6.2: Factor by Grouping

section 6.2 Learning Objectives

6.2: Factoring by Grouping

  • Factor a four-term polynomial by grouping

 

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?

Thought bubble with the words .....and i should care why?

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:

[latex]\begin{array}{l}\left(x+4\right)\left(x+2\right)\\=x^{2}+2x+4x+8\\=x^2+6x+8\end{array}[/latex]

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 1

Factor [latex]a^2+3a+5a+15[/latex]

Notice that with this process, we have been able to factor a four-term polynomial into the product of two binomials.

This process is called the grouping technique . Broken down into individual steps, here’s how we apply this technique to a four-term polynomial (you can also follow this process in the example below).

  • Group the terms of the polynomial into pairs.
  • Find the greatest common factor of each pair and factor it out.
  • 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

It is worth noting that grouping can be used for polynomials with more than four terms, but the steps above and our examples will focus on four-term polynomials.  Let’s try factoring a few more. 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.

Example 2

Factor [latex]2x^{2}+4x+5x+10[/latex].

Another example follows that contains subtraction. Note how we choose a positive GCF from each group of terms, and the negative signs stay.

Example 3

Factor [latex]2x^{2}–3x+8x–12[/latex].

The video that follows provides another example of factoring by grouping.

In the next example, we will factor out a negative with the GCF. So what is the difference here compared to the last example. Recall that we looked at an example of this in Section 6.1, where we decided it can be useful to factor out the negative when it is part of the leading coefficient. As we group in the next example, the negative will be “in front” of the second pair. Factoring out the negative plays an important role in grouping as the resulting binomials will not match if we fail to factor out the negative.

Example 4

Factor [latex]3x^{2}+3x–2x–2[/latex].

In the following video we present another example of factoring by grouping when the GCF of one of the pairs of terms is negative.

So far, every polynomial we have factored has had degree 2. The next example shows that we can still try applying the grouping technique to a four-term polynomial regardless of degree.

Example 5

Factor [latex]4x^{3}+12x^{2}+x+3[/latex].

Another feature we may encounter is a polynomial with more than one variable, as shown in the next two examples.

Example 6

Factor [latex]6pq+9qr-2p-3r[/latex]

Example 7

Factor [latex]x^2y-xy^2-2x+2y[/latex]

 

Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

Example 8

Factor [latex]7x^{2}–21x+5x–5[/latex].

In the example above, each pair can be factored, but then there is no common factor between the pairs! Despite factoring each pair, that does not result in a factoring of the original polynomial. Don’t forget that “to factor” means to “express as a product.” Where we got stuck, the two separate pieces were still attached by addition. That is why the final answer is “prime.”

Now that you are an expert on grouping, we will see in the next two sections how this technique can help us to factor trinomials.