Factoring Trinomials

Now that we know how to factor by grouping, we are going to factor a trinomial whose leading coefficient is [latex]1[/latex].  In particular, we will look at trinomials of the form [latex]x^2+bx+c[/latex], where [latex]b, c[/latex] are real numbers.  Polynomials whose leading coefficients are 1 can be factored using the grouping method that we used in the previous section.  For example, [latex]x^2+5x+4[/latex] can be written as [latex]x^2+x+4x+4[/latex], which can then be factored using grouping: [latex]x^2+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)[/latex].  The question is, how do we know how to split up the [latex]x[/latex]-term?  If we multiply the generic binomials [latex](x+r)(x+s)[/latex], we get [latex]x(x+s)+r(x+s)=x^2+sx+rx+rs=x^2+(r+s)x+rs[/latex].

If we now compare this to the generic trinomial [latex]x^2+bx+c[/latex], we can equate the [latex]x[/latex]-terms so that [latex]r+s=b[/latex] and we can equate the constants so that [latex]c=rs[/latex].  In other words, we split up the [latex]x[/latex]-term [latex]bx[/latex] into [latex]rx+sx[/latex] where [latex]r+s=b[/latex] and [latex]rs=c[/latex].

Let’s factor the trinomial [latex]x^{2}+5x+6[/latex]. In this polynomial, [latex]b=5[/latex] and [latex]c=6[/latex]. We look for two numbers that add to 5 and multiply to 6. A chart will help us organize possibilities. On the left, list all possible factors of the [latex]c[/latex]-term, [latex]6[/latex]; on the right we’ll find the sums.

Factors whose product is  [latex]6[/latex]	Sum of the factors
[latex]1\cdot6=6[/latex]	[latex]1+6=7[/latex]
[latex]2\cdot3=6[/latex]	[latex]2+3=5[/latex]
[latex]-1\cdot -6=6[/latex]	[latex]-1+(-6)=-7[/latex]
[latex]-2\cdot -3=6[/latex]	[latex]-2+(-3)=-5[/latex]

There are only two possible factor combinations with positive factors, [latex]1[/latex] and [latex]6[/latex], or [latex]2[/latex] and [latex]3[/latex], but there are also two possible factor combinations with negative factors, [latex]-1[/latex] and [latex]-6[/latex], or [latex]-2[/latex] and [latex]-3[/latex] . We can see that [latex]2+3=5[/latex]. So [latex]2x+3x=5x[/latex], giving us the correct [latex]x[/latex]-term.

Note that if we wrote [latex]x^{2}+5x+6[/latex] as [latex]x^{2}+3x+2x+6[/latex] and grouped the pairs as [latex]\left(x^{2}+3x\right)+\left(2x+6\right)[/latex]; then factored, [latex]x\left(x+3\right)+2\left(x+3\right)[/latex], and factored out [latex]x+3[/latex], the answer would be [latex]\left(x+3\right)\left(x+2\right)[/latex]. Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.

The following video presents another example of how to use grouping to factor a quadratic polynomial.

Let’s take a look at the trinomial [latex]x^{2}+x–12[/latex]. In this trinomial, the c term is [latex]−12[/latex]. So we look at all of the combinations of factors whose product is [latex]−12[/latex]. Then see which of these combinations will give us the correct [latex]x[/latex]-term, where b is [latex]1[/latex].

Factors whose product is [latex]−12[/latex]	Sum of the factors
[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]

There is only one combination where the product is [latex]−12[/latex] and the sum is [latex]1[/latex], and that is when [latex]r=4[/latex], and [latex]s=−3[/latex]. Let’s use these to factor our original trinomial.

In this example, we could also rewrite [latex]x^{2}+x-12[/latex] as [latex]x^{2}– 3x+4x–12[/latex] first. Then factor [latex]x\left(x – 3\right)+4\left(x–3\right)[/latex], and factor out [latex]\left(x–3\right)[/latex] getting [latex]\left(x–3\right)\left(x+4\right)[/latex]. Since multiplication is commutative, this is the same answer.

 

Factoring Tips

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored. In this case the trinomial is said to be prime.

Here are some tips that can ease the way.

After you have factored a number of trinomials in the form [latex]x^{2}+bx+c[/latex], you may notice that the numbers you identify for r and s end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews three examples.

Trinomial	[latex]x^{2}+7x+10[/latex]	[latex]x^{2}+5x+6[/latex]	[latex]x^{2}+x-12[/latex]
r and s values	[latex]r=+5,s=+2[/latex]	[latex]r=+2,s=+3[/latex]	[latex]r=+4,s=–3[/latex]
Factored form	[latex]\left(x+5\right)\left(x+2\right)[/latex]	[latex]\left(x+2\right)\left(x+3\right)[/latex]	[latex](x+4)(x–3)[/latex]

This is only true when the leading coefficient of the trinomial is 1.

The Shortcut

Notice that in each of the examples above, the r and s values are repeated in the factored form of the trinomial. So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[/latex] (where the coefficient in front of [latex]x^{2}[/latex] is [latex]1[/latex]), if we can identify the correct r and s values, we can effectively skip the grouping steps and go straight to the factored form. The idea is that we can build factors for a trinomial in this form: [latex]x^2+bx+c[/latex] by finding [latex]r[/latex] and [latex]s[/latex], then placing them in two binomial factors like this:

[latex]\left(x+r\right)\left(x+s\right)\text{ OR }\left(x+s\right)\left(x+r\right)[/latex]

In the next example, the leading coefficient is negative.

The following video presents two more examples of factoring a trinomial with a leading coefficient of 1.

To summarize our process, consider the following steps:

 

In the last example, [latex]b[/latex] was negative and [latex]c[/latex] was positive. This will always mean that if it can be factored, [latex]r[/latex] and [latex]s[/latex] will both be negative.

The more examples you work through, the more likely you are to spot patterns. This will lead to faster factoring, which is useful in later courses.