Factor Trinomials Part I – Using Grouping

In the last section we introduced the technique of factoring by grouping as a means to be able to factor a four-term polynomial. However, many of the polynomials we encountered were actually trinomials in disguise. Now we will get to the work of starting with a three-term polynomial, and rewriting it as a four term polynomial so it can be factored using our grouping process.

We will start with factoring trinomials of the form [latex]x^{2}+bx+c[/latex], and hence with a leading coefficient of 1.

Remember that when [latex]\left(x+2\right)[/latex] and [latex]\left(x+5\right)[/latex], are multiplied, the result is a four term polynomial and then it is simplified into a trinomial:

 [latex]\left(x+2\right)\left(x+5\right)=x^2+5x+2x+10=x^2+7x+10[/latex]

Factoring is the reverse of multiplying, so let’s go in reverse and factor the trinomial [latex]x^{2}+7x+10[/latex]. The individual terms [latex]x^{2}[/latex], [latex]7x[/latex], and 10 share no common factors. If we rewrite the middle term as the sum of the two terms [latex]7x=5x+2x[/latex] then we can use the grouping technique:

[latex](x^{2}+5x)+(2x+10)[/latex]

Factor each pair: [latex]\begin{array}{l}x\underbrace{\left(x+5\right)}+2\underbrace{\left(x+5\right)}\\\text{common binomial factor}\end{array}[/latex]

Then pull out the common binomial factor: [latex]\left(x+5\right)\left(x+2\right)[/latex]

But how would we know to rewrite [latex]7x[/latex] as [latex]5x+2x[/latex]? What would have happened if we had rewritten [latex]7x[/latex] as [latex]6x+x[/latex]?

[latex](x^{2}+6x)+(x+10)[/latex]

Factor each pair: [latex]x\left(x+6\right)+1\left(x+10\right)[/latex]

Then we don’t have a common binomial! There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.

Method to the Madness

The following is a summary of the method, then we will show some examples of how to use it.

For example, to factor [latex]x^{2}+7x+10[/latex], you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).

Look at factor pairs of 10: 1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite [latex]7x[/latex] as [latex]2x+5x[/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[/latex] as [latex]5x+2x[/latex]. Both will work.

With all terms positive in the last example, we did not really need to consider any negative factors. 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 look at all of the combinations of factors whose product is [latex]−12[/latex]. Then see which of these combinations will give you the correct middle term, where b is 1.

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

In the above example, you 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.

In the following video, we present another example of how to use grouping to factor a trinomial of this form.

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.

While there is no foolproof way to find the right combination on the first guess, there 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 includes some of the problems you have seen so far.

Factoring Trinomials II – The Shortcut (Product and Sum Method)

Notice that in all of the examples, 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 1), if you can identify the correct r and s values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let’s look at some examples where we use this idea.

Shortcut This Way

In the next two examples, we will show how you can skip the step of factoring by grouping and move directly to the factored form of a product of two binomials with the r and s values that you find. The idea is that you can build factors for a trinomial in this form: [latex]x^2+bx+c[/latex] by finding r and s, 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]

This technique is sometime referred to as the “Product and Sum Method.”

Let us revisit the problem [latex]x^2+7x+10[/latex]. We determined that the factors [latex]2[/latex] and [latex]5[/latex] multiply to [latex]10[/latex] while simultaneously adding to [latex]7[/latex]. We proceeded by rewriting the problem and transforming it into a grouping problem. Alternatively, we could have jumped to the final answer immediately using Product and Sum, factoring the polynomial as

[latex]x^2+7x+10=(x+2)(x+5)[/latex].

Note that the order does not matter, and we could have also written [latex]x^2+7x+10=(x+5)(x+2)[/latex].

We give several more examples demonstrating the Product and Sum Method.

Let us try another.

In the following video, we present two more examples of factoring a trinomial using the shortcut presented here.

Another situation we sometime encounter is a trinomial with two variables of the form [latex]x^2+bxy+cy^2[/latex]. Applying the same logic behind beginning each parenthetical expression with [latex]x[/latex] and [latex]x[/latex] to produce the leading term of [latex]x^2[/latex], we will similarly end each expression with [latex]y[/latex].  We will then be able to deal with the coefficients using the same approach as previous examples.

Its Okay To Be Prime

Many students become frustrated when searching for the proper pair of factors, only to discover that no such pair existed! Remember, not every polynomial can be factored.

Factor Trinomials Part III – Factoring Out A GCF First

Not all trinomials look like [latex]x^{2}+5x+6[/latex], where the coefficient in front of the [latex]x^{2}[/latex] term is 1. In these cases, your first step should be to look for common factors for the three terms.

Notice that once you have identified and pulled out the common factor, if the resulting trinomial has a leading coefficient of 1, you can factor the remaining trinomial as usual.

Let’s begin with an example where the leading coefficient is [latex]-1[/latex]. Recall that in Section 6.1, we discussed that it could be useful to factor out the negative in this case. Here, it is imperative in order to directly apply the shortcut we have introduced.

In the next example, we must first factor out a more substantial GCF.

The following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF.

The general form of trinomials with a leading coefficient of a is [latex]ax^{2}+bx+c[/latex]. Sometimes the factor of a can be factored out as you saw above; this happens when a is in common to all three terms. The remaining trinomial that still needs factoring (if possible) will then be simpler, with the leading term only being an [latex]x^{2}[/latex] term, instead of an [latex]ax^{2}[/latex] term.

But what do we do if a is not 1 and yet is not a common factor? We explore this in the next section.