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 [latex]x^2[/latex]) is not 1 is called the ac-method. This method splits the [latex]x[/latex]-term into the sum of two terms so that we can use grouping to factor the function. Determining how to split up the [latex]x[/latex]-term is done by finding two numbers whose product is [latex]ac[/latex] and whose sum is [latex]b[/latex], where [latex]a, b, c[/latex] are real number coefficients of the function [latex]f(x)=ax^2+bx+c[/latex].
Suppose we wish to factor [latex]f(x)=2x^2+5x+3[/latex]. The first step is to figure out how to write the [latex]x[/latex]-term as the sum of two new terms. We are looking for two numbers, [latex]s, t[/latex], with a product of [latex]ac=6[/latex] and a sum of [latex]b=5[/latex]. We start by listing factors of 6, then look at their sum. A table can be useful to keep everything organized.
| Factors: [latex]st=6[/latex] | Sum: [latex]s+t=5[/latex] |
|---|---|
| [latex]1,6[/latex] | [latex]7[/latex] |
| [latex]-1,-6[/latex] | [latex]-7[/latex] |
| [latex]2,3[/latex] | [latex]5[/latex] |
| [latex]-2,-3[/latex] | [latex]-5[/latex] |
The pair [latex]s=2[/latex] and [latex]t=3[/latex] gives the correct [latex]x[/latex]-term of [latex]5x[/latex], so we will write [latex]5x=2x+3x[/latex]:
[latex]\begin{aligned}f(x)&=2{x}^{2}+\color{blue}{5x}+3\\&=2x^2+\color{blue}{2x+3x}+3\end{aligned}[/latex]
Now we can group the polynomial into two binomials,
[latex]f(x)=(2x^2+2x)+(3x+3)[/latex]
then factor by grouping:
[latex]\begin{aligned}f(x)&=(2x^2+2x)+(3x+3)\\&=2x(x+1)+3(x+1)\\&=(x+1)(2x+3)\end{aligned}[/latex]
Factoring using the ac-method
To factor a trinomial function of the form [latex]f(x)=a{x}^{2}+bx+c[/latex] by grouping, we find two numbers with a product of [latex]ac[/latex] and a sum of [latex]b[/latex]. We use these numbers to divide the [latex]x[/latex]-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 [latex]f(x)=5{x}^{2}+7x - 6[/latex].
Solution
We have a trinomial with [latex]a=5,b=7[/latex], and [latex]c=-6[/latex].
[latex]ac=-30[/latex] so we need to find two numbers [latex]s, t[/latex], with a product of [latex]-30[/latex] and a sum of [latex]7[/latex]. In the table, we list factors until we find a pair with the desired sum.
| Factors: [latex]st=-30[/latex] | Sum: [latex]s+t=7[/latex] |
|---|---|
| [latex]1,-30[/latex] | [latex]-29[/latex] |
| [latex]-1,30[/latex] | [latex]29[/latex] |
| [latex]2,-15[/latex] | [latex]-13[/latex] |
| [latex]-2,15[/latex] | [latex]13[/latex] |
| [latex]3,-10[/latex] | [latex]-7[/latex] |
| [latex]-3,10[/latex] | [latex]7[/latex] |
So [latex]s=-3[/latex] and [latex]t=10[/latex].
Now write [latex]7x=-3x+10x[/latex] in the function and factor by grouping:
[latex]\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}[/latex]
We can check our work by multiplying. Use the distributive property to confirm that [latex]\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6[/latex].
In this example, when we wrote [latex]7x=-3x+10x[/latex], we can just as easily write [latex]7x=10x-3x[/latex]. We will end up with the same factorization:
[latex]\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}[/latex]
The commutative property of multiplication says [latex]ab=ba[/latex], 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 [latex]f(x)=a{x}^{2}+bx+c[/latex],
- List factors of [latex]ac[/latex].
- Find [latex]s[/latex] and [latex]t[/latex], factors of [latex]ac[/latex] with a sum of [latex]b[/latex].
- Write the original expression as [latex]a{x}^{2}+sx+tx+c[/latex].
- Factor by grouping.
Factoring trinomials whose leading coefficient is not [latex]1[/latex] becomes quick and kind of fun once you get the idea.
Example 2
Factor [latex]g(x)=2{x}^{2}+9x+9[/latex].
Solution
Find two numbers, [latex]s, t[/latex], such that [latex]st=18[/latex] and [latex]s + t = 9[/latex].
[latex]9[/latex] and [latex]18[/latex] are both positive, so we will only consider positive factors.
| Factors of [latex]st=18[/latex] | [latex]s+t=9[/latex] |
|---|---|
| [latex]1, 18[/latex] | [latex]19[/latex] |
| [latex]3,6[/latex] | [latex]9[/latex] |
We can stop because we have found our factors.
Write the [latex]9x[/latex] as [latex]3x+6x[/latex], then factor by grouping:
[latex]\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}[/latex]
Answer
[latex]g(x)=(x+3)(2x+3)[/latex]
Try It 1
Factor [latex]f(x)=6{x}^{2}+x - 1[/latex].
If the leading term is negative, we can pull out –1 as a GCF before we factor the trinomial.
Example 3
Factor [latex]f(x)=-4x^2+8x+21[/latex].
Solution
Start by pulling –1 as a GCF. This will change all of the signs of the coefficients: [latex]f(x)=-1(4x^2-8x-21)[/latex]
Now, [latex]ac=4(-21)=-84[/latex] so we need two numbers, [latex]s, t[/latex], whose product is –84 and whose sum is –8.
| Product: [latex]st=-84[/latex] | Sum: [latex]s+t=-8[/latex] |
| [latex]-12, 7[/latex] | [latex]-12+7=-5[/latex] |
| [latex]-28, 3[/latex] | [latex]-28+3=-25[/latex] |
| [latex]-14, 6[/latex] | [latex]-14+6=-8[/latex] |
Write [latex]8x[/latex] as [latex]-14x+6x[/latex] then factor by grouping:
[latex]\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}[/latex]
Try It 2
Factor [latex]f(x)=-6x^2-7x+20[/latex].
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 [latex]f(x)=7x^{2}-16x–5[/latex].
Solution
[latex]ac=7(-5)=-35[/latex], [latex]b=-16[/latex]. We need to find two numbers [latex]s, t[/latex], where [latex]st=-35[/latex] and [latex]s+t=-16[/latex].
| Factors: [latex]st=-35[/latex] | Sum: [latex]s+t=-16[/latex] |
|---|---|
| [latex]-1, 35[/latex] | [latex]34[/latex] |
| [latex]1, -35[/latex] | [latex]-34[/latex] |
| [latex]-5, 7[/latex] | [latex]2[/latex] |
| [latex]-7,5[/latex] | [latex]-2[/latex] |
None of the factors add up to [latex]-16[/latex] so the trinomial cannot be factored.
[latex]f(x)=7x^{2}-16x–5[/latex] 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 [latex]f(x)=6x^3+21x^2-27x[/latex].
Solution
The terms in the function have a GCF of 3x, so we will pull out the GCF first:
[latex]\begin{aligned}f(x)&=6x^3+21x^2-27x\\&=3x(2x^2+7x-9)\end{aligned}[/latex]
To factor the trinomial, [latex]ac=2(-9)=-18[/latex] and [latex]b=7[/latex] so, we need two numbers that multiply to -18 and add to 7: 9 and –2.
Write [latex]7x=9x-2x[/latex] then factor by grouping:
[latex]\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}[/latex]
Try It 3
Factor [latex]f(x)=-6x^3+14x^2+40x[/latex]
Remember to always look for a GCF before jumping in to factoring any trinomial.
Example 6
Factor [latex]f(x)=72x^3+168x^2-144x[/latex].
Solution
First notice that [latex]x[/latex] 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:
[latex]\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}[/latex]
Factor out the GCF of [latex]24x[/latex] from the trinomial:
[latex]\begin{aligned}f(x)&=72x^3+168x^2-144x\\&=24x(3x^2+7x-6)\end{aligned}[/latex]
Now use the [latex]ac[/latex]-method:
[latex]ac=-18[/latex] and [latex]b=+7[/latex]: [latex]9\cdot -2=-18[/latex] and [latex]9-2=7[/latex]
Write [latex]7x=9x-2x[/latex] then factor by grouping:
[latex]\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}[/latex]
Try It 4
Factor [latex]g(x)=140x^3-161x^2+42x[/latex]
Candela Citations
- Revision and Adaptation. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
- Try It: hjm583; hjm108; hjm877; hjm647 Examples 5 and 6. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
- Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
- Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: http://nrocnetwork.org/dm-opentext. License: CC BY: Attribution
