When we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this example: [latex]\left(x+4\right)\left(x+2\right)=x^{2}+2x+4x+8[/latex].
We can apply what we have learned about factoring out a common monomial to return a four term polynomial to the product of two binomials. Why would we even want to do this?
Why Should I Care?
Because it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:
Additionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don’t all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.
Example
Factor [latex]a^2+3a+5a+15[/latex]
Show Solution
There isn’t a common factor between all four terms, so we will group the terms into pairs that will enable us to find a GCF for them. For example, we wouldn’t want to group [latex]a^2\text{ and }15[/latex] because they don’t share a common factor.
Notice that when you factor a two term polynomial, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials. As we noted before, this is an important middle step in learning how to factor a three term polynomial.
This process is called the grouping technique. Broken down into individual steps, here’s how to do it (you can also follow this process in the example below).
Group the terms of the polynomial into pairs that share a GCF.
Find the greatest common factor and then use the distributive property to pull out the GCF
Look for the common binomial between the factored terms
Factor the common binomial out of the groups, the other factors will make the other binomial
Let’s try factoring a few more four-term polynomials. Note how there is a now a constant in front of the [latex]x^2[/latex] term. We will just consider this another factor when we are finding the GCF.
Factor out the common factor, [latex]\left(2x–3\right)[/latex], from both terms.
[latex]\left(x+4\right)\left(2x–3\right)[/latex]
Answer
[latex]\left(x+4\right)\left(2x-3\right)[/latex]
The video that follows provides another example of factoring by grouping.
In the next example, we will have a GCF that is negative. It is important to pay attention to what happens to the resulting binomial when the GCF is negative.
Factor the common factor [latex]−2[/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]−2[/latex] is factored out.
The two groups [latex]7x\left(x–3\right)[/latex] and [latex]5\left(x–1\right)[/latex] do not have any common factors, so this polynomial cannot be factored any further.
Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: . License: CC BY: Attribution
