Learning Outcomes
- Factor a trinomial function using the trial and error method
- Explain “prime polynomials”
The Trial and Error Method
So far we have seen that we can factor polynomial functions by “pulling out” the greatest common factor from all terms, or by “pulling out” the greatest common factor from pairs of terms using factoring by grouping. In this section we are going to concentrate on factoring trinomials; polynomials with three terms.
Recall how we used the distributive property to multiply two binomials:
[latex]\begin{aligned}\left(x+2\right)\left(x+3\right) &= x^2+3x+2x+6\\&=x^2+5x+6\end{aligned}[/latex]
The product of two binomial generally results in a trinomial. To factor a trinomial of the form [latex]f(x)=x^2+bx+c[/latex], we want to reverse the distributive property and return [latex]x^2+5x+6[/latex] to [latex]\left(x+2\right)\left(x+3\right) [/latex].
Suppose the polynomial function [latex]f(x)=x^2+bx+c[/latex] factors into [latex](x+s)(x+t)[/latex].
Multiplied out using the distributive property,
[latex]\begin{aligned}(x+s)(x+t)&=x^2+sx+tx+st\\&=x^2+(s+t)x+st\end{aligned}[/latex].
This means that [latex]s+t=b[/latex] and [latex]st=c[/latex]
So, to factor a polynomial function of the form [latex]f(x)=x^2+bx+c[/latex] we will have to find two numbers [latex]s[/latex] and [latex]t[/latex] such that [latex]st=c[/latex] and [latex]s+t=b[/latex].
Factoring a Trinomial with Leading Coefficient 1
To factor a polynomial function of the form [latex]f(x)=x^2+bx+c[/latex] we have to find two numbers [latex]s[/latex] and [latex]t[/latex] such that [latex]st=c[/latex] and [latex]s+t=b[/latex]. Then [latex]f(x)=x^2+bx+c=(x+s)(x+t)[/latex]
Polynomial functions that do not factor are referred to as prime.
Let us put this idea into practice.
Example 1
Factor [latex]f(x)={x}^{2}+2x - 15[/latex].
Solution
We have a trinomial with leading coefficient [latex]a=1,b=2[/latex], and [latex]c=-15[/latex]. We need to find two numbers, [latex]s, t[/latex] with a product of [latex]st =-15[/latex] and a sum of [latex]s+t=2[/latex].
In the table, we list factors of the product –15 until we find a pair with the desired sum of 2.
| Product: [latex]st=-15[/latex] | Sum: [latex](s+t)=2[/latex] |
|---|---|
| [latex]1,-15[/latex] | [latex]-14[/latex] |
| [latex]-1,15[/latex] | [latex]14[/latex] |
| [latex]3,-5[/latex] | [latex]-2[/latex] |
| [latex]-3,5[/latex] | [latex]2[/latex] |
Now that we have identified [latex]s[/latex] and [latex]t[/latex] as [latex]-3[/latex] and [latex]5[/latex], write the factored form as [latex]\left(x - 3\right)\left(x+5\right)[/latex].
So, [latex]f(x)=(x-3)(x+5)[/latex].
To summarize our process, consider the following steps:
factoring a trinomial function
For a trinomial in the form [latex]f(x)=x^2+bx+c[/latex] (i.e.,[latex]a=1[/latex]),
- List factors of [latex]c[/latex].
- Find [latex]s[/latex] and [latex]t[/latex], whose product is [latex]c[/latex] and sum [latex]b[/latex].
- Write the factored expression [latex]\left(x+s\right)\left(x+t\right)[/latex].
Try It 1
Factor [latex]g(x)=x^2+7x+6[/latex].
The next example shows that when [latex]c[/latex] is negative, either s or t (but not both) will be negative.
Example 2
Factor [latex]x^{2}+x–12[/latex].
Solution
Consider all the combinations of numbers whose product is [latex]-12[/latex] and list their sum.
| Factors: [latex]st=−12[/latex] | Sum: [latex]s+t=1[/latex] |
|---|---|
| [latex]1\cdot−12=−12[/latex] | [latex]1+−12=−11[/latex] |
| [latex]2\cdot−6=−12[/latex] | [latex]2+−6=−4[/latex] |
| [latex]3\cdot−4=−12[/latex] | [latex]3+−4=−1[/latex] |
| [latex]4\cdot−3=−12[/latex] | [latex]4+−3=1[/latex] |
| [latex]6\cdot−2=−12[/latex] | [latex]6+−2=4[/latex] |
| [latex]12\cdot−1=−12[/latex] | [latex]12+−1=11[/latex] |
Choose the values whose sum is [latex]+1[/latex]: [latex]s=4[/latex] and [latex]t=−3[/latex], and place them into a product of binomials.
[latex]\left(x+4\right)\left(x-3\right)[/latex]
Try It 2
Factor [latex]h(x)=x^2+6x-16[/latex].
In our last example, we will show how to factor a trinomial whose b term is negative.
Example 3
Factor [latex]{x}^{2}-7x+6[/latex].
Solution
List the factors of [latex]6[/latex]. Note that [latex]b=-7[/latex], so we will need to consider negative numbers in our list.
| Factors: [latex]st=6[/latex] | Sum: [latex]s+t=-7[/latex] |
|---|---|
| [latex]1,6[/latex] | [latex]7[/latex] |
| [latex]2, 3[/latex] | [latex]5[/latex] |
| [latex]-1, -6[/latex] | [latex]-7[/latex] |
| [latex]-2, -3[/latex] | [latex]-5[/latex] |
Choose the pair that sum to [latex]-7[/latex], which is [latex]-1, -6[/latex]
Write the pair as constant terms in a product of binomials.
[latex]\left(x-1\right)\left(x-6\right)[/latex]
In the last example, the [latex]b[/latex] was negative and [latex]c[/latex] was positive. This will always mean that if it can be factored, s and t will both be negative.
Try It 3
Factor [latex]f(x)=x^2-4x+4[/latex].
Try It 4
Factor [latex]g(x)=x^2+3x+4[/latex].
Trinomials with [latex]a\ne1[/latex]
When the leading coefficient of a trinomial is not equal to 1, factoring by trial and error becomes more difficult as there are more pairs of numbers that could work.
For example, suppose we want to factor [latex]f(x)=12x^2-x-20[/latex]. First we need two numbers that multiply to 12: 1(12); 2(6); 3(4); 4(3); 6(2); 12(1). Then we need two numbers that multiply to –20: 1(–20); –20(1); –1(20); 20(–1); 2(–10); –10(2); –2(10); 10(–2); 4(–5); –5(4); 5(–4); –4(5). Then from all these possible combinations we need the [latex]x[/latex]-terms to add up to –1. There are 84 different options to try! Let’s find a more efficient method.
