# Polynomials: Factoring by Grouping

Factoring polynomials is an important step in solving polynomial equations. Getting a polynomial into binomials is one of the easiest ways to solve for x-intercepts as you can set each binomial equal to zero and find multiple answers.

We have learned how to factor out a constant and how to factor different types of trinomials into binomials, but what about when you have four terms in a polynomial?

One possibility is factoring by grouping. Sometimes you can factor a number or a variable out of every term in a polynomial. (So, $x^3+3x^2+4x\rightarrow x(3x^2+3x+4)$). When you factor out a term, you group the other terms in parentheses and put the factored term in front, setting up a distributive property expression.

Often, however, you can only factor a number or variable out of some of the terms in a polynomial. In these cases, it can help to "factor by grouping." The associative property says that, in addition problems (and all polynomials are actually long addition problems with positive and negative numbers), you can group the terms however you like. So, try grouping pairs of terms from which you can factor out a number, variable, or term.

*Example*:

$\eqalign{3x^2+6x-8x-16&=0\\(3x^2+6x)+(-8x-16)&=0\quad\text{Break polynomial into groups}}$

In this example, you will know that you can pull a 3x out of the first terms and a -8 out of the net two terms, so group those pairs together. (Note, the negative sign stuck to the 8 when we grouped it, if we had left it outside the group, distributive property would have also affected the 16, so best to join your two groups with an addition sign and just keep the signs as they are on the original numbers).

Factoring by grouping gets really interesting when, after you factor a different term out of the different groups, you see that the binomial you have left, in both groups, is the same!

*Example*:

$\eqalign{3x^2+6x-8x-16&=0\\(3x^2+6x)+(-8x-16)&=0\quad\text{Break polynomial into groups}\\3x(x+2)-8(x+2)&=0\quad\text{Factor what you can from each group}\\3x\mathbf{(x+2)}-8\mathbf{(x+2)}&=0\quad\text{The binomial is identical}}$

As you can see here, both of the groups contain the terms you factored out (3x and -8) and a common binomial (x+2).

The identical binomial in each group means that you could have actually* factored out* that binomial, you just couldn't recognize it! So, now that you can see it, you can factor the binomial out, put it in the front of the expression (just like you would if you pulled out a single term and wanted to put it in front like a distributive term) and multiply that binomial times the remaining terms. Put those remaining terms together as a binomial and, voila, you have two binomials.

*Example*:

$\eqalign{3x^2+6x-8x-16&=0\\(3x^2+6x)+(-8x-16)&=0\quad\text{Break polynomial into groups}\\3x(x+2)-8(x+2)&=0\quad\text{Factor what you can from each group}\\3x\mathbf{(x+2)}-8\mathbf{(x+2)}&=0\quad\text{The binomial is identical, factor it out and put it in front}\\(x+2)(3x-8)&=0\quad\text{Polynomial factored into binomials}}$

Not every polynomial will factor nicely by grouping, but it's always worth checking out!

#### Practice Problems:

## Polynomials: Factoring by Grouping

Factor the following polynomials into binomials by grouping:

- $2x^2+6x+7x+21$
- $3x^2-9x-8x+24$
- $4x^2-28x+12x-84$
- $2x^2+26x-3x-39$
- $16x^2+48x-14x-42$
- $x^3-3x^2-4x+12$
- $x^3+2x^2+3x+6$
- $2x^3+2x^2-2x-2$
- $3x^3+27x^2-x-9$
- $2x^5-4x^4-3x+6$

#### Answer Key: