Factor Trinomials

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

Figure shows the equation open parentheses x plus 2 close parentheses open parentheses x plus 3 close parentheses equals x squared plus 5 x plus 6. The left side of the equation is labeled factors and the right is labeled product. An arrow pointing right is labeled multiply. An arrow pointing left is labeled factor.

To figure out how we would factor a trinomial of the form

x^{2} + b x + c,

| {: valign=”top”}{: .unnumbered .unstyled summary=”We have open parentheses x plus m close parentheses open parentheses x plus n close parentheses. Foil to find the product x squared plus mx plus nx plus mn. Factor the GCF from the middle terms x squared plus open parentheses m plus n close parentheses x plus mn. Now our trinomial is of the form x squared plus bx plus c, where b is m plus n and c is mn” data-label=””}

In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

How do you get a positive product and a negative sum? We use two negative numbers.

Factors of $28$	Sum of factors
$− 1, −28$ $− 2, −14$ $− 4, −7$	$− 1 + (− 28) = −29$ $− 2 + (− 14) = −16$ $− 4 + (− 7) = {−11}^{*}$

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

How do you get a negative product and a positive sum? We use one positive and one negative number.

When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.

Factors of $−48$	Sum of factors
$−1, 48$ $−2, 24$ $−3, 16$ $−4, 12$ $−6, 8$	$−1 + 48 = 47$ $−2 + 24 = 22$ $−3 + 16 = 13$ $−4 + 12 = 8$ $−6 + 8 = 2^{*}$

Sometimes you’ll need to factor trinomials of the form

x^{2} + b x y + c y^{2}

in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in How To Factor trinomials.

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Factors of $− 12$	Sum of factors
$1, −12$ $−1, 12$ $2, −6$ $−2, 6$ $3, −4$ $−3, 4$	$1 + (−12) = −11$ $−1 + 12 = 11$ $2 + (−6) = −4$ $−2 + 6 = 4$ $3 + (−4) = −1$ $−3 + 4 = 1$

Let’s summarize the method we just developed to factor trinomials of the form

x^{2} + b x + c .

Factor Trinomials of the form ax² + bx + c using Trial and Error

Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form

a x^{2} + b x + c .

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Factors of $− 5$	Sum of factors
$− 1, 5$ $1, −5$	$− 1 + 5 = 4^{*}$ $1 + (− 5) = −4$

We know the first terms of the binomial factors will multiply to give us

3 x^{2} .

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.

Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses.

Which factors are correct? To decide that, we multiply the inner and outer terms.

How to Factor a Trinomial Using Trial and Error

Factor completely using trial and error: $3 y^{2} + 22 y + 7 .$

![Step 1 is to write the trinomial in descending order. The trinomial 3 y squared plus 22y plus 7 is already in descending order.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_007a.jpg) ![Step 2 is to factor the GCF. Here, there is none.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_007b.jpg) ![Step 3 is Find all the factor pairs of the first term. The only factors here are 1y and 3y. Since there is only one pair, we can put each as the first term in the parentheses.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_007c.jpg) ![Step 4 is to find all the factor pairs of the third term. Here, the only pair is 1 and 7.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_007d.jpg) ![Step 5 is to test all the possible combinations of the factors until the correct product is found. For possible factors open parentheses y plus 1 close parentheses open parentheses 37 plus 7 close parentheses, the product is 3 y squared plus 10y plus 7. For the possible factors open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses, the product is 3 y squared plus 22y plus 7, which is the correct product. Hence, the correct factors are open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_007e.jpg) ![Step 6 is to check by multiplying.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_007f.jpg)

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.

$6 b^{2} - 13 b + 5$
$(b - 1) (6 b - 5)$	$6 b^{2} - 11 b + 5$
$(b - 5) (6 b - 1)$	$6 b^{2} - 31 b + 5$
$(2 b - 1) (3 b - 5)$	$6 b^{2} - 13 b + 5^{*}$
$(2 b - 5) (3 b - 1)$	$6 b^{2} - 17 b + 5$

Factor Trinomials of the Form

a x^{2} + b x + c

is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

How to Factor Trinomials using the “ac” Method

Factor using the ‘ac’ method: $6 x^{2} + 7 x + 2 .$

![Step 1 is to factor the GCF. There is none in 6 x squared plus 7x plus 2.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_011a.jpg) ![Step 2 is to find the product of a and c. The product of 6 and 2 is 12.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_011b.jpg) ![Step 3 is to find 2 numbers m and n such that mn is ac and m plus n is b. So we need to numbers that multiply to 12 and add to 7. Both factors must be positive. 3 times 4 is 12 and 3 plus 4 is 7.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_011c.jpg) ![Step 4 is to split the middle term using m and n. So we rewrite 7 x as 3x plus 4x. It would give the same result if we used 4x plus 3x. Rewriting, we get 6 x squared plus 3x plus 4x plus 2. Notice that this is the same as the original polynomial. We just split the middle term to get a more useful form](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_011d.jpg) ![Step 5 is to factor by grouping. So, we get, 3x open parentheses 2x plus 1 close parentheses plus 2 open parentheses 2x plus 1 close parentheses. This is equal to 2x plus 1, 3x plus 2.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_011e.jpg) ![Step 6 is to check by multiplying the factors.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_06_02_011f.jpg)

Factor Using Substitution

form. However, we can often make a thoughtful substitution that will allow us to make it fit the

a x^{2} + b x + c

form. This is called factoring by substitution. It is standard to use u for the substitution.

is the variable part of the first term. Look for this relationship as you try to find a substitution.

Factor by substitution: ${(x - 2)}^{2} + 7 (x - 2) + 12$

The binomial in the middle term, $(x - 2)$

is squared in the first term. If we let $u = x - 2$

and substitute, our trinomial will be in $a x^{2} + b x + c$

form.


{: valign=”top”}	Rewrite the trinomial to prepare for the substitution.
{: valign=”top”}	Let $u = x - 2$

and substitute.
{: valign=”top”}	Factor the trinomial.
{: valign=”top”}	Replace u with $x - 2 .$

| | {: valign=”top”}| Simplify inside the parentheses. | | {: valign=”top”}{: .unnumbered .unstyled summary=”The polynomial is open parentheses x minus 2 close parentheses squared plus 7 open parentheses x minus 2 close parentheses plus 12. Substituting x minus 2 with u, we get u squared plus 7u plus 12. We factor this to get u plus 3, u plus 4. Replacing u with x minus 2, we get open parentheses x minus 2 plus 3 close parentheses open parentheses x minus 2 plus 4 close parentheses. Simplifying, we get x plus 1, x plus 2.” data-label=””}

This could also be factored by first multiplying out the ${(x - 2)}^{2}$

and the $7 (x - 2)$

and then combining like terms and then factoring. Most students prefer the substitution method.

Key Concepts

Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.

Practice Makes Perfect

**Factor Trinomials of the Form $x^{2} + b x + c$

In the following exercises, factor each trinomial of the form $x^{2} + b x + c .$

p^{2} + 11 p + 30

(p + 5) (p + 6)

w^{2} + 10 x + 21

n^{2} + 19 n + 48

(n + 3) (n + 16)

b^{2} + 14 b + 48

a^{2} + 25 a + 100

(a + 5) (a + 20)

u^{2} + 101 u + 100

x^{2} - 8 x + 12

(x - 2) (x - 6)

q^{2} - 13 q + 36

y^{2} - 18 y + 45

(y - 3) (y - 15)

m^{2} - 13 m + 30

x^{2} - 8 x + 7

(x - 1) (x - 7)

y^{2} - 5 y + 6

5 p - 6 + p^{2}

(p - 1) (p + 6)

6 n - 7 + n^{2}

8 - 6 x + x^{2}

(x - 4) (x - 2)

7 x + x^{2} + 6

x^{2} - 12 - 11 x

(x - 12) (x + 1)

−11 - 10 x + x^{2}

In the following exercises, factor each trinomial of the form $x^{2} + b x y + c y^{2} .$

x^{2} - 2 x y - 80 y^{2}

(x + 8 y) (x - 10 y)

p^{2} - 8 p q - 65 q^{2}

m^{2} - 64 m n - 65 n^{2}

(m + n) (m - 65 n)

p^{2} - 2 p q - 35 q^{2}

a^{2} + 5 a b - 24 b^{2}

(a + 8 b) (a - 3 b)

r^{2} + 3 r s - 28 s^{2}

x^{2} - 3 x y - 14 y^{2}

Prime

u^{2} - 8 u v - 24 v^{2}

m^{2} - 5 m n + 30 n^{2}

Prime

c^{2} - 7 c d + 18 d^{2}

**Factor Trinomials of the Form $a x^{2} + b x + c$

Using Trial and Error**

In the following exercises, factor completely using trial and error.

p^{3} - 8 p^{2} - 20 p

p (p - 10) (p + 2)

q^{3} - 5 q^{2} - 24 q

3 m^{3} - 21 m^{2} + 30 m

3 m (m - 5) (m - 2)

11 n^{3} - 55 n^{2} + 44 n

5 x^{4} + 10 x^{3} - 75 x^{2}

5 x^{2} (x - 3) (x + 5)

6 y^{4} + 12 y^{3} - 48 y^{2}

2 t^{2} + 7 t + 5

(2 t + 5) (t + 1)

5 y^{2} + 16 y + 11

11 x^{2} + 34 x + 3

(11 x + 1) (x + 3)

7 b^{2} + 50 b + 7

4 w^{2} - 5 w + 1

(4 w - 1) (w - 1)

5 x^{2} - 17 x + 6

4 q^{2} - 7 q - 2

(4 q + 1) (q - 2)

10 y^{2} - 53 y - 11

6 p^{2} - 19 p q + 10 q^{2}

(2 p - 5 q) (3 p - 2 q)

21 m^{2} - 29 m n + 10 n^{2}

4 a^{2} + 17 a b - 15 b^{2}

(4 a - 3 b) (a + 5 b)

6 u^{2} + 5 u v - 14 v^{2}

−16 x^{2} - 32 x - 16

−16 (x + 1) (x + 1)

−81 a^{2} + 153 a + 18

−30 q^{3} - 140 q^{2} - 80 q

10 q (3 q + 2) (q + 4)

−5 y^{3} - 30 y^{2} + 35 y

**Factor Trinomials of the Form $a x^{2} + b x + c$

using the ‘ac’ Method**

In the following exercises, factor using the ‘ac’ method.

5 n^{2} + 21 n + 4

(5 n + 1) (n + 4)

8 w^{2} + 25 w + 3

4 k^{2} - 16 k + 15

(2 k - 3) (2 k - 5)

5 s^{2} - 9 s + 4

6 y^{2} + y - 15

(3 y + 5) (2 y - 3)

6 p^{2} + p - 22

2 n^{2} - 27 n - 45

(2 n + 3) (n - 15)

12 z^{2} - 41 z - 11

60 y^{2} + 290 y - 50

10 (6 y - 1) (y + 5)

6 u^{2} - 46 u - 16

48 z^{3} - 102 z^{2} - 45 z

3 z (8 z + 3) (2 z - 5)

90 n^{3} + 42 n^{2} - 216 n

16 s^{2} + 40 s + 24

8 (2 s + 3) (s + 1)

24 p^{2} + 160 p + 96

48 y^{2} + 12 y - 36

12 (4 y - 3) (y + 1)

30 x^{2} + 105 x - 60

Factor Using Substitution

In the following exercises, factor using substitution.

x^{4} - x^{2} - 12

(x^{2} + 3) (x^{2} - 4)

x^{4} + 2 x^{2} - 8

x^{4} - 3 x^{2} - 28

(x^{2} - 7) (x^{2} + 4)

x^{4} - 13 x^{2} - 30

{(x - 3)}^{2} - 5 (x - 3) - 36

(x - 12) (x + 1)

{(x - 2)}^{2} - 3 (x - 2) - 54

{(3 y - 2)}^{2} - (3 y - 2) - 2

(3 y - 4) (3 y - 1)

{(5 y - 1)}^{2} - 3 (5 y - 1) - 18

Mixed Practice

In the following exercises, factor each expression using any method.

u^{2} - 12 u + 36

(u - 6) (u - 6)

x^{2} - 14 x - 32

r^{2} - 20 r s + 64 s^{2}

(r - 4 s) (r - 16 s)

q^{2} - 29 q r - 96 r^{2}

12 y^{2} - 29 y + 14

(4 y - 7) (3 y - 2)

12 x^{2} + 36 y - 24 z

6 n^{2} + 5 n - 4

(2 n - 1) (3 n + 4)

3 q^{2} + 6 q + 2

13 z^{2} + 39 z - 26

13 (z^{2} + 3 z - 2)

5 r^{2} + 25 r + 30

3 p^{2} + 21 p

3 p (p + 7)

7 x^{2} - 21 x

6 r^{2} + 30 r + 36

6 (r + 2) (r + 3)

18 m^{2} + 15 m + 3

24 n^{2} + 20 n + 4

4 (2 n + 1) (3 n + 1)

4 a^{2} + 5 a + 2

x^{4} - 4 x^{2} - 12

(x^{2} + 2) (x^{2} - 6)

x^{4} - 7 x^{2} - 8

{(x + 3)}^{2} - 9 (x + 3) - 36

(x - 9) (x + 6)

{(x + 2)}^{2} - 25 (x + 2) - 54

Writing Exercises

Many trinomials of the form $x^{2} + b x + c$

factor into the product of two binomials $(x + m) (x + n) .$

Explain how you find the values of m and n.

Answers will vary.

Tommy factored $x^{2} - x - 20$

as $(x + 5) (x - 4) .$

Sara factored it as $(x + 4) (x - 5) .$

Ernesto factored it as $(x - 5) (x - 4) .$

Who is correct? Explain why the other two are wrong.

List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form $a x^{2} + b x + c .$

Answers will vary.

How is the “ac” method similar to the “undo FOIL” method? How is it different?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Is there a greatest common factor?
Yes. The GCF is 5.
Factor it.
The trinomial inside the parentheses has a leading coefficient that is not 1.
Find the product $a c .$	$a c = 28$
Find two numbers that multiply to $a c$	$(−4) (−7) = 28$
and add to b.	$−4 + (−7) = −11$
Split the middle term.

Factor the trinomial by grouping.

Check by multiplying all three factors. $\begin{matrix} 5 (y - 2) (2 y - 7) \\ 5 (2 y^{2} - 7 y - 4 y + 14) \\ 5 (2 y^{2} - 11 y + 14) \\ 10 y^{2} - 55 y + 70 ✓ \end{matrix}$


Rewrite the trinomial to prepare for the substitution.
Let $u = x^{2}$ and substitute.
Factor the trinomial.
Replace u with $x^{2} .$
Check: $\begin{matrix} (x^{2} + 1) (x^{2} - 5) \\ x^{4} - 5 x^{2} + x^{2} - 5 \\ x^{4} - 4 x^{2} - 5 ✓ \end{matrix}$


{: valign=”top”}	Foil to find the product.
{: valign=”top”}	Factor the GCF from the middle terms.
{: valign=”top”}	Our trinomial is of the form $x^{2} + b x + c .$

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since the last term, 5, is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

$6 b^{2} - 13 b + 5$
Possible factors	Product
$(b - 1) (6 b - 5)$	$6 b^{2} - 11 b + 5$
$(b - 5) (6 b - 1)$	$6 b^{2} - 31 b + 5$
$(2 b - 1) (3 b - 5)$	$6 b^{2} - 13 b + 5^{*}$
$(2 b - 5) (3 b - 1)$	$6 b^{2} - 17 b + 5$