Factoring Polynomials: GCF and Factoring by Grouping

In a blog post last week, we talked about how to use the FOIL method to multiply out an expression like (3x+4)(2y-7) into a simpler form like 6xy-21x+8y-28. But what happens if you need to go the other direction - what if you have something like 8xy-10x+4y-5 and need to figure out what would multiply to get that?

Finding which things multiply together to get a given result is called factoring. When you're factoring a polynomial, you're basically undoing the distributive property: finding what could have been distributed through parentheses to get the terms we currently have.

The easiest type of factoring is to find some amount that is common to all the terms, which means it can successfully be divided out from each term. This is called "Greatest Common Factor factoring" or "GCF Factoring" for short.

For example, if you have $15xy + 10x^2 - 25xyz$, all three terms have a multiple of 5 and have an x in them, so everything is divisible by 5x. We can therefore "undo the distributive property" by pulling 5x to the outside of parentheses: 5x(__+__+__). To find what goes in the blanks, we will divide each of the individual terms by 5x. $15xy ÷ 5x$ gives us 3y, $10x^2 ÷ 5x$ gives us 2x, and $-25xyz ÷ 5x$ gives us -5yz. Therefore, with 5x factored out, we have $5x(3y+2x-5yz)$.

Unfortunately, GCF factoring doesn't always work, particularly in situations where the polynomial came from "double distributing." For example, last week we found that multiplying out $(5x-7)(8y+4)$ would result in $40xy - 56y + 20x - 28.$. If we wanted to factor $40xy - 56y + 20x - 28$ to get back to $(5x-7)(8y+4)$ , factoring out the GCF wouldn't work, because each of the original terms in the multiplication problem is only used to get two of the four terms in the result: for example, 5x is only in 40xy and 20x, not in -56y or -28.

To compensate for double-distributing, we can use the factoring by grouping method. Since each original factor is included in exactly two of the result terms, we can split the result into two groups and do the GCF factoring on each group separately. Let's see how that looks on $40xy - 56y + 20x - 28$.

First, we'll split the polynomial into two groups by using parentheses: one group with the first two terms, and another group with the last two terms: $(40xy - 56y) + (20x - 28).$ Next, we factor out the greatest common factor of each group.

On the first group, 40 and -56 are each a multiple of 8, and both terms have a y in them, so we can factor out a GCF of 8y, which leaves behind 5x on the first term and -7 on the second term: 8y(5x-7). We repeat the process on group 2: 20x and -28 are both multiples of 4, and pulling it out gives us 4(5x-7). Putting the polynomial back together, we have 8y(5x-7)+4(5x-7). Notice that we ended up having (5x-7) on both - that's why factoring by grouping works! Since both groups are a multiple of (5x-7), that means (5x-7) is a common factor that can be factored out, giving us $(5x-7)(8y+4)$, exactly as it should have been.

Let's try another factoring by grouping problem where we don't already know the answer from last week: $8x² - 5xy - 24xz +15zy$. As before, we'll start by splitting it into two groups: $(8x² - 5xy) + (- 24xz +15zy)$. For the first group, the only thing they have in common is an x, so it becomes $x(8x -5y)$. On the second group, they both have a z, and -24 and 15 are both multiples of 3, so we could pull out 3z. However, that would give us $3z(-8x+5y)$, which means the "left behind" portions in parentheses don't match on the two groups! Instead, let's factor out a negative 3z, giving us $-3z(8x-5y)$, so the "leftovers" are matching like we need them to be. The key here is that the GCF for a group should have the same sign as the first term in the group - if the first term of the group is negative, the GCF is also negative. Putting the groups together, we now have $x(8x -5y)-3z(8x-5y)$, so we can factor out $(8x-5y)$ to get $(8x-5y)(x-3z)$.

To sum up: to do factoring by grouping, you first split the polynomial into two equally-sized groups. Next, factor out the GCF of each group, giving each the same sign as the first term in its group. If you did it correctly, the "leftovers" in parentheses from both groups should be the same. Finally, factor out the matching "leftovers," putting the two GCF's together into a second set of parentheses.