Consider the trinomial

\begin{equation*} x^2 + 10x + 16 \end{equation*}

Can we find two binomial factors,

\begin{equation*} (x + a)(x + b) \end{equation*}

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

\begin{equation*} (x + a)(x + b) = x^2 + (a + b)x + ab \end{equation*}

Thus, we are looking for two numbers, $a$ and $b\text{,}$ that satisfy

\begin{equation*} (x + a)(x + b) = x^2 + (\alert{a + b})x + \blert{ab}= x^2 + \alert{10}x + \blert{16} \end{equation*}

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

To find the numbers, we list all the possible integer factorizations of $16\text{:}$

\begin{equation*} 1 \cdot 16, ~~~2 \cdot 8, ~~~\text{ and }~~~ 4 \cdot 4 \end{equation*}

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

\begin{equation*} x^2 + 10x + 16 = (x + 8)(x + 2) \end{equation*}

In Example A.57 we factor quadratic trinomials in which one or more of the coefficients is negative.

###### ExampleA.57

Factor.

1. $x^2-7x+12$

2. $x^2-x-12$

Solution
1. Find two numbers whose product is $12$ and whose sum is $-7\text{.}$ Because the product is positive and the sum is negative, the two numbers must both be negative. The possible factors of $12$ are $-1$ and $-12\text{,}$ $-2$ and $-6\text{,}$ or $-3$ and $-4\text{.}$ Only $-4$ and $-3$ have the correct sum, $-7\text{.}$ Hence,

\begin{equation*} x^2 - 7x + 12 = (x - 4) (x - 3) \end{equation*}
2. Find two numbers whose product is $-12$ and whose sum is $-1\text{.}$ Because the product is negative, the two numbers must be of opposite sign and their sum must be $-1\text{.}$ By listing the possible factors of $-12\text{,}$ we find that the two numbers are $-4$ and $3\text{.}$ Hence,

\begin{equation*} x^2 - x - 12 = (x - 4) (x + 3) \end{equation*}

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

###### ExampleA.58

Factor $~8x^2 - 9 - 21x$

Solution
1. Write the trinomial in decreasing powers of $x\text{.}$

\begin{equation*} 8x^2-21x-9 \end{equation*}
2. List the possible factors for the quadratic term.

\begin{align*} (8x\hphantom{00000})\amp(x\hphantom{000000})\\ (4x\hphantom{00000})\amp(2x\hphantom{00000}) \end{align*}
3. Consider possible factors for the constant term: $9$ may be factored as $9\cdot 1$ or as $3\cdot 3\text{.}$ Form all possible pairs of binomial factor using these factorizations. 4. Select the combinations of the products ① and ② whose sum or difference could be the linear term, $-21x\text{.}$

\begin{equation*} (8x \hphantom{000} 3) (x \hphantom{000} 3) \end{equation*}
5. Insert the proper signs:

\begin{equation*} (8x + 3) (x - 3) \end{equation*}

With practice, you can usually factor trinomials of the form $Ax^2 + Bx + C$ mentally. The following observations may help.

1. If $A\text{,}$ $B$ and $C$ are all positive, both signs in the factored form are positive. For example, as a first step in factoring $6x^2 + 11x + 4\text{,}$ we could write

\begin{equation*} (\hphantom{000} + \hphantom{000} ) (\hphantom{000} + \hphantom{000} ) \end{equation*}
2. 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 $6x^2 - 11x + 4\text{,}$ we could write

\begin{equation*} (\hphantom{000} - \hphantom{000} ) (\hphantom{000} - \hphantom{000} ) \end{equation*}
3. If $C$ is negative, the signs in the factored form are opposite. Thus as a first step in factoring $6x^2 - 5x - 4\text{,}$ we could write

\begin{equation*} (\hphantom{000} + \hphantom{000} ) (\hphantom{000} - \hphantom{000} ) ~~\text{ or }~~ (\hphantom{000} - \hphantom{000} ) (\hphantom{000} + \hphantom{000} ) \end{equation*}
###### ExampleA.59
1. \begin{aligned}[t] 6x^2 + 5x + 1 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} + \hphantom{00} )\\ \amp = (3x + 1) (2x + 1) \end{aligned}

2. \begin{aligned}[t] 6x^2 - 5x + 1 \amp = (\hphantom{00} - \hphantom{00} ) (\hphantom{00} - \hphantom{00} )\\ \amp = (3x - 1) (2x - 1) \end{aligned}

3. \begin{aligned}[t] 6x^2 - x - 1 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} - \hphantom{00} )\\ \amp = (3x + 1) (2x - 1) \end{aligned}

4. \begin{aligned}[t] 6x^2 - xy - y^2 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} - \hphantom{00} )\\ \amp = (3x + y) (2x - y) \end{aligned}

### SubsectionSpecial 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.

###### Special Products
1. $(a + b)^2 = (a + b) (a + b) = a^2 + 2ab + b^2$

2. $(a - b)^2 = (a - b) (a - b) = a^2 - 2ab + b^2$

3. $(a + b)(a-b) = a^2 - b^2$

###### CautionA.60

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

\begin{align*} (x + 4)^2 \amp\ne x^2 + 16, \amp\amp\text{ instead } \amp (x + 4)^2 \amp = x^2 + 8x + 16\\ (t-5)^2 \amp\ne t^2 - 16, \amp\amp\text{ instead } \amp (t -5)^2 \amp= t^2 -10t + 25 \end{align*}
###### ExampleA.61
1. \begin{aligned}[t] 3(x + 4)^2 \amp = 3(x^2 + 2 \cdot 4x + 4^2) \\ \amp = 3x^2 + 24x + 48 \end{aligned}

2. \begin{aligned}[t] (y + 5) (y - 5) \amp = y^2-5^2 \\ \amp = y^2-25 \end{aligned}

3. \begin{aligned}[t] (3x - 2y)^2 \amp = (3x)^2 - 2(3x)(2y) + (2y)^2 \\ \amp = 9x^2-12xy+4y^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.

###### Special Factorizations
1. $a^2 + 2ab + b^2=(a + b)^2$

2. $a^2 - 2ab + b^2=(a - b)^2$

3. $a^2 - b^2=(a + b)(a-b)$

4. $a^2+b^2 ~~\text{ cannot be factored}$

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, $a^2 + b^2\text{,}$ cannot be factored.

###### ExampleA.62

Factor.

1. $x^2 + 8x + 16$

2. $y^2 -10y + 25$

3. $4a^2 - 12ab + 9b^2$

4. $25m^2n^2 + 20mn + 4$

Solution
1. Because $16$ is equal to $4^2$ and $8$ is equal to $2\cdot 4\text{,}$

\begin{align*} x^2 + 8x + 16 \amp = x^2 - 2 \cdot 4x + 4^2\\ \amp = (x + 4)^2 \end{align*}
2. Because $25=5^2$ and $10=2\cdot 5\text{,}$

\begin{align*} y^2 - 10y + 25 \amp = y^2 - 2 \cdot 5y + 5^2\\ \amp = (y-5)^2 \end{align*}
3. Because $4a^2=(2a)^2\text{,}$ $9b^2=(3b)^2\text{,}$ and $2ab=2(2a)(3b)\text{,}$

\begin{align*} 4a^2 - 12ab + 9b^2 \amp = (2a)^2 - 2(2a)(3b) + (3b)^2\\ \amp = (2a-3b)^2 \end{align*}
4. Because $25m^2n^2=(5mn)^2\text{,}$ $4=2^2\text{,}$ and $20mn=2(5mn)(2)\text{,}$

\begin{align*} 25m^2n^2 + 20mn + 4 \amp = (5mn)^2 + 2(5mn)(2) + 2^2\\ \amp = (5mn+2)^2 \end{align*}

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

###### ExampleA.63

Factor if possible.

1. $x^2 - 81$

2. $4x^2 - 9y^2$

3. $x^2 + 81$

Solution
1. The expression $x^2 - 81$ is the difference of two squares, $x^2 - 9^2\text{,}$ and thus can be factored according to Special Factorization (III) above.

\begin{align*} x^2-81 \amp = x^2-9^2\\ \amp = (x+9)(x-9) \end{align*}
2. Because $4x^2 - 9y^2$ can be written as $(2x)^2 -(3y)^2\text{,}$

\begin{align*} 4x^2 - 9y^2 \amp =(2x^2)-(3y)^2 \\ \amp =(2x+3y)(2x-3y) \end{align*}
3. The expression $x^2 + 81\text{,}$ or $x^2 + 0x + 81\text{,}$ is not factorable, because no two real numbers have a product of $81$ and a sum of $0\text{.}$

###### CautionA.64

$x^2 + 81\ne (x + 9) (x + 9)\text{,}$ which you can verify by multiplying

\begin{equation*} (x + 9) (x + 9) = x^2 + 18x + 8 \end{equation*}

The factors $x + 9$ and $x - 9$ in Example A.63a are called conjugates of each other. In general, any binomials of the form $a+b$ and $a-b$ are called a conjugate pair.

### SubsectionSection Summary

#### SubsubsectionVocabulary

Look up the definitions of new terms in the Glossary.

• Perfect-square trinomial

• Difference of squares

• Conjugate

#### SubsubsectionSKILLS

Practice each skill in the exercises listed.

2. Expand special products: #37–48

3. Factor special quadratic expressions: #49–68

### SubsectionExercises A.8

For Problems 1-36, factor completely.

###### 1

$x^2+5x+6$

###### 2

$x^2+5x+4$

###### 3

$y^2-7y+12$

###### 4

$y^2-7y+10$

###### 5

$x^2-6-x$

###### 6

$x^2-15-2x$

###### 7

$2x^2+3x-2$

###### 8

$3x^2-7x+2$

###### 9

$7x+4x^2-2$

###### 10

$1-5x+6x^2$

###### 11

$9y^2-21y-8$

###### 12

$10y^2-3y-18$

###### 13

$10u^2-3-u$

###### 14

$8u^2-3+5u$

###### 15

$21x^2-43x-14$

###### 16

$24x^2-29x+5$

###### 17

$5a+72a^2-25$

###### 18

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

###### 19

$12-53x+30x^2$

###### 20

$39x+80x^2-20$

###### 21

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

###### 22

$48t^2-122t+39$

###### 23

$3x^2-7ax+2a^2$

###### 24

$9x^2+9ax-10a^2$

###### 25

$15x^2-4xy-4y^2$

###### 26

$12x^2+7xy-12y^2$

###### 27

$18u^2+20v^2-39uv$

###### 28

$24u^2-20v^2+17uv$

###### 29

$12a^2-14b^2-13ab$

###### 30

$24a^2-15b^2-2ab$

###### 31

$10a^2b^2-19ab+6$

###### 32

$12a^2b^2-ab-20$

###### 33

$56x^2y^2-2xy-4$

###### 34

$54x^2y^2+3xy-2$

###### 35

$22a^2z^2-21-19az$

###### 36

$26a^2z^2-24+23az$

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

###### 37

$(x+3)^2$

###### 38

$(y-4)^2$

###### 39

$(2y-5)^2$

###### 40

$(3x+2)^2$

###### 41

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

###### 42

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

###### 43

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

###### 44

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

###### 45

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

###### 46

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

###### 47

$(8xz+3)(8xz+3)$

###### 48

$(7yz-2)(7yz-2)$

For Problems 49-68, factor completely.

###### 49

$x^2-25$

###### 50

$x^2-36$

###### 51

$x^2-24x+144$

###### 52

$x^2+26x+169$

###### 53

$x^2-4y^2$

###### 54

$9x^2-y^2$

###### 55

$4x^2+12x+9$

###### 56

$4y^2+4y+1$

###### 57

$9u^2-30uv+25v^2$

###### 58

$16s^2-56st+49t^2$

###### 59

$4a^2-25b^2$

###### 60

$16a^2-9b^2$

###### 61

$x^2y^2-81$

###### 62

$x^2y^2-64$

###### 63

$9x^2y^2+6xy+1$

###### 64

$4x^2y^2+12xy+9$

###### 65

$16^2y^2-1$

###### 66

$64x^2y^2-1$

###### 67

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

###### 68

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