7.2 Factor Trinomials of the Form x²+bx+c

Factor Trinomials of the Form x² + 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.

This figure shows the steps of multiplying the factors (x + 2) times (x + 3). The multiplying is completed using FOIL to demonstrate. The first term is x squared and is below F. The second term is 3 x below “O”. The third term is 2 x below “I”. The fourth term is 6 below L. The simplified product is then given as x 2 plus 5 x + 6.

To factor the trinomial means to start with the product, $x^{2} + 5 x + 6$ , and end with the factors, $(x + 2) (x + 3)$ . 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.

$\begin{matrix} x^{2} + 5 x + 6 \\ (x) (x) \end{matrix}$

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.

**Table 7.1**
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 $x^{2} + 5 x + 6$ . They are $(x + 2) (x + 3)$ .

$\begin{matrix} x^{2} + 5 x + 6 & product \\ (x + 2) (x + 3) & factors \end{matrix}$

You should check this by multiplying.

Looking back, we started with $x^{2} + 5 x + 6$ , which is of the form $x^{2} + b x + c$ , where $b equals 5$ and $c equals 6$ . We factored it into two binomials of the form $(x + m) and (x + n)$ .

$\begin{matrix} x^{2} + 5 x + 6 & x^{2} + b x + c \\ (x + 2) (x + 3) & (x + m) (x + n) \end{matrix}$

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

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

Example 2

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

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

$\begin{matrix} u^{2} + 11 u + 24 \\ Write the factors as two binomials with first terms u . & (u) (u) \end{matrix}$

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

Factors of $24$	Sum of factors
$1, 24$	$1 + 24 = 25$
$2, 12$	$2 + 12 = 14$
$3, 8$	$3 + 8 = 11 *$
$4, 6$	$4 + 6 = 10$

$\begin{matrix} Use 3 and 8 as the last terms of the binomials. & (u + 3) (u + 8) \\ Check. \\ (u + 3) (u + 8) \\ u^{2} + 3 u + 8 u + 24 \\ u^{2} + 11 u + 24 ✓ \end{matrix}$

Example 3

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

$\begin{matrix} y^{2} + 17 y + 60 \\ Write the factors as two binomials with first terms y . & (y) (y) \end{matrix}$

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

Factors of $60$	Sum of factors
$1, 60$	$1 + 60 = 61$
$2, 30$	$2 + 30 = 32$
$3, 20$	$3 + 20 = 23$
$4, 15$	$4 + 15 = 19$
$5, 12$	$5 + 12 = 17 *$
$6, 10$	$6 + 10 = 16$

$\begin{matrix} Use 5 and 12 as the last terms. & (y + 5) (y + 12) \\ Check. \\ (y + 5) (y + 12) \\ (y^{2} + 12 y + 5 y + 60) \\ (y^{2} + 17 y + 60) ✓ \end{matrix}$

Factor Trinomials of the Form x² + 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 4

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

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

$\begin{matrix} t^{2} - 11 t + 28 \\ Write the factors as two binomials with first terms t . & (t) (t) \end{matrix}$

Find two numbers that: multiply to 28 and add to $−11$ .

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

$\begin{matrix} Use −4, −7 as the last terms of the binomials. & (t - 4) (t - 7) \\ Check. \\ (t - 4) (t - 7) \\ t^{2} - 7 t - 4 t + 28 \\ t^{2} - 11 t + 28 ✓ \end{matrix}$

Factor Trinomials of the Form $x^{2} + b x + c$ 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 5

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

To get a negative last term, multiply one positive and one negative. We need factors of $−5$ that add to positive 4.

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

Notice: We listed both $1, −5 and minus 1, 5$ to make sure we got the sign of the middle term correct.

	$z^{2} + 4 z - 5$
Factors will be two binomials with first terms z.	$(z) (z)$
Use −1, 5 as the last terms of the binomials.	$(z - 1) (z + 5)$
Check. $\begin{matrix} (z - 1) (z + 5) \\ z^{2} + 5 z - 1 z - 5 \\ z^{2} + 4 z - 5 ✓ \end{matrix}$

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

Example 6

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

This time, we need factors of $−5$ that add to $−4$ .

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

	$z^{2} - 4 z - 5$
Factors will be two binomials with first terms z.	$(z) (z)$
Use 1, −5 as the last terms of the binomials.	$(z + 1) (z - 5)$
Check. $\begin{matrix} (z + 1) (z - 5) \\ z^{2} - 5 z + 1 z - 5 \\ z^{2} - 4 z - 5 ✓ \end{matrix}$

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

Example 7

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

$\begin{matrix} q^{2} - 2 q - 15 \\ Factors will be two binomials with first terms q . & (q) (q) \\ You can use 3, −5 as the last terms of the & (q + 3) (q - 5) \\ binomials. \end{matrix}$

Factors of $−15$	Sum of factors
$1, −15$	$1 + (−15) = −14$
$−1, 15$	$−1 + 15 = 14$
$3, −5$	$3 + (−5) = −2 *$
$−3, 5$	$−3 + 5 = 2$

Check.

$(q + 3) (q - 5)$

$q^{2} - 5 q + 3 q - 15$

$q^{2} - 2 q - 15 ✓$

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 8

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

	$y^{2} - 6 y + 15$
Factors will be two binomials with first terms y.	$(y) (y)$

Factors of 15	Sum of factors
$−1, −15$	$−1 + (−15) = −16$
$−3, −5$	$−3 + (−5) = −8$

As shown in the table, none of the factors add to $−6$ ; therefore, the expression is prime.

Example 9

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

	$2 x + x^{2} - 48$
First we put the terms in decreasing degree order.	$x^{2} + 2 x - 48$
Factors will be two binomials with first terms x.	$(x) (x)$

As shown in the table, you can use $−6, 8$ as the last terms of the binomials.

$(x - 6) (x + 8)$

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

Check.

$(x - 6) (x + 8)$

$x^{2} - 6 x + 8 x - 48$

$x^{2} + 2 x - 48 ✓$

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

Factor Trinomials of the Form x² + bxy + cy²

Sometimes you’ll need to factor trinomials of the form $x^{2} + b x y + c y^{2}$ with two variables, such as $x^{2} + 12 x y + 36 y^{2} .$ The first term, $x^{2}$ , is the product of the first terms of the binomial factors, $x \cdot x$ . 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 10

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

	$x^{2} + 12 x y + 36 y^{2}$
Note that the first terms are x, last terms contain y.	$open parenthesis x _ y close parenthesis open parenthesis x _ y close parenthesis$

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	$6 + 6 = 12 *$

Use 6 and 6 as the coefficients of the last terms.	$(x + 6 y) (x + 6 y)$
Check your answer. $\begin{matrix} (x + 6 y) (x + 6 y) \\ x^{2} + 6 x y + 6 x y + 36 y^{2} \\ x^{2} + 12 x y + 36 y^{2} ✓ \end{matrix}$

Example 11

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

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.

$\begin{matrix} r^{2} - 8 r s - 9 s^{2} \\ Note that the first terms are r, last terms contain s . & (r_s) (r_s) \end{matrix}$

Find the numbers that multiply to $−9$ and add to $−8$ .

Factors of $−9$	Sum of factors
$1, −9$	$1 + (−9) = −8 *$
$−1, 9$	$−1 + 9 = 8$
$3, −3$	$3 + (−3) = 0$

$\begin{matrix} Use 1, −9 as coefficients of the last terms. & (r + s) (r - 9 s) \\ Check your answer. \\ (r - 9 s) (r + s) \\ r^{2} + r s - 9 r s - 9 s^{2} \\ r^{2} - 8 r s - 9 s^{2} ✓ \end{matrix}$

Example 12

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

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.

	$u^{2} - 9 u v - 12 v^{2}$
Note that the first terms are u, last terms contain v.	$open parenthesis u _ v close parenthesis open parenthesis u _ v close parenthesis$

Find the numbers that multiply to $−12$ and add to $−9$ .

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

Note there are no factor pairs that give us $−9$ as a sum. The trinomial is prime.

Key Concepts

Factor trinomials of the form $x^{2} + b x + c$
1. Write the factors as two binomials with first terms x: $(x) (x)$ .
2. Find two numbers m and n that
  Multiply to c, $m \cdot n = c$
  Add to b, $m + n = b$
3. Use m and n as the last terms of the factors: $(x + m) (x + n)$ .
4. Check by multiplying the factors.

Practice Makes Perfect

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

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

$x^{2} + 4 x + 3$

$(x + 1) (x + 3)$

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

$m^{2} + 12 m + 11$

$(m + 1) (m + 11)$

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

$a^{2} + 9 a + 20$

$(a + 4) (a + 5)$

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

$p^{2} + 11 p + 30$

$(p + 5) (p + 6)$

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

$n^{2} + 19 n + 48$

$(n + 3) (n + 16)$

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

$a^{2} + 25 a + 100$

$(a + 5) (a + 20)$

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

$x^{2} - 8 x + 12$

$(x - 2) (x - 6)$

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

$y^{2} - 18 y + 45$

$(y - 3) (y - 15)$

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

$x^{2} - 8 x + 7$

$(x - 1) (x - 7)$

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

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

$(p - 1) (p + 6)$

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

$y^{2} - 6 y - 7$

$(y + 1) (y - 7)$

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

$x^{2} - x - 12$

$(x - 4) (x + 3)$

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

$a^{2} - 3 a - 28$

$(a - 7) (a + 4)$

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

$w^{2} - 5 w - 36$

$(w - 9) (w + 4)$

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

$x^{2} + x + 5$

prime

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

$8 - 6 x + x^{2}$

$(x - 4) (x - 2)$

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

$x^{2} - 12 - 11 x$

$(x - 12) (x + 1)$

$−11 - 10 x + x^{2}$

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

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

$p^{2} + 3 p q + 2 q^{2}$

$(p + q) (p + 2 q)$

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

$r^{2} + 15 r s + 36 s^{2}$

$(r + 3 s) (r + 12 s)$

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

$m^{2} - 12 m n + 20 n^{2}$

$(m - 2 n) (m - 10 n)$

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

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

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

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

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

$(m + n) (m - 65 n)$

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

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

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

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

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

prime

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

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

prime

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

Mixed Practice

In the following exercises, factor each expression.

$u^{2} - 12 u + 36$

$(u - 6) (u - 6)$

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

$x^{2} - 14 x - 32$

$(x + 2) (x - 16)$

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

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

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

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

$k^{2} + 34 k + 120$

$(k + 4) (k + 30)$

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

$y^{2} + 10 y + 15$

prime

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

$m^{2} + m n - 56 n^{2}$

$(m + 8 n) (m - 7 n)$

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

$u^{2} - 17 u v + 30 v^{2}$

$(u - 15 v) (u - 2 v)$

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

$c^{2} - 8 c d + 26 d^{2}$

prime

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

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.

$(x + 8) (x - 7)$

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

Answers may vary

How do you determine whether to use plus or minus signs in the binomial factors of a trinomial of the form $x^{2} + b x + c$ 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.

Answers may vary

Look at Example 3, where we factored $y^{2} + 17 y + 60$ . 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.

This table has the following statements all to be preceded by “I can…”. The first is “factor trinomials of the form x ^ 2 +b x + c”. The second is “factor trinomials of the form x^2 + b x y + c y ^ 2”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

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

7.2 Factor Trinomials of the Form x2+bx+c

Factor Trinomials of the Form x2 + bx + c

Factor Trinomials of the Form x2 + bx + c with b Negative, c Positive

Factor Trinomials of the Form x2+bx+c with c Negative

Factor Trinomials of the Form x2 + bxy + cy2

Key Concepts

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

7.2 Factor Trinomials of the Form x²+bx+c

Factor Trinomials of the Form x² + bx + c

Factor Trinomials of the Form x² + bx + c with b Negative, c Positive

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

Factor Trinomials of the Form x² + bxy + cy²