## Learning Objectives

By the end of this section, you will be able to:

- Factor trinomials of the form ${x}^{2}+bx+c$
- Factor trinomials of the form ${x}^{2}+bxy+c{y}^{2}$

## Be Prepared 7.4

Before you get started, take this readiness quiz.

Multiply: $(x+4)(x+5).$

If you missed this problem, review Example 6.38.

## Be Prepared 7.5

Simplify: ⓐ $\mathrm{-9}+(\mathrm{-6})$ ⓑ $\mathrm{-9}+6.$

If you missed this problem, review Example 1.37.

## Be Prepared 7.6

Simplify: ⓐ $\mathrm{-9}(6)$ ⓑ $\mathrm{-9}(\mathrm{-6}).$

If you missed this problem, review Example 1.46.

## Be Prepared 7.7

Simplify: ⓐ $\left|\mathrm{-5}\right|$ ⓑ $\left|3\right|.$

If you missed this problem, review Example 1.33.

## Factor Trinomials of the Form *x*^{2} + *bx* + *c*

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication—to start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

To factor the trinomial means to start with the product, ${x}^{2}+5x+6$, and end with the factors, $\left(x+2\right)\left(x+3\right)$. You need to think about where each of the terms in the trinomial came from.

The *first term* came from multiplying the first term in each binomial. So to get ${x}^{2}$ in the product, each binomial must start with an *x*.

The *last term* in the trinomial came from multiplying the last term in each binomial. So the last terms must multiply to 6.

What two numbers multiply to 6?

The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use?

Consider the *middle term*. It came from adding the outer and inner terms.

So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in Table 7.1—the table will be very helpful when you work with numbers that can be factored in many different ways.

Factors of $6$ | Sum of factors |
---|---|

$1,6$ | $1+6=7$ |

$2,3$ | $2+3=5$ |

We see that 2 and 3 are the numbers that multiply to 6 and add to 5. So we have the factors of ${x}^{2}+5x+6$. They are $\left(x+2\right)\left(x+3\right)$.

You should check this by multiplying.

Looking back, we started with ${x}^{2}+5x+6$, which is of the form ${x}^{2}+bx+c$, where $b=5$ and $c=6$. We factored it into two binomials of the form $\left(x+m\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(x+n\right)$.

To get the correct factors, we found two numbers *m* and *n* whose product is *c* and sum is *b*.

## Example 7.17

### How to Factor Trinomials of the Form ${x}^{2}+bx+c$

Factor: ${x}^{2}+7x+12$.

### Solution

## Try It 7.33

Factor: ${x}^{2}+6x+8$.

## Try It 7.34

Factor: ${y}^{2}+8y+15$.

Let’s summarize the steps we used to find the factors.

## How To

### Factor trinomials of the form ${x}^{2}+bx+c$.

- Step 1.
Write the factors as two binomials with first terms
*x*: $\phantom{\rule{1em}{0ex}}\left(x\phantom{\rule{1.5em}{0ex}}\right)\left(x\phantom{\rule{1.5em}{0ex}}\right)$. - Step 2.
Find two numbers
*m*and*n*that

Multiply to*c*, $m\xb7n=c$

Add to*b*, $m+n=b$ - Step 3.
Use
*m*and*n*as the last terms of the factors: $\phantom{\rule{4em}{0ex}}(x+m)(x+n)$. - Step 4. Check by multiplying the factors.

## Example 7.18

Factor: ${u}^{2}+11u+24$.

### Solution

Notice that the variable is *u*, so the factors will have first terms *u*.

$\begin{array}{cccc}& & & {u}^{2}+11u+24\hfill \\ \text{Write the factors as two binomials with first terms}\phantom{\rule{0.2em}{0ex}}u.\hfill & & & \left(u\phantom{\rule{1.5em}{0ex}}\right)\left(u\phantom{\rule{1.5em}{0ex}}\right)\hfill \end{array}$

Find two numbers that: multiply to 24 and add to 11.

Factors of $24$ | Sum of factors |
---|---|

$\phantom{\rule{0.4em}{0ex}}1,24$ | $1+24=25$ |

$\phantom{\rule{0.4em}{0ex}}2,12$ | $2+12=14$ |

$3,8$ | $\phantom{\rule{0.8em}{0ex}}3+8=11\text{*}$ |

$4,6$ | $\phantom{\rule{0.4em}{0ex}}4+6=10$ |

$\begin{array}{cccc}\text{Use 3 and 8 as the last terms of the binomials.}\hfill & & & \hfill \phantom{\rule{2.8em}{0ex}}\left(u+3\right)\left(u+8\right)\hfill \\ \\ \text{Check.}\hfill & & & \\ \\ \\ \phantom{\rule{2.5em}{0ex}}\left(u+3\right)\left(u+8\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{u}^{2}+3u+8u+24\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{u}^{2}+11u+24\phantom{\rule{0.2em}{0ex}}\u2713\hfill & & & \end{array}$

## Try It 7.35

Factor: ${q}^{2}+10q+24$.

## Try It 7.36

Factor: ${t}^{2}+14t+24$.

## Example 7.19

Factor: ${y}^{2}+17y+60$.

### Solution

$\begin{array}{cccc}& & & {y}^{2}+17y+60\hfill \\ \text{Write the factors as two binomials with first terms}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \left(y\phantom{\rule{1.5em}{0ex}}\right)\left(y\phantom{\rule{1.5em}{0ex}}\right)\hfill \end{array}$

Find two numbers that multiply to 60 and add to 17.

Factors of $60$ | Sum of factors |
---|---|

$1,60$ | $1+60=61$ |

$2,30$ | $2+30=32$ |

$3,20$ | $3+20=23$ |

$4,15$ | $4+15=19$ |

$5,12$ | $\phantom{\rule{0.4em}{0ex}}5+12=17\text{*}$ |

$6,10$ | $6+10=16$ |

$\begin{array}{cccc}\text{Use}\phantom{\rule{0.2em}{0ex}}5\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}12\phantom{\rule{0.2em}{0ex}}\text{as the last terms.}\hfill & & & \hfill \phantom{\rule{9em}{0ex}}\left(y+5\right)\left(y+12\right)\hfill \\ \\ \text{Check.}\hfill & & & \\ \\ \\ \phantom{\rule{2.5em}{0ex}}(y+5)(y+12)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}({y}^{2}+12y+5y+60)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}({y}^{2}+17y+60)\u2713\hfill & & & \end{array}$

## Try It 7.37

Factor: ${x}^{2}+19x+60$.

## Try It 7.38

Factor: ${v}^{2}+23v+60$.

### Factor Trinomials of the Form *x*^{2} + *bx* + *c* with *b* Negative, *c* Positive

In the examples so far, 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.

Remember: To get a negative sum and a positive product, the numbers must both be negative.

Again, think about FOIL and where each term in the trinomial came from. Just as before,

- the first term, ${x}^{2}$, comes from the product of the two first terms in each binomial factor,
*x*and*y*; - the positive last term is the product of the two last terms
- the negative middle term is the sum of the outer and inner terms.

How do you get a *positive product* and a *negative sum*? With two negative numbers.

## Example 7.20

Factor: ${t}^{2}-11t+28$.

### Solution

Again, with the positive last term, 28, and the negative middle term, $\mathrm{-11}t$, we need two negative factors. Find two numbers that multiply 28 and add to $\mathrm{-11}$.

$\begin{array}{cccc}& & & {t}^{2}-11t+28\hfill \\ \\ \text{Write the factors as two binomials with first terms}\phantom{\rule{0.2em}{0ex}}t.\hfill & & & \left(t\phantom{\rule{1.5em}{0ex}}\right)\left(t\phantom{\rule{1.5em}{0ex}}\right)\hfill \end{array}$

Find two numbers that: multiply to 28 and add to $\mathrm{-11}$.

Factors of $28$ | Sum of factors |
---|---|

$\phantom{\rule{0.4em}{0ex}}\mathrm{-1},\mathrm{-28}$ | $\mathrm{-1}+\left(\mathrm{-28}\right)=\mathrm{-29}$ |

$\phantom{\rule{0.4em}{0ex}}\mathrm{-2},\mathrm{-14}$ | $\mathrm{-2}+\left(\mathrm{-14}\right)=\mathrm{-16}$ |

$\mathrm{-4},\mathrm{-7}$ | $\phantom{\rule{0.8em}{0ex}}\mathrm{-4}+\left(\mathrm{-7}\right)=\mathrm{-11}\text{*}$ |

$\begin{array}{cccc}\text{Use}\phantom{\rule{0.2em}{0ex}}\mathrm{-4},\mathrm{-7}\phantom{\rule{0.2em}{0ex}}\text{as the last terms of the binomials.}\hfill & & & \hfill \phantom{\rule{2.8em}{0ex}}\left(t-4\right)\left(t-7\right)\hfill \\ \text{Check.}\hfill & & & \\ \\ \\ \phantom{\rule{2.5em}{0ex}}(t-4)(t-7)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{t}^{2}-7t-4t+28\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{t}^{2}-11t+28\phantom{\rule{0.2em}{0ex}}\u2713\hfill & & & \end{array}$

## Try It 7.39

Factor: ${u}^{2}-9u+18$.

## Try It 7.40

Factor: ${y}^{2}-16y+63$.

### Factor Trinomials of the Form ${x}^{2}+bx+c$ with *c* Negative

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.

Remember: To get a negative product, the numbers must have different signs.

## Example 7.21

Factor: ${z}^{2}+4z-5$.

### Solution

To get a negative last term, multiply one positive and one negative. We need factors of $\mathrm{-5}$ that add to positive 4.

Factors of $\mathrm{-5}$ | Sum of factors |
---|---|

$\phantom{\rule{1.2em}{0ex}}1,\mathrm{-5}$ | $1+\left(\mathrm{-5}\right)=\mathrm{-4}$ |

$\mathrm{-1},5$ | $\phantom{\rule{0.5em}{0ex}}\mathrm{-1}+5=4\text{*}$ |

Notice: We listed both $1,\mathrm{-5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-1,5$ to make sure we got the sign of the middle term correct.

${z}^{2}+4z-5$ | |

Factors will be two binomials with first terms z. |
$\left(z\phantom{\rule{1.5em}{0ex}}\right)\left(z\phantom{\rule{1.5em}{0ex}}\right)$ |

Use −1, 5 as the last terms of the binomials. | $\left(z-1\right)\left(z+5\right)$ |

Check. $\begin{array}{}\\ \phantom{\rule{2.5em}{0ex}}\left(z-1\right)\left(z+5\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}{z}^{2}+5z-1z-5\hfill \\ \phantom{\rule{2.5em}{0ex}}{z}^{2}+4z-5\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ |

## Try It 7.41

Factor: ${h}^{2}+4h-12$.

## Try It 7.42

Factor: ${k}^{2}+k-20$.

Let’s make a minor change to the last trinomial and see what effect it has on the factors.

## Example 7.22

Factor: ${z}^{2}-4z-5$.

### Solution

This time, we need factors of $\mathrm{-5}$ that add to $\mathrm{-4}$.

Factors of $\mathrm{-5}$ | Sum of factors |
---|---|

$\phantom{\rule{1.2em}{0ex}}1,\mathrm{-5}$ | $\phantom{\rule{0.4em}{0ex}}1+(\mathrm{-5})=\mathrm{-4}\text{*}$ |

$\mathrm{-1},5$ | $\mathrm{-1}+5=4$ |

${z}^{2}-4z-5$ | |

Factors will be two binomials with first terms z. |
$\left(z\phantom{\rule{1.5em}{0ex}}\right)\left(z\phantom{\rule{1.5em}{0ex}}\right)$ |

Use 1, −5 as the last terms of the binomials. | $\left(z+1\right)\left(z-5\right)$ |

Check. $\begin{array}{}\\ \phantom{\rule{2.5em}{0ex}}\left(z+1\right)\left(z-5\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}{z}^{2}-5z+1z-5\hfill \\ \phantom{\rule{2.5em}{0ex}}{z}^{2}-4z-5\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ |

Notice that the factors of ${z}^{2}-4z-5$ are very similar to the factors of ${z}^{2}+4z-5$. It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.

## Try It 7.43

Factor: ${x}^{2}-4x-12$.

## Try It 7.44

Factor: ${y}^{2}-y-20$.

## Example 7.23

Factor: ${q}^{2}-2q-15$.

### Solution

$\begin{array}{cccc}& & & {q}^{2}-2q-15\hfill \\ \text{Factors will be two binomials with first terms}\phantom{\rule{0.2em}{0ex}}q.\hfill & & & \left(q\phantom{\rule{1.5em}{0ex}}\right)\left(q\phantom{\rule{1.5em}{0ex}}\right)\hfill \\ \text{You can use}\phantom{\rule{0.2em}{0ex}}3,\mathrm{-5}\phantom{\rule{0.2em}{0ex}}\text{as the last terms of the}\hfill & & & \left(q+3\right)\left(q-5\right)\hfill \\ \text{binomials.}\hfill & & & \end{array}$

Factors of $\mathrm{-15}$ | Sum of factors |
---|---|

$\phantom{\rule{0.6em}{0ex}}1,\mathrm{-15}$ | $1+(\mathrm{-15})=\mathrm{-14}$ |

$\mathrm{-1},15$ | $\phantom{\rule{0.7em}{0ex}}\mathrm{-1}+15=14$ |

$\phantom{\rule{0.6em}{0ex}}3,\mathrm{-5}$ | $\phantom{\rule{0.6em}{0ex}}3+(\mathrm{-5})=\mathrm{-2}\text{*}$ |

$\mathrm{-3},5$ | $\phantom{\rule{1.2em}{0ex}}\mathrm{-3}+5=2$ |

Check.

$\phantom{\rule{2.5em}{0ex}}\left(q+3\right)\left(q-5\right)$

$\phantom{\rule{2.5em}{0ex}}{q}^{2}-5q+3q-15$

$\phantom{\rule{2.5em}{0ex}}{q}^{2}-2q-15\phantom{\rule{0.2em}{0ex}}\u2713$

## Try It 7.45

Factor: ${r}^{2}-3r-40$.

## Try It 7.46

Factor: ${s}^{2}-3s-10$.

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.

## Example 7.24

Factor: ${y}^{2}-6y+15$.

### Solution

${y}^{2}-6y+15$ | |

Factors will be two binomials with first terms y. |
$\left(y\phantom{\rule{1.5em}{0ex}}\right)\left(y\phantom{\rule{1.5em}{0ex}}\right)$ |

Factors of 15 | Sum of factors |
---|---|

$\mathrm{-1},\mathrm{-15}$ | $\mathrm{-1}+(\mathrm{-15})=\mathrm{-16}$ |

$\mathrm{-3},\mathrm{-5}$ | $\phantom{\rule{0.4em}{0ex}}\mathrm{-3}+(\mathrm{-5})=\mathrm{-8}$ |

As shown in the table, none of the factors add to $\mathrm{-6}$; therefore, the expression is prime.

## Try It 7.47

Factor: ${m}^{2}+4m+18$.

## Try It 7.48

Factor: ${n}^{2}-10n+12$.

## Example 7.25

Factor: $2x+{x}^{2}-48$.

### Solution

$2x+{x}^{2}-48$ | |

First we put the terms in decreasing degree order. | ${x}^{2}+2x-48$ |

Factors will be two binomials with first terms x. |
$\left(x\phantom{\rule{1.5em}{0ex}}\right)\left(x\phantom{\rule{1.5em}{0ex}}\right)$ |

As shown in the table, you can use $\mathrm{-6},8$ as the last terms of the binomials.

Factors of $\mathrm{-48}$ | Sum of factors |
---|---|

$\mathrm{-1},48$ | $\mathrm{-1}+48=47$ |

$\mathrm{-2},24$ $\mathrm{-3},16$ $\mathrm{-4},12$ $\mathrm{-6},8$ |
$\mathrm{-2}+24=22$ $\mathrm{-3}+16=13$ $\mathrm{-4}+12=8$ $\phantom{\rule{0.5em}{0ex}}\mathrm{-6}+8=2$ |

Check.

$\phantom{\rule{2.5em}{0ex}}(x-6)(x+8)$

$\phantom{\rule{2.5em}{0ex}}{x}^{2}-6x+8x-48$

$\phantom{\rule{2.5em}{0ex}}{x}^{2}+2x-48\phantom{\rule{0.2em}{0ex}}\u2713$

## Try It 7.49

Factor: $9m+{m}^{2}+18$.

## Try It 7.50

Factor: $\mathrm{-7}n+12+{n}^{2}$.

Let’s summarize the method we just developed to factor trinomials of the form ${x}^{2}+bx+c$.

## How To

### Factor trinomials.

When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.

**When c is positive, m and n have the same sign.**

**When c is negative, m and n have opposite signs.**

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*.

## Factor Trinomials of the Form *x*^{2} + *bxy* + *cy*^{2}

Sometimes you’ll need to factor trinomials of the form ${x}^{2}+bxy+c{y}^{2}$ with two variables, such as ${x}^{2}+12xy+36{y}^{2}.$ The first term, ${x}^{2}$, is the product of the first terms of the binomial factors, $x\xb7x$. The ${y}^{2}$ 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 the previous objective.

## Example 7.26

Factor: ${x}^{2}+12xy+36{y}^{2}$.

### Solution

${x}^{2}+12xy+36{y}^{2}$ | |

Note that the first terms are x, last terms contain y. |
$\left(x\_y\right)\left(x\_y\right)$ |

Find the numbers that multiply to 36 and add to 12.

Factors of $36$ | Sum of factors |
---|---|

1, 36 | $1+36=37$ |

2, 18 | $2+18=20$ |

3, 12 | $3+12=15$ |

4, 9 | $4+9=13$ |

6, 6 | $6+6=12\text{*}$ |

Use 6 and 6 as the coefficients of the last terms. | $\left(x+6y\right)\left(x+6y\right)$ |

Check your answer. $\begin{array}{}\\ \\ \phantom{\rule{3.5em}{0ex}}\left(x+6y\right)\left(x+6y\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}{x}^{2}+6xy+6xy+36{y}^{2}\hfill \\ \phantom{\rule{2.5em}{0ex}}{x}^{2}+12xy+36{y}^{2}\u2713\hfill \end{array}$ |

## Try It 7.51

Factor: ${u}^{2}+11uv+28{v}^{2}$.

## Try It 7.52

Factor: ${x}^{2}+13xy+42{y}^{2}$.

## Example 7.27

Factor: ${r}^{2}-8rs-9{s}^{2}$.

### Solution

We need $r$ in the first term of each binomial and $s$ in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

$\begin{array}{cccc}& & & {r}^{2}-8rs-9{s}^{2}\hfill \\ \text{Note that the first terms are}\phantom{\rule{0.2em}{0ex}}r,\phantom{\rule{0.2em}{0ex}}\text{last terms contain}\phantom{\rule{0.2em}{0ex}}s.\hfill & & & \left(r\_s\right)\left(r\_s\right)\hfill \end{array}$

Find the numbers that multiply to $\mathrm{-9}$ and add to $\mathrm{-8}$.

Factors of $\mathrm{-9}$ | Sum of factors |
---|---|

$\phantom{\rule{1em}{0ex}}1,\mathrm{-9}$ | $\phantom{\rule{0.7em}{0ex}}1+(\mathrm{-9})=\mathrm{-8}\text{*}$ |

$\mathrm{-1},9$ | $\phantom{\rule{0.4em}{0ex}}\mathrm{-1}+9=8$ |

$\phantom{\rule{1em}{0ex}}3,\mathrm{-3}$ | $3+(\mathrm{-3})=0$ |

$\begin{array}{cccc}\text{Use}\phantom{\rule{0.2em}{0ex}}1,\mathrm{-9}\phantom{\rule{0.2em}{0ex}}\text{as coefficients of the last terms.}\hfill & & & \hfill \phantom{\rule{3.4em}{0ex}}\left(r+s\right)\left(r-9s\right)\hfill \\ \text{Check your answer.}\hfill & & & \\ \\ \\ \phantom{\rule{2.5em}{0ex}}\left(r-9s\right)\left(r+s\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{r}^{2}+rs-9rs-9{s}^{2}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{r}^{2}-8rs-9{s}^{2}\phantom{\rule{0.2em}{0ex}}\u2713\hfill & & & \end{array}$

## Try It 7.53

Factor: ${a}^{2}-11ab+10{b}^{2}$.

## Try It 7.54

Factor: ${m}^{2}-13mn+12{n}^{2}$.

## Example 7.28

Factor: ${u}^{2}-9uv-12{v}^{2}$.

### Solution

We need *u* in the first term of each binomial and $v$ in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

${u}^{2}-9uv-12{v}^{2}$ | |

Note that the first terms are u, last terms contain v. |
$\left(u\_v\right)\left(u\_v\right)$ |

Find the numbers that multiply to $\mathrm{-12}$ and add to $\mathrm{-9}$.

Factors of $\mathrm{-12}$ | Sum of factors |
---|---|

$\phantom{\rule{2em}{0ex}}1,\mathrm{-12}$ | $1+\left(\mathrm{-12}\right)=\mathrm{-11}$ |

$\phantom{\rule{0.5em}{0ex}}\mathrm{-1},12$ | $\mathrm{-1}+12=11$ |

$\phantom{\rule{1.2em}{0ex}}2,\mathrm{-6}$ | $2+\left(\mathrm{-6}\right)=\mathrm{-4}$ |

$\mathrm{-2},6$ | $\mathrm{-2}+6=4$ |

$\phantom{\rule{1.2em}{0ex}}3,\mathrm{-4}$ | $3+\left(\mathrm{-4}\right)=\mathrm{-1}$ |

$\mathrm{-3},4$ | $\mathrm{-3}+4=1$ |

Note there are no factor pairs that give us $\mathrm{-9}$ as a sum. The trinomial is prime.

## Try It 7.55

Factor: ${x}^{2}-7xy-10{y}^{2}$.

## Try It 7.56

Factor: ${p}^{2}+15pq+20{q}^{2}$.

## Section 7.2 Exercises

### Practice Makes Perfect

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

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

${y}^{2}+8y+7$

${b}^{2}+14b+13$

${m}^{2}+7m+12$

${w}^{2}+10w+21$

${b}^{2}+14b+48$

${u}^{2}+101u+100$

${q}^{2}-13q+36$

${m}^{2}-13m+30$

${y}^{2}-5y+6$

${n}^{2}+6n-7$

${v}^{2}-2v-3$

${r}^{2}-2r-8$

${b}^{2}-13b-30$

${t}^{2}-3t-54$

${x}^{2}-3x-9$

$7x+{x}^{2}+6$

$\mathrm{-11}-10x+{x}^{2}$

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

In the following exercises, factor each trinomial of the form ${x}^{2}+bxy+c{y}^{2}$. If the trinomial cannot be factored, answer “Prime.”

${m}^{2}+6mn+5{n}^{2}$

${u}^{2}+10uv+24{v}^{2}$

${p}^{2}-16pq+63{q}^{2}$

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

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

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

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

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

**Mixed Practice**

In the following exercises, factor each expression.

${w}^{2}+4w-32$

${y}^{2}+41y+40$

${x}^{2}-16xy+64{y}^{2}$

${m}^{2}+29m+120$

${z}^{2}-3z+28$

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

${m}^{2}-31mn+30{n}^{2}$

${r}^{2}+11rs+36{s}^{2}$

### Everyday Math

**Consecutive integers** Deirdre is thinking of two consecutive integers whose product is 56. The trinomial ${x}^{2}+x-56$ describes how these numbers are related. Factor the trinomial.

**Consecutive integers** Deshawn is thinking of two consecutive integers whose product is 182. The trinomial ${x}^{2}+x-182$ describes how these numbers are related. Factor the trinomial.

### Writing Exercises

Many trinomials of the form ${x}^{2}+bx+c$ factor into the product of two binomials $(x+m)(x+n)$. Explain how you find the values of *m* and *n*.

How do you determine whether to use plus or minus signs in the binomial factors of a trinomial of the form ${x}^{2}+bx+c$ where $b$ and $c$ may be positive or negative numbers?

Will factored ${x}^{2}-x-20$ as $(x+5)(x-4)$. Bill factored it as $(x+4)(x-5)$. Phil factored it as $(x-5)(x-4)$. Who is correct? Explain why the other two are wrong.

Look at Example 7.19, where we factored ${y}^{2}+17y+60$. We made a table listing all pairs of factors of 60 and their sums. Do you find this kind of table helpful? Why or why not?

### 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 goals?