. 2b2 – 18
5. 5ax4 – 5ax3 – 20ax2
6. 12y3 – 75y
7. 4x2 – 6x – 70
8. 6w4 + 60w3 + 126w2
9. 9z3 – 25z
10. 7x3 + 28x2 + 28x
11. x4 – 10x2 + 9
12. 2x3 – 4x2 + 2x
13. 3ax5 – 78ax3 + 75ax
14. 4a3 + 46a2 + 22a
15. 11a2 – 11
Fill in the blanks.
16. If an expression has a monomial that is raised to a power then we should put in
the ___________________ coefficients, __________________ the power times
all the exponents and simplify the ____________________.
Simplify the following
17. (a2)6
21. (.4z3)2
18. (4xy3)2
2
22. ( a4)2
3
19. (-3x5)3
20. (-4ab2)3
23. (-.5xy)3
24. (2xy3)-2
List the slope and intercept for each of the equations below.
25. y = 3x + 12
26. y = -2x + 1
27. y = -x + 1
28. y = x
29. x = -8
30. y =
2
x-3
3
Using the slope/intercept form of a linear equation ( y = mx + b ) graph the following
equations. List the slope and intercept before making your graph.
1
x-3
2
31.
y = 2x + 1
32. y = -3x + 5
33.
y=
34.
y=2
35. x = -3
36.
y=-
4
x +6
3
37. Which of the equations below best represents the graph.
a) y = 2(x - 4)
b) y = 2x - 4
c) 2y = -2x + 4
1
d) y =
(x + 4)
2
4
2
-4
-2
2
4
-2
-4
Exercises Unit 7 Section 7
Set B
Factor the following polynomials completely
1. 7x2 + 28x + 21
2. a4 + 15a3 – 34a2
3. x2z2 – 2x2z – 8x2
4. 2b2 – 18b + 36
5. 5ax4 – 20ax2
6. 12y2 – 36y + 24
7. x4 – 20x2 + 64
8. 6w2 – 32w + 26
9. 50z3 – 2z
10. 2x5 – 10x3 + 8x
11. 5az3 + 30az2 + 45az
12. 8x3 + 40x2 + 50x
13. 8a – 32a3
14. y3 + 14y2 + 49y
15. x2y2 – z2
16. 2x3 +32x2 + 126x
17. yz3 + 5yz2 – 24yz
19. x4 – 16
20. b5 + 2b3 – 15b
18. 2a3 – 10a2 – 72a
1
21. y2 –
25
22. Which of the expressions below is the complete factorization of 8x3 - 50x
a. 2x(4x2 -25)
b. 2(x3 - 25x)
c. 2x(2x + 5)(2x - 5)
d. 2x(4x + 5)(x - 5)
23. The expression 2az5 - 100az3 + 98az factored completely is
a. a(2z5 - 100z3 + 98z)
b. 2a(z5 - 50z3 + 49z)
c. 2az(z2 + 48)(z2 + 1)
d. 2az(z + 7)(z - 7)(z + 1)(z - 1)
24. Factor the expression 4ax4 - 37ax2 + 9a completely and explain how you are
doing the factoring at each step.
25. Complete the definitions of slope
a.
m
run
b.
m


c.
m
vertical change
d.
m
change in
change in
Given the point and the slope find the equation of the line they determine in
Slope-Intercept form.
26. m = 3 (2, 7)
27.
m = -2 (-4, -3)
28. m =
2
(9, -1)
3
Given the slope and point find the equation of the line Point-Slope form.
29. m = -6 (-4, 0)
30. m = 2.7 (8, -1.5)
The roster and table below represent lines. Find the slope of the line that passes
through the points.
31. { (-5, 6), (-2, 5), (1, 4), (4, 3) }
32.
x
y
-1
0
-3
-1
1
1
2
3
Unit 7 Section 8
Objective

The student will solve polynomial equations using the Zero Product
Theorem.
Solving linear equations is done using the Properties of Equality. Solving higher order
polynomial functions can be done using the Zero Product Theorem and factoring.
The Zero Product Theorem
If the product of two numbers is zero, then one or both of the numbers
must be zero.
When we see the theorem in symbols it is easier to interpret. Below is a symbolic
representation of the zero product theorem.
If ab = 0, then a = 0 or b = 0
We can explore this by picking some numbers.
If
If
If
If
we
we
we
we
pick
pick
pick
pick
a
a
a
a
=
=
=
=
5
0
5
0
and
and
and
and
b
b
b
b
=
=
=
=
4
4
0
0
then
then
then
then
ab
ab
ab
ab
=
=
=
=
20
0
0
0
The first problem show us what happens when neither ‘a’ nor ‘b’ is zero; our answer cannot
be a zero. The next three problems show us what happens when either, or both,
‘a’ and ‘b’ are zero; the answer is zero.
In the problems above ‘a’ and ‘b’ represent single variable values but the theorem
also applies to binomials. Since (x – 5) when simplified is one single value it can
be treated as one of our variables like ‘a’ or ‘b’ in the theorem. So, if we should see
problems similar to those given below we should be able to find the answer quickly.
(x – 5) = 0
(y + 2) = 0
(a – .5) = 0
(z) = 0
a = .5
z=0
The solutions to these equations are:
x=5
y = -2
We can check the answers by substituting as below.
(5 – 5) = 0
(-2 + 2) = 0
(.5 – .5) = 0
(0) = 0
This method for solving polynomials requires that the equation is equal to zero. Once
the equation is equal to zero we can factor it so that we have parenthetical expressions
similar to our previous examples. The examples below show us how equations in
factored form can be solved.
Example A
Example B
(x + 4)(x – 7) = 0
(x)(x – 2)
x = -4, 7
In example A the ‘-4’ will
make the first parentheses
equal to 0. So it is one of
our solutions. The ‘7’ will
make the second parentheses
equal to 0. So this is our
other solution.
x = 0, 2
Example C
(3x)(x + 2)(x + 2)(z – 1.5)
x = 0,-2, -2, 1.5
In example B the ‘0’ will
make the first parentheses
equal to 0. So it is one of
our solutions. The ‘2’ will
make the second parentheses
equal to 0. So this is our
other solution. This example
could be written as
x(x – 2)
But the ‘x’ is still a factor in
either form.
In example C the ‘0’ will
make the first parentheses
equal to 0. So it is one of
our solutions. The ‘-2’ will
make the second and third
parentheses equal to 0. So
this is another solution. The
fourth factor gives us a
‘1.5’ as a solution.
The solution could be listed as
x = 0, -2, 1.5 since we
don’t have to repeat the -2.
These solutions could be listed as sets as well. For the three either
examples
form. above, the
solutions could be listed as shown below. The symbol ‘  ’ stands for “element of”.
Example A
Example B
Example C
x  {-4, 7}
x  {0, 2}
x  {0, -2, 1.5}
Our parenthetical expression may also have coefficients in them. The examples below
show us how to find the solution when the coefficient is present.
Example A
Example B
Example C
(2x + 4) = 0
(3x – 5) = 0
(4x – 1) = 0
To solve theses we just need to remember how to solve a two-step equation.
2x + 4 = 0
-4 = -4
2x = -4
2
2
x = -2
3x – 5 = 0
+5 +5
3x = 5
3
3
x=
5
3
4x – 1 = 0
+1 +1
4x = 1
4
4
x=
1
4
Comparing the problem with the solution we can draw a conclusion that will help
us find solutions to two-step equations quickly.
Example A
Example B
Example C
(2x + 4) = 0
(3x – 5) = 0
(4x – 1) = 0
x=
4
= -2
2
x=
5
3
x =
1
4
To find the solutions quickly we can take the opposite of the constant in the parenthetical
expression and divide it by the coefficient of the variable. We can do this because the
process of the solving a two-step equation will use the Addition Property of Equality with
the opposite of the constant and using the Division Property of Equality with the coefficient.
Finding the solutions to the two-step equations below can be done mentally using this
principle.
2
5
This is the opposite of the constant ‘-2’ divided by the coefficient ‘5’.
Example A
(5x + 2) = 0
x=
3
2
This is the opposite of the constant ‘3’ divided by the coefficient ‘-2’.
Example B
(-2x - 3) = 0
x=
11
6
This is the opposite of the constant ‘11’ divided by the coefficient ‘6’.
Example C
(6x – 11) = 0
x=
We can use this principle when we are solving polynomials in factored form. The examples
below demonstrate this process.
Example A
(2x + 9)(x – 4) = 0
x=-
9
,4
2
Example B
Example C
2x(3x – 4)(4x – 7)
-3x2(x + 5)(7x – 1)
x = 0,
4 7
,
3 4
x = 0, -5,
1
7
4 7
, }
3 4
x  {0, -5,
Represented as sets our solutions would be
x  {-
9
, 4}
2
x  {0,
1
}
7
Solving polynomial equations in factored form can be done quickly. We need to be able
to start with polynomials in exponential forms and find the factored form. We will use
all the techniques we have learned in this unit. The examples that follow show the
complete process.
Example A
Solve x2 + 17x + 42 = 0
x2 + 17x + 42 = 0
(x + 14)(x + 3) = 0
We factored the trinomial (14)(3) = 42 and 14 + 3 = 17.
x = -14, -3
x  {-14, -3}
Example B
Solve a2 – 18a + 32 = 0
a2 – 18a + 32 = 0
(a – 16)(a – 2) = 0
We factored the trinomial (-16)(-2)=32 and -16+-2=-18.
a = 16, 2
a  {16, 2}
Example C
Solve y2 – 49 = 0
y2 – 49 = 0
(y – 7)(y + 7) = 0
We used the difference of perfect squares.
y = -7, 7
y  {-7, 7}
Example D
Solve z2 – 11z = 0
z2 – 11z = 0
z(z – 11) = 0
We used common factoring here.
z = 0, 11
z  {0, 11}
Example E
Solve 0 = x2 + 22x + 121
0 = x2 + 22x + 121
0 = (x + 11)(x + 11)
x = -11
x  {-11}
or
(x + 11)2
The standard form for a polynomial is in descending order of exponents. The standard
form for a polynomial equation must be equal to zero. In our previous examples, the
equations were all equal to zero. The side of the equation which is zero does not
matter. If the equation is not equal to zero initially, then we must use the Properties
of Equality to make the equation equal to zero. The examples below show us how to
get the equation into standard from and solve the equation.
Example A
Solve
x3 = 12x2
x3 = 12x2
-12x2 -12x2
x3 – 12x2 = 0
x2(x – 12) = 0
The equation is not in standard form.
We use the Subtraction Property of Equality to get the
equation into standard form.
We perform common factoring.
x = 0, 12
x  {0, 12}
Example B
Solve
x2 = 16x – 28
x2 = 16x – 28
-16x + 28 -16x + 28
x2 – 16x + 28 = 0
(x – 2)(x – 14) = 0
The equation is not in standard form, and we want the
square term to be positive. So we will subtract 16x from
both sides and add 28 to both sides.
We factor the trinomial.
x = 2, 14
x  {2, 14}
We will also need to use multiple types of factoring to solve some of the equations. The
examples that follow show us how to use our combined types of factoring.
Example A
Solve
10x3 -12x2 – 22x = 0
10x3 -12x2 – 22x = 0
2x(5x2 – 6x – 11) = 0
2x(5x – 11)(x + 1) = 0
11
x = 0,
, -1
5
11
x  {0,
, -1}
5
First we perform common factoring.
This is the result of “trial and error” factoring.
Example B
Solve
3x4 - 75x2 = -432
3x4 - 75x2 = -432
+432
+432
3x4 - 75x2 + 432 = 0
3(x4 – 25x2 + 144) = 0
3(x2 – 16)(x2 - 9) = 0
We must get this equation into standard form.
Second we perform common factoring.
Next we factor the trinomial.
3(x + 4)(x – 4)(x + 3)(x – 3) = 0
Each of the parenthetical expressions is a difference
of perfect squares so we factor these expressions.
x = -4, 4, -3, 3
x  {-4, 4, -3, 3}
VIDEO LINK: Khan Academy Solving a Quadratic Equation by Factoring
Exercises Unit 7 Section 8
Set A
Solve the following equations
1. x2 + 8x + 15 = 0
2. a2 – 12a + 35 = 0
3. x2 + 20x = 0
4.
a2 + 11a – 26 = 0
5. x2 – x – 30 = 0
6. a2 + 9a + 18 = 0
7.
y2 – 7y = 0
8. a2 + 13a + 42 = 0
9. x2 – 36 = 0
10.
x2 + 6x + 5 = 0
11. x2 + x = 0
12. a2 – 14a + 40 = 0
13.
y3 + 16y = 0
14. z2 + 13z – 30 = 0
15. x2 + 15x + 44 = 0
16.
y2 + y - 20 = 0
17. t2 + 20t + 100 = 0
18. w2 – 10w + 25 = 0
19.
v2 + 2v + 1 = 0
20. b2 – 4 = 0
21. u2 – 26u + 105 = 0
Find the equation of the line that would pass through the two given points. List your
answer in the slope-intercept form of a linear equation.
22.
(4, 7) (5, 9)
23. ( -1, 8) ( 3, -4)
Answer the following
24. Alex and Joe are painting a fence around a property. In the first two hours
they painted 82 ft of fence. After 6 hours they had finished 246 ft of fence.
a. List the two data points in this problem.
b. Find the rate of change or slope and list the units.
c. Write an equation that models this situation.
Exercises Unit 7 Section 8
Set B
Factor the following and solve.
1. x2 + 19x + 34 = 0
2. a2 – 21a = 0
3. x2 – 19x + 18 = 0
5. x2 – 17x + 70 = 0
6. a2 + 8a + 16 = 0
7. 3a2 + 17a + 10 = 0
8. 5z2 + 8z = 0
9. 6y2 – 25y + 14 = 0
10. a2 – 2a – 80 = 0
11. 15a2 + 19x + 6 = 0
12. 0 = z2 – 16
13. 0 = 5x2 + 16x + 3
14. 36x2 – 25 = 0
15. x2 – 14x – 72 = 0
16. 22x2 + x – 5 = 0
17. 0 = x2 – 12x +36
18. a2 – 49 = 0
19. x2 – 12x + 11 = 0
20. a3 + 13a2 = 0
21. 0 = 2x2 + 9x – 11
22. x3 + 3x2 + 2x = 0
23. a3 – 25a = 0
24. 4z3 – 28z2 + 48 = 0
25. 0 = 5x2 + 20x – 25
26. x4 – 81 = 0
27. 3y3 + 18y2 + 27y = 0
28. 4x3 + 30x2 + 26x = 0
29. x4 – 37x2 + 36 = 0
30. w4 – 10w3 + 20w2 = 0
31. 7x2 – 28x + 28 = 0
32. 0 = 2a3 – 18a
33. 0 = b3 – 14b2 + 40b
4.
0 = x2 + 9x – 10
Factor completely and solve.
34. Sally is saving to buy a tablet computer. She put $60 into a savings account.
Her plan is to add $15 a week to the account.
a. Write an equation to model this situation.
b. Use the equation to find how much will be in the account in 7 weeks.
c. If the tablet she wants costs $330 how many weeks will it take her to
save at least this amount.
35. When a single point can make two different equations true we know
this is a point at which the two lines __________________.
For each given system of equations check to see if the listed point is the solution.
Answer yes or no and show your work.
36.
y = 4x - 1
y = 2x + 5
(3, 11)
37.
y = -2x + 4
y=x-1
(-1, 2)
Exercises Unit 7 Section 8
Set C
Factor the following and solve.
1. x2 + x = 30
2. a2 = 2a
3. x2 + 18 = 11x
4. -10 = x2 + 11x
5. x2 = 16x – 60
6. a2 + 12a + 27 = 0
7. 3a2 + 34a = -11
8. 2z2 = -8z
9. y2 = 144
10. a2 – a = -90
11. 5x2 – 14x – 3 = 0
12. 0 = z2 – 1
13. 0 = 5x2 + 36x + 7
14. 9x2 = 16
15. x2 – 15x = -56
16. 0 = x2 + x – 6
17. x2 = 20x – 36
18. a3 = 64a
19. x3 – 13x2 = -12
20. z3 = z2
21. 0 = x2 + 9x – 22
22. x3 = 2x2 – x
23. 49a = a3
24. 12z3 = 12z2 – 4z
25. 12x2 = 20x + x3
26. x3 – 16x = 0
27. 3y3 + 72y = -33y2
28. 4x + x2 = -5
29. x4 + 4 = 5x2
30. –18w + 81 = - w2
31. 0 = x2 – 24x + 144
32. 15y = 3y2
33. x5 = 20x3 – 48x
34. 2x2 – 36 = x2
35. x2 + 11x = 4x – 10
37. 3x2 - 9x + 2 = 2x2 + 4x – 28
36. 4x2 + 3x = 2x
38. x2 + x – 1 = 3x2 + 4x
Fill in the blanks
39. If an expression has a monomial that is raised to a power then we should put in
the understood ___________________ then __________________ the power
times all the exponents and simplify the ____________________.
Simplify the following
40. (z2)3
41. (a8b2)3
42. (-2x5)3
43. (-6a5)2
1
44. ( x-3)2
2
3
45. (- a4)2
5
46. (-.5abc)2
47. (2.1y3)3
Decide from the two listed equations if the lines are parallel, perpendicular or neither.
1
48. y = 4x + 2
49. y =
x+3
50. y = 5x + 1
2
y = -4x - 1
y = -2x + 3
y = 5x + 4
5
51. y = - x + 2
2
2
y= x-1
5
1
52. y = - x + 3
3
y = -3x + 3
2
x+1
7
2
y= x-1
7
53. y =
54. Given the expression (4x2 - 25)(2x2 + 5x + 3) = 0
a. Factor the left side of the equation completely. Explain the method of
factoring you used for each quantity.
b. Use your result from part 'a' to find the solutions to the equation.
c. Choose one of your solutions and check it algebraically. Show yyour work.
55. Jose solved the equation y2 + 5y = 24 correctly by factoring. Which of the values
below are his solutions?
a. 3, -8
b. -3, 8
c. 4, -6
d. 3 only
56. The solutions to the equation 6 = 5x2 + 7x are
a. -2, 
5
3
56. The correct factors of
b. 2,
5
3
c. -2,
3
5
d. 2, 
5
3
1 2
9
z 
are
4
25
a. (
1
3
z - )2
2
5
b. (
1
3 1
3
z - )( z + )
2
5 2
5
c. (
1
1
3
z - )(z + )
4
5
5
c. (
1
1
z)(z + 9)
4
25
57. Solve the following equation and explain thoroughly how you found the solutions.
14x3 = 3x2 + 2x