# Factoring by Grouping

## Factoring by Grouping

Factoring by grouping is a technique used in a very specific case, and is likely the technique you'll use the least in this section. However, when it's applicable, it gets the job done, and fast.

It works best when you're given a polynomial whose terms don't *all* have a greatest common factor, but perhaps *some* of them do have factors in common. Basically, you'll group the large polynomial into two smaller parts, both of which contain terms with common factors. Then, you'll factor both of those smaller groups individually. Sound confusing? Probably. However, when you see it at work in an example, it's easier to understand what's going on.

**Example 2**: Factor the polynomial 2*x*^{3} - 4*x*^{2} - 3*x* + 6.

**Solution**: Unfortunately, these terms don't have any factor in common (except 1, and that's true of any group of terms). However, look at the polynomial as two groups, each containing two terms.

- (2
*x*^{3}- 4*x*^{2}) + (-3*x*+ 6)

##### Kelley's Cautions

The key to ensuring that factoring by grouping works right is getting the factored binomials to match; in Example 2, both binomials were (*x* - 2). Basically, this means you should decide whether to factor out the GCF or its opposite from the second group to make them match; if you don't, you won't be able to finish the problem.

Notice that the left group has a GCF of 2*x*^{2}; go ahead and factor that out and pay special attention to the binomial that results.

- 2
*x*^{2}(*x*- 2) + (-3*x*+ 6)

##### You've Got Problems

Problem 2: Factor the polynomial 12*x*^{4} + 6*x*^{3} + 14*x* + 7.

You can factor a 3 out of the second group of terms, but it's more useful to factor out a -3. Why? If you do, the binomial that results *exactly matches* the binomial you got when you factored the first group.

- 2
*x*^{2}(*x*- 2) - 3(*x*- 2)

Here's the part that sometimes confuses students. You should now factor out the binomial (*x* - 2) from both terms. At first it feels weird factoring out something other than a monomial, but it's definitely allowed. When you pull that (*x* - 2) out, all you're left with in the first term is the 2*x*^{2}, which was hanging out front, and in the second term, just the -3 remains. So, write the binomial, and pop the leftovers in a set of parentheses after it, like so:

- (
*x*- 2)(2*x*^{2}- 3)

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with **Alpha Books**, a member of Penguin Group (USA) Inc.

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