10.8 Factoring Quadratic Trinomials

Consider the trinomial

$x^{2} + 10 x + 16$

Can we find two binomial factors,

$(x + a) (x + b)$

whose product is the given trinomial? The product of the binomials is

$(x + a) (x + b) = x^{2} + (a + b) x + a b$

Thus, we are looking for two numbers, $a$ and $b$ , that satisfy

$(x + a) (x + b) = x^{2} + (a + b) x + a b = x^{2} + 10 x + 16$

By comparing the coefficients of the terms in the two trinomials, we see that $a + b = 10$ and $a b equals 16$ . That is, the sum of the two numbers is the coefficient of the linear term, $10$ , and their product is the constant term, $16$ .

To find the numbers, we list all the possible integer factorizations of $16$ :

$1 \cdot 16, 2 \cdot 8, and 4 \cdot 4$

We see that only one combination gives the correct linear term: $8$ and $2$ . These are the numbers $a$ and $b$ , so

$x^{2} + 10 x + 16 = (x + 8) (x + 2)$

In Example we factor quadratic trinomials in which one or more of the coefficients is negative.

Example 1

Factor.

$x^{2} - 7 x + 12$
$x^{2} - x - 12$

Find two numbers whose product is $12$ and whose sum is $minus 7$ . Because the product is positive and the sum is negative, the two numbers must both be negative. The possible factors of $12$ are $minus 1$ and $minus 12$ , $minus 2$ and $minus 6$ , or $minus 3$ and $minus 4$ . Only $minus 4$ and $minus 3$ have the correct sum, $minus 7$ . Hence,
$x^{2} - 7 x + 12 = (x - 4) (x - 3)$
Find two numbers whose product is $minus 12$ and whose sum is $minus 1$ . Because the product is negative, the two numbers must be of opposite sign and their sum must be $minus 1$ . By listing the possible factors of $minus 12$ , we find that the two numbers are $minus 4$ and $3$ . Hence,
$x^{2} - x - 12 = (x - 4) (x + 3)$

If the coefficient of the quadratic term is not $1$ , we must also consider its factors.

Example 2

Factor $8 x^{2} - 9 - 21 x$

Write the trinomial in decreasing powers of $x$ .
$8 x^{2} - 21 x - 9$
List the possible factors for the quadratic term.
$open parenthesis 8 x close parenthesis open parenthesis x close parenthesis; open parenthesis 4 x close parenthesis open parenthesis 2 x close parenthesis$
Consider possible factors for the constant term: $9$ may be factored as $9 times 1$ or as $3 times 3$ . Form all possible pairs of binomial factor using these factorizations.
Select the combinations of the products $①$ and $②$ whose sum or difference could be the linear term, $minus 21 x$ .
$open parenthesis 8 x 3 close parenthesis open parenthesis x 3 close parenthesis$
Insert the proper signs:
$(8 x + 3) (x - 3)$

With practice, you can usually factor trinomials of the form $A x^{2} + B x + C$ mentally. The following observations may help.

Note

If $A$ , $B$ and $C$ are all positive, both signs in the factored form are positive. For example, as a first step in factoring $6 x^{2} + 11 x + 4$ , we could write
$(+) (+)$
If $A$ and $C$ are positive and $B$ is negative, both signs in the factored form are negative. Thus as a first step in factoring $6 x^{2} - 11 x + 4$ , we could write
$(-) (-)$
If $C$ is negative, the signs in the factored form are opposite. Thus as a first step in factoring $6 x^{2} - 5 x - 4$ , we could write
$(+) (-) or (-) (+)$

Example 3

$\begin{aligned} 6 x^{2} + 5 x + 1 & = (+) (+) \\ = (3 x + 1) (2 x + 1) \end{aligned}$
$\begin{aligned} 6 x^{2} - 5 x + 1 & = (-) (-) \\ = (3 x - 1) (2 x - 1) \end{aligned}$
$\begin{aligned} 6 x^{2} - x - 1 & = (+) (-) \\ = (3 x + 1) (2 x - 1) \end{aligned}$
$\begin{aligned} 6 x^{2} - x y - y^{2} & = (+) (-) \\ = (3 x + y) (2 x - y) \end{aligned}$

Special Products and Factors

The products below are special cases of the multiplication of binomials. They occur so often that you should learn to recognize them on sight.

Caution

Notice that in (I) $(a + b)^{2} \neq a^{2} + b^{2}$ , and that in (II) $(a - b)^{2} \neq a^{2} - b^{2}$ . For example,

$\begin{aligned} (x + 4)^{2} & \neq x^{2} + 16, & instead & (x + 4)^{2} & = x^{2} + 8 x + 16 \\ (t - 5)^{2} & \neq t^{2} - 16, & instead & (t - 5)^{2} & = t^{2} - 10 t + 25 \end{aligned}$

Example 4

$\begin{aligned} 3 (x + 4)^{2} & = 3 (x^{2} + 2 \cdot 4 x + 4^{2}) \\ = 3 x^{2} + 24 x + 48 \end{aligned}$
$\begin{aligned} (y + 5) (y - 5) & = y^{2} - 5^{2} \\ = y^{2} - 25 \end{aligned}$
$\begin{aligned} (3 x - 2 y)^{2} & = (3 x)^{2} - 2 (3 x) (2 y) + (2 y)^{2} \\ = 9 x^{2} - 12 x y + 4 y^{2} \end{aligned}$

Each of the formulas for special products, when viewed from right to left, also represents a special case of factoring quadratic polynomials.

The trinomials in (I) and (II) are sometimes called perfect-square trinomials because they are squares of binomials. Note that the sum of two squares, $a^{2} + b^{2}$ , cannot be factored.

Example 5

Factor.

$x^{2} + 8 x + 16$
$y^{2} - 10 y + 25$
$4 a^{2} - 12 a b + 9 b^{2}$
$25 m^{2} n^{2} + 20 m n + 4$

Because $16$ is equal to $4 squared$ and $8$ is equal to $2 times 4$ ,
$\begin{aligned} x^{2} + 8 x + 16 & = x^{2} - 2 \cdot 4 x + 4^{2} \\ = (x + 4)^{2} \end{aligned}$
Because $25 equals 5 squared$ and $10 = 2 \cdot 5$ ,
$\begin{aligned} y^{2} - 10 y + 25 & = y^{2} - 2 \cdot 5 y + 5^{2} \\ = (y - 5)^{2} \end{aligned}$
Because $4 a squared equals open parenthesis 2 a close parenthesis squared$ , $9 b squared equals open parenthesis 3 b close parenthesis squared$ , and $2 a b equals 2 open parenthesis 2 a close parenthesis open parenthesis 3 b close parenthesis$ ,
$\begin{aligned} 4 a^{2} - 12 a b + 9 b^{2} & = (2 a)^{2} - 2 (2 a) (3 b) + (3 b)^{2} \\ = (2 a - 3 b)^{2} \end{aligned}$
Because $25 m squared n squared equals open parenthesis 5 m n close parenthesis squared$ , $4 equals 2 squared$ , and $20 m n equals 2 open parenthesis 5 m n close parenthesis open parenthesis 2 close parenthesis$ ,
$\begin{aligned} 25 m^{2} n^{2} + 20 m n + 4 & = (5 m n)^{2} + 2 (5 m n) (2) + 2^{2} \\ = (5 m n + 2)^{2} \end{aligned}$

Binomials of the form $a^{2} - b^{2}$ are often called the difference of two squares.

Example 6

Factor if possible.

$x squared minus 81$
$4 x squared minus 9 y squared$
$x squared plus 81$

The expression $x squared minus 81$ is the difference of two squares, $x squared minus 9 squared$ , and thus can be factored according to Special Factorization (III) above.
$\begin{aligned} x^{2} - 81 & = x^{2} - 9^{2} \\ = (x + 9) (x - 9) \end{aligned}$
Because $4 x squared minus 9 y squared$ can be written as $open parenthesis 2 x close parenthesis squared minus open parenthesis 3 y close parenthesis squared$ ,
$\begin{aligned} 4 x^{2} - 9 y^{2} & = (2 x^{2}) - (3 y)^{2} \\ = (2 x + 3 y) (2 x - 3 y) \end{aligned}$
The expression $x squared plus 81$ , or $x^{2} + 0 x + 81$ , is not factorable, because no two real numbers have a product of $81$ and a sum of $0$ .

The factors $x plus 9$ and $x minus 9$ in Examplea are called conjugates of each other. In general, any binomials of the form $a + b$ and $a - b$ are called a conjugate pair.

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Perfect-square trinomial
Difference of squares
Conjugate

SKILLS

Practice each skill in the exercises listed.

Factor quadratic trinomials: #1–36
Expand special products: #37–48
Factor special quadratic expressions: #49–68

Exercises A.8

For Problems 1-36, factor completely.

$x^{2} + 5 x + 6$

$(x + 2) (x + 3)$

$x^{2} + 5 x + 4$

$y^{2} - 7 y + 12$

$(y - 3) (y - 4)$

$y^{2} - 7 y + 10$

$x^{2} - 6 - x$

$(x - 3) (x + 2)$

$x^{2} - 15 - 2 x$

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

$(2 x - 1) (x + 2)$

$3 x^{2} - 7 x + 2$

$7 x + 4 x^{2} - 2$

$(4 x - 1) (x + 2)$

$1 - 5 x + 6 x^{2}$

$9 y^{2} - 21 y - 8$

$(3 y + 1) (3 y - 8)$

$10 y^{2} - 3 y - 18$

$10 u^{2} - 3 - u$

$(2 u + 1) (5 u - 3)$

$8 u^{2} - 3 + 5 u$

$21 x^{2} - 43 x - 14$

$(3 x - 7) (7 x + 2)$

$24 x^{2} - 29 x + 5$

$5 a + 72 a^{2} - 25$

$(9 a + 4) (8 a - 3)$

$- 30 a + 72 a^{2} - 25$

$12 - 53 x + 30 x^{2}$

$(2 x - 3) (15 x - 4)$

$39 x + 80 x^{2} - 20$

$- 30 t - 44 + 54 t^{2}$

$2 (3 t + 2) (9 t - 11)$

$48 t^{2} - 122 t + 39$

$3 x^{2} - 7 a x + 2 a^{2}$

$(x - 2 a) (3 x - a)$

$9 x^{2} + 9 a x - 10 a^{2}$

$15 x^{2} - 4 x y - 4 y^{2}$

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

$12 x^{2} + 7 x y - 12 y^{2}$

$18 u^{2} + 20 v^{2} - 39 u v$

$(3 u - 4 v) (6 u - 5 v)$

$24 u^{2} - 20 v^{2} + 17 u v$

$12 a^{2} - 14 b^{2} - 13 a b$

$(3 a + 2 b) (4 a - 7 b)$

$24 a^{2} - 15 b^{2} - 2 a b$

$10 a^{2} b^{2} - 19 a b + 6$

$(5 a b - 2) (2 a b - 3)$

$12 a^{2} b^{2} - a b - 20$

$56 x^{2} y^{2} - 2 x y - 4$

$2 (4 x y + 1) (7 x y - 2)$

$54 x^{2} y^{2} + 3 x y - 2$

$22 a^{2} z^{2} - 21 - 19 a z$

$(2 a z - 3) (11 a z + 7)$

$26 a^{2} z^{2} - 24 + 23 a z$

For Problems 37-48, write the expression as a polynomial and simplify.

$open parenthesis x plus 3 close parenthesis squared$

$x^{2} + 6 x + 9$

$open parenthesis y minus 4 close parenthesis squared$

$open parenthesis 2 y minus 5 close parenthesis squared$

$4 y^{2} - 20 y + 25$

$open parenthesis 3 x plus 2 close parenthesis squared$

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

$x squared minus 9$

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

$(3 t - 4 s) (3 t + 4 s)$

$9 t squared minus 16 s squared$

$(2 x + a) (2 x - a)$

$(5 a - 2) (5 a - 2)$

$25 a^{2} - 20 a b + 4 b^{2}$

$(4 u + 5 v) (4 u + 5 v)$

$(8 x z + 3) (8 x z + 3)$

$64 x^{2} z^{2} + 48 x z + 9$

$(7 y z - 2) (7 y z - 2)$

For Problems 49-68, factor completely.

$x squared minus 25$

$(x + 5) (x - 5)$

$x squared minus 36$

$x^{2} - 24 x + 144$

$open parenthesis x minus 12 close parenthesis squared$

$x^{2} + 26 x + 169$

$x squared minus 4 y squared$

$(x + 2 y) (x - 2 y)$

$9 x squared minus y squared$

$4 x^{2} + 12 x + 9$

$open parenthesis 2 x plus 3 close parenthesis squared$

$4 y^{2} + 4 y + 1$

$9 u^{2} - 30 u v + 25 v^{2}$

$open parenthesis 3 u minus 5 v close parenthesis squared$

$16 s^{2} - 56 s t + 49 t^{2}$

$4 a squared minus 25 b squared$

$(2 a + 5 b) (2 a - 5 b)$

$16 a squared minus 9 b squared$

$x squared y squared minus 81$

$(x y + 9) (x y - 9)$

$x squared y squared minus 64$

$9 x^{2} y^{2} + 6 x y + 1$

$open parenthesis 3 x y plus 1 close parenthesis squared$

$4 x^{2} y^{2} + 12 x y + 9$

$16 squared y squared minus 1$

$(4 x y - 1) (4 x y + 1)$

$64 x squared y squared minus 1$

$(x + 2)^{2} - y^{2}$

$(x + 2 - y) (x + 2 + y)$

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

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.