Factoring by grouping is a valuable strategy for factoring expressions. This process can be applied to a variety of problems and is popular in college mathematics textbooks. This is definitely a strategy you need to remember. You will see in the Polynomial Equations section that factoring by grouping may also be used when solving polynomial equations.

In Algebra 1, factoring by grouping was introduced in relation to quadratic expressions (ax² + bx + c).
In Algebra 2, factoring by grouping will be applied to more diverse expressions with usually four terms.

Steps for Factoring by Grouping:
1. Look for a Greatest Common Factor (GCF) among all the terms. If a GCF exists, factor it out. Remember to include the GCF in your final answer. Proceed.
2. Create smaller groups within the problem. This may be as simple as grouping the first two terms and grouping the last two terms, or it may require rearranging the terms. The goal is to create equal expressions within the factored parentheses from each grouping.
3. Factor out the GCF from both groupings. It may be necessary to factor out a negative value from the second term. Be careful of the signs.
4. At this point, the values inside the grouped parentheses should be the same. Factor out this value. If you are not finding this "same" parentheses value, rearrange the four terms and try again.
5. Determine if the remaining factors can be factored further. If no further factoring is possible, you have arrived at the final answer.
Factoring by grouping is not applicable in all situations.

Let's take a look at some examples:

	Factor: cx + cy + dx + dy
There is no GCF. The first two terms contain a common factor of "c". The second two terms contain a common factor of "d".	cx + cy + dx + dy
Group the terms to factor out c and d.	(cx + cy) + (dx + dy)
Factor out c and d.	c(x + y) + d(x + y)
Notice that the parentheses in both groups contain the same values, (x + y). Factor out this common parentheses.	(x + y)(c + d)

	Factor: 5x3 + 15x2 - 10x - 30
Factor out the GCF, greatest common factor, of 5.	5[x3 + 3x2 - 2x - 6]
Group the first two terms. Group the second two terms after factoring out a -1.	5[(x3 + 3x2) -1(2x + 6)]
Factor out x2 from the first two grouped terms. Factor out -2 from the second two grouped terms.	5[x2(x + 3) -1•2(x + 3 )] 5[x2(x + 3) - 2(x + 3)]
Notice that the parentheses in both groups contain the same value of (x + 3). Factor out the common parentheses.	5(x + 3)(x2 - 2)

It may be the case in the second grouping, that you may be able to factor out either a positive value or a negative value. To determine which is needed in the problem, look at the signs before the second and fourth terms. If the two signs are the same (either both positive or both negative), factor out a positive value. If the two signs are different, factor out a negative value.

	Factor: x2 + mn - mx - nx
There is no GCF. Group the first two terms and the second two terms.	(x2 + mn) + (-mx - nx)
Factor each grouping. Whoa!! The values within the parentheses are not the same. We will not be able to factor this problem with this approach.	1(x2 + mn) - x(m + n)
You may need to rearrange the terms from the original problem to form different groupings. Remember the goal is to create equal expressions within the parentheses.	Start again! Rearrange the terms in the original problem. Then form groups. (x2 - mx) + ( - nx + mn)
Factor each new grouping.	x(x - m) - n(x - m)
Factor out the common parentheses of (x - m).	(x - m)(x - n)

It may be the case that if you group the first two terms and group the second two terms, you will NOT arrive at a common expression within the parentheses after factoring the groups. If this happens, "rearrange" the terms in the original expression. Remember, the goal is to create equal expressions within the factored parentheses.

	Factor: 5x3 + 7x2 - 20x - 28
There is no GCF, so group the terms. Be careful of the negative sign in front of the second grouping.	(5x3 + 7x2) - (20x + 28)
Factor the groupings.	x2(5x + 7) - 4(5x + 7)
Factor out the common parentheses of (5x + 7).	(5x + 7)(x2 - 4)
Be careful! You are not finished. The (x2 - 4) can be factored further!	(5x + 7)(x - 2)(x + 2)

Be careful! After applying the factoring by grouping strategy, it may be necessary to factor your answer even further. Always LOOK at your final answer to be sure it has been factored completely.

	Factor: a4b - 2a3 + a4 - 2a3b
Factor out the GCF, greatest common factor, of a3.	a3(ab - 2 + a - 2b)
Look ahead! Re-group to create the same value in the grouped parentheses..	a3(ab - 2b + a - 2) a3[(ab - 2b) + (a - 2)]
Factor the groupings.	a3[b(a - 2) + 1(a - 2)]
Factor out the common parentheses.	a3[(a - 2)(b + 1)]
Final answer.	a3(a - 2)(b + 1)

To see how factoring by grouping can be used to solve polynomial equations,
see the section on Polynomial Equations.