# Factoring by Grouping

When we use FOIL to multiply two binomials, we multiply the first, outside, inside, and last terms to get four products:

(*x* + *y*)(*a* + *b*) =* ax* + *bx* + *ay* + *by*.

Sometimes we have no choice but to use this method, but other times we can simplify and arrive at a final answer with a smaller number of terms instead. Since we know how big you are on smaller numbers of things, we will let you know how.

Factoring by grouping is like "undistributing" or "unFOILing." You pictured a baked potato too, huh?

The simplest situation in which we can factor by grouping is when we have a 4-term polynomial whose first two terms have their own common factor, and whose last two terms have their own common factor. Confused? Check out the examples...they will unwrap the mystery. Then all that will be left to do is to slap on a pat of butter and sprinkle on some bacon bits.

### Sample Problem

Factor by grouping: 6*x*^{2} + 3*x* + 20*x* + 10.

This is a polynomial written with four terms that do not have a single common factor among them. However, the first two terms have a common factor (3*x*), and the last two terms have a common factor (10). This situation doesn't answer all of our wildest factoring dreams, but we'll take it.

By pulling out the common factors for each pair of terms, we can rewrite the original polynomial as

3*x*(2*x* + 1) + 10(2*x* + 1).

These two terms now have a common factor of (2*x* + 1). Seems like we should be able to do something with that information, don't you think? In fact, we can pull out this common factor and rewrite the polynomial as

(3*x* + 10)(2*x* + 1).

### Sample Problem

Factor the polynomial *x*^{2} + 3*x* + 2*x* + 6 by grouping.

Once again, the terms do not all have a common factor, but the first two terms have a common factor of *x* and the last two terms have a common factor of 2. Looks like we'll be doing the same thing here that we did above. Will all this grouping we're doing, we wonder if we should look for a Groupon.

We can rewrite the polynomial as

*x*(*x* + 3) + 2(*x* + 3)

and then pull out the factor (*x* + 3) to find

(*x* + 2)(*x* + 3).

In some cases the terms are out of order. ("Bailiff, arrest those terms!") The first and third term might have a common factor, and the second and fourth term might have a common factor. In this case we need to rearrange the terms before we factor. That's fine; we were in the mood to redecorate anyway. Now which one would look good over here by the mantel?

We can also use grouping to factor polynomials that don't necessarily have degree 2. We'll show you how to do this. It's our way of doing you a "degree solid."

### Sample Problem

Use grouping to factor the polynomial 2*y*^{3} + *y*^{2} + 8*y*^{2} + 4*y*.

We pull out the common factor *y*^{2} from the first two terms and the common factor 4*y* from the second two terms to get

*y*^{2}(2*y* + 1) + 4*y*(2*y* + 1).

Then we factor out (2*y* + 1) from each term and obtain

(*y*^{2} + 4*y*)(2*y* + 1).

When the polynomial isn't written nicely with 4 terms but is instead written as a trinomial, we can still factor by grouping. The trick is to split up the middle term appropriately so we can factor by grouping like we've been doing. If you have trouble splitting the middle term appropriately, King Solomon should be able to step in and give you a hand.

### Sample Problem

Factor the trinomial 2*x*^{2} + 17*x* + 30 by grouping.

This isn't written with 4 terms, so we need to split the term 17*x* into two terms. Let's use

17*x* = 5*x* + 12*x*.

Don't worry, we'll explain how we got this in a moment. You sure? You look worried...

We can now rewrite the trinomial as

2*x*^{2} + 5*x* + 12*x* + 30

and factor as

*x*(2*x* + 5) + 6(2*x* + 5) = (*x* + 6)(2*x* + 5).

You still look worried, so we'll tell you now how we arrived at those two particular terms. You can unclench.

For a trinomial of the form a*x*^{2} + *bx* + *c*, find two numbers whose product is *ac* and whose sum is *b*, and use those numbers to split the middle term. Sounds like a riddle, but this rule actually does work.

Applying this rule to the previous example, we needed two numbers whose product was 2 × 30 = 60 and whose sum was 17. By going through all the possible ways to produce 60 as a product of 2 numbers and looking at the sum of those two numbers, we found that 5 and 12 were the numbers we wanted. No need to be overwhelmed, there weren't *that* many possible ways. This process won't take up your whole afternoon. You'll still be able to make it to the mall before it closes.

The work went like this:

1 + 60 | = | 61 | |

2 + 30 | = | 32 | |

3 + 20 | = | 23 | |

4 + 15 | = | 19 | |

5 + 12 | = | 17 | Bingo! |

### Sample Problem

Factor by grouping: 4*x*^{2} + 13*x* + 9.

We need to split the term 13*x* into two terms. Using the handy-dandy tool we picked up, the coefficients of the two terms will be two numbers whose product is 4 × 9 = 36 and whose sum is 13. If you haven't figured it out already, you're thinking too hard. We don't need to do much work for this one: since 4 + 9 = 13, we already have our factors before we've even begun. Dear algebra, please try to be more like this example. Thank you.

Rewrite the original polynomial as

4*x*^{2} + 4*x* + 9*x* + 9

and then factor as

4*x*(*x* + 1) + 9(*x* + 1) = (4*x* + 9)(*x* + 1).

## Factoring by Grouping: The Real Story

The textbook rule says that, to factor a polynomial of the form *ax*^{2} + *bx* + *c*, we find two numbers whose product is *ac* and whose sum is *b*, and use these two numbers to break up the term *bx*. Why does this work? While we're at it, what is the meaning of life? Whoa...we'll start with polynomials and work our way up.

The explanation will require a lot of variables, so bear with us. If we use FOIL to multiply the binomials

(*px* + *s*)(*rx* + *t*)

where *p*, *r*, *s* and *t* are integers, we find

(*pr*)*x*^{2} + (*pt*)*x* + (*sr*)*x* + (*st*)

Need proof? By letting

*a* = *pr**b* = *pt* + *sr*, and*c* = *st*

We can abbreviate the polynomial (*pr*)*x*^{2} + (*pt*)*x* + (*sr*)*x* + *st* as a*x*^{2} + *bx* + *c*.

So far, all we've done is multiply two binomials and abbreviate the coefficients in the *pr*oduct so we can write it as a trinomial. You know what they say: three nomials are better than two.

To factor by grouping, we need to think about doing the above process backwards. Instead of using *b* to abbreviate something, we're given the polynomial a*x*^{2} + *bx* + *c* and asked to find out what *b* is abbreviating. We want to break *b* up into two numbers, but not any two numbers. We want two numbers that we can use to factor by grouping. The Oracle is closed on Tuesdays, so we need to find a different way to figure out which two to use.

We assume that the polynomial a*x*^{2} + *bx* + *c* came from multiplying two binomials of the form (*px* + *s*)(*rx* + *t*), but we don't know what *p*, *s*, *r* or *t* are.

Here's what the *ac* trick is for. Since multiplication is commutative,

*ac* = (*pr*)(*st*) = (*pt*)(*sr*).

The factors *pt* and *sr* multiply to give *ac*, and add to give *b*. Those are the numbers we'll find when we follow the textbook rule for factoring by grouping, which says to look for the numbers that multiply to give *ac* and add to give *b*.

After we've found those numbers, we can rewrite the polynomial

a*x*^{2} + *bx* + *c*

as

*ax*^{2} + (*pt*)*x* + (*sr*)*x* + *c*.

Since *a* is (*pr*) and *c* is (*st*) (they couldn't fool us...we knew all along), we've now rewritten the original polynomial as

(*pr*)*x*^{2} + (*pt*)x + (*sr*)x + (*st*).

This we can factor by grouping:

*px*(*rx* + *t*) + *s*(*rx* + t) = (*px* + *s*)(*rx* + *t*).

We have now factored the polynomial.