3.5.3: Factoring Trinomials – The ac-Method

Learning Outcome

Factor a trinomial using the ac-method

A method that is often used when the coefficient of the leading term (i.e., the coefficient of $x^2$ ) is not 1 is called the ac-method. This method splits the $x$ -term into the sum of two terms so that we can use grouping to factor the function. Determining how to split up the $x$ -term is done by finding two numbers whose product is $ac$ and whose sum is $b$ , where $a, b, c$ are real number coefficients of the function $f(x)=ax^2+bx+c$ .

Suppose we wish to factor $f(x)=2x^2+5x+3$ . The first step is to figure out how to write the $x$ -term as the sum of two new terms. We are looking for two numbers, $s, t$ , with a product of $ac=6$ and a sum of $b=5$ . We start by listing factors of 6, then look at their sum. A table can be useful to keep everything organized.

Factors: $st=6$	Sum: $s+t=5$
$1,6$	$7$
$-1,-6$	$-7$
$2,3$	$5$
$-2,-3$	$-5$

The pair $s=2$ and $t=3$ gives the correct $x$ -term of $5x$ , so we will write $5x=2x+3x$ :

$\begin{aligned}f(x)&=2{x}^{2}+\color{blue}{5x}+3\\&=2x^2+\color{blue}{2x+3x}+3\end{aligned}$

Now we can group the polynomial into two binomials,

$f(x)=(2x^2+2x)+(3x+3)$

then factor by grouping:

$\begin{aligned}f(x)&=(2x^2+2x)+(3x+3)\\&=2x(x+1)+3(x+1)\\&=(x+1)(2x+3)\end{aligned}$

Factoring using the ac-method

To factor a trinomial function of the form $f(x)=a{x}^{2}+bx+c$ by grouping, we find two numbers with a product of $ac$ and a sum of $b$ . We use these numbers to divide the $x$ -term into the sum of two terms and factor each portion of the expression separately. Then we factor out the GCF of the entire expression.

Example 1

Factor $f(x)=5{x}^{2}+7x - 6$ .

Solution

We have a trinomial with $a=5,b=7$ , and $c=-6$ .

$ac=-30$ so we need to find two numbers $s, t$ , with a product of $-30$ and a sum of $7$ . In the table, we list factors until we find a pair with the desired sum.

Factors: $st=-30$	Sum: $s+t=7$
$1,-30$	$-29$
$-1,30$	$29$
$2,-15$	$-13$
$-2,15$	$13$
$3,-10$	$-7$
$-3,10$	$7$

So $s=-3$ and $t=10$ .

Now write $7x=-3x+10x$ in the function and factor by grouping:

$\begin{aligned}f(x)&=5{x}^{2}+7x - 6\\&=5{x}^{2}-3x+10x - 6 \hfill & \text{Write the original trinomial as }a{x}^{2}+sx+tx+c\hfill \\ &=x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each pair}\hfill \\ &=\left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}\hfill \end{aligned}$

We can check our work by multiplying. Use the distributive property to confirm that $\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6$ .

In this example, when we wrote $7x=-3x+10x$ , we can just as easily write $7x=10x-3x$ . We will end up with the same factorization:

$\begin{aligned}f(x)&=5{x}^{2}+7x - 6\\&=5{x}^{2}+10x-3x - 6 \\ &=5x\left(x + 2\right)-3\left(x + 2\right)\\ &=\left(x+2\right)\left(5x - 3\right)\end{aligned}$

The commutative property of multiplication says $ab=ba$ , so we have the same result as before.

We can summarize our process in the following way:

factoring a trinomial function by the ac-method

To factor a trinomial function of the form $f(x)=a{x}^{2}+bx+c$ ,

List factors of $ac$ .
Find $s$ and $t$ , factors of $ac$ with a sum of $b$ .
Write the original expression as $a{x}^{2}+sx+tx+c$ .
Factor by grouping.

Factoring trinomials whose leading coefficient is not $1$ becomes quick and kind of fun once you get the idea.

Example 2

Factor $g(x)=2{x}^{2}+9x+9$ .

Solution

Find two numbers, $s, t$ , such that $st=18$ and $s + t = 9$ .

$9$ and $18$ are both positive, so we will only consider positive factors.

Factors of $st=18$	$s+t=9$
$1, 18$	$19$
$3,6$	$9$

We can stop because we have found our factors.

Write the $9x$ as $3x+6x$ , then factor by grouping:

$\begin{aligned}g(x)&=2x^2+\color{blue}{9x}+9\\&=2x^2+\color{blue}{3x+6x}+9\\&=(2x^2+3x)+(6x+9)\\&=x(2x+3)+3(2x+3)\\&=(2x+3)(x+3)\end{aligned}$

Answer

$g(x)=(x+3)(2x+3)$

Try It 1

Factor $f(x)=6{x}^{2}+x - 1$ .

Show Answer

$f(x)=(2x+1)(3x-1)$

If the leading term is negative, we can pull out –1 as a GCF before we factor the trinomial.

Example 3

Factor $f(x)=-4x^2+8x+21$ .

Solution

Start by pulling –1 as a GCF. This will change all of the signs of the coefficients: $f(x)=-1(4x^2-8x-21)$

Now, $ac=4(-21)=-84$ so we need two numbers, $s, t$ , whose product is –84 and whose sum is –8.

Product: $st=-84$	Sum: $s+t=-8$
$-12, 7$	$-12+7=-5$
$-28, 3$	$-28+3=-25$
$-14, 6$	$-14+6=-8$

Write $8x$ as $-14x+6x$ then factor by grouping:

$\begin{aligned}f(x)&=-4x^2+8x+21\\&=-1(4x^2-8x-21)\\&=-1\left (4x^2-14x+6x-21\right )\\&=-1\left [2x(2x-7)+3(2x-7)\right ]\\&=-1(2x-7)(2x+3)\end{aligned}$

Try It 2

Factor $f(x)=-6x^2-7x+20$ .

Show Answer

$f(x)=-1(3x-4)(2x+5)$

Sometimes we encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials. Such polynomials are said to be prime.

Example 4

Factor $f(x)=7x^{2}-16x–5$ .

Solution

$ac=7(-5)=-35$ , $b=-16$ . We need to find two numbers $s, t$ , where $st=-35$ and $s+t=-16$ .

Factors: $st=-35$	Sum: $s+t=-16$
$-1, 35$	$34$
$1, -35$	$-34$
$-5, 7$	$2$
$-7,5$	$-2$

None of the factors add up to $-16$ so the trinomial cannot be factored.

$f(x)=7x^{2}-16x–5$ is prime.

Before we jump head first into factoring a trinomial, we should always look to see if there is a GCF that can be “pulled out”.

Example 5

Factor $f(x)=6x^3+21x^2-27x$ .

Solution

The terms in the function have a GCF of 3x, so we will pull out the GCF first:

$\begin{aligned}f(x)&=6x^3+21x^2-27x\\&=3x(2x^2+7x-9)\end{aligned}$

To factor the trinomial, $ac=2(-9)=-18$ and $b=7$ so, we need two numbers that multiply to -18 and add to 7: 9 and –2.

Write $7x=9x-2x$ then factor by grouping:

$\begin{aligned}f(x)&=3x(2x^2+7x-9)\\&=3x(2x^2+9x-2x-9)\\&=3x\left [x(2x+9)-1(2x+9)\right ]\\&=3x(2x+9)(x-1)\end{aligned}$

Try It 3

Factor $f(x)=-6x^3+14x^2+40x$

Show Answer

$f(x)=-2x(x-4)(3x+5)$

Remember to always look for a GCF before jumping in to factoring any trinomial.

Example 6

Factor $f(x)=72x^3+168x^2-144x$ .

Solution

First notice that $x$ is common to all terms in the trinomial. Also 72, 168, and 144 have at least 2 in common. Let’s factor them to find the GCF:

$\begin{aligned}72 &= 2^3\cdot 3^2\\168&=2^3\cdot 3\cdot 7\\144&=2^4\cdot 3^2\\\hline GCF&=2^3\cdot 3\\&=24\end{aligned}$

Factor out the GCF of $24x$ from the trinomial:

$\begin{aligned}f(x)&=72x^3+168x^2-144x\\&=24x(3x^2+7x-6)\end{aligned}$

Now use the $ac$ -method:

$ac=-18$ and $b=+7$ : $9\cdot -2=-18$ and $9-2=7$

Write $7x=9x-2x$ then factor by grouping:

$\begin{aligned}f(x)&=24x(3x^2+7x-6)\\&=24x(3x^2+9x-2x-6)\\&=24x[3x(x+3)-2(x+3)]\\&=24x(x+3)(3x-2)\end{aligned}$

Try It 4

Factor $g(x)=140x^3-161x^2+42x$

Show Answer

$g(x)=7x(5x-2)(4x-3)$

Chapter 3: Polynomial Functions