Print-friendly page

Simple Polynomial Factoring (page 1 of 2)

Sections: Simple factoring, Factoring "in pairs"

Factoring polynomial expressions is not quite the same as factoring numbers, but the concept is very similar. When factoring numbers or factoring polynomials, you are finding numbers or polynomials that divide out evenly from the original numbers or polynomials. But in the case of polynomials, you are dividing numbers and variables out of expressions, not just dividing numbers out of numbers.

Previously, you have simplified expressions by distributing through parentheses, such as:

2(x + 3) = 2(x) + 2(3) = 2x + 6

Simple factoring in the context of polynomial expressions is backwards from distributing. That is, instead of multiplying something through a parentheses, you will be seeing what you can take back out and put in front of a parentheses, such as:

2x + 6 = 2(x) + 2(3) = 2(x + 3)

The trick is to see what can be factored out of every term in the expression. Warning: Don't make the mistake of thinking that "factoring" means "dividing something off and making it magically disappear". Remember that "factoring" means "dividing out and putting in front of the parentheses". Nothing "disappears" when you factor; things merely get rearranged.

• Factor 3x – 12.

The only thing common between the two terms (that is, the only thing that can be divided out of each term and then moved up front) is a "3". So I'll factor this number out to the front:

3x – 12 = 3(          )

When I divided the "3" out of the "3x", I was left with only the "x" remaining. I'll put that "x" as my first term inside the parentheses:

3x – 12 = 3(x         )

When I divided the "3" out of the "–12", I left a "–4" behind, so I'll put that in the parentheses, too:

3x – 12 = 3(x – 4)

This is my final answer:  3(x – 4)

Warning: Be careful not to drop "minus" signs when you factor.

Some books teach this topic by using the concept of the Greatest Common Factor, or GCF. In that case, you would methodically find the GCF of all the terms in the expression, put this in front of the parentheses, and then divide each term by the GCF and put the resulting expression inside the parentheses. The result will be the same. But this seems like an awful lot of work to me, so I just go straight to the factoring.

Here are some more examples:

• Factor 7x – 7.

A "7" can come out of each term, so I'll factor this out front:

7x – 7 = 7(         )

Dividing the 7 out of "7x" leaves just an "x":

7x – 7 = 7(x       )

What am I left with when I divide the 7 out of the second term? Well, if "nothing" is left, then "1" is left. (Remember: 7 ÷ 7 = 1.) So I get:

7x – 7 = 7(x – 1)

Take careful note: When "nothing" is left after factoring, a "1" is left.

• Factor 12y² – 5y.

In this case, no number is a common factor between the two terms (specifcally, the 12 and the 5 share no common numerical factor), but I can still divide out a common variable factor of "y" from each of the two terms.

12y² – 5y = y(          )

In the first term, I have the "12" and the other "y" factor left over:

12y² – 5y = y(12y     )

In the second term, I have the "5" left over:

12y² – 5y = y(12y – 5)

Don't forget the "minus" sign in the middle!

• Factor x²y³ + xy

I can factor an "x" and a "y" out of each term: x²y³ = xy(x¹y²) = xy(xy²) and xy = xy(1).

x²y³ + xy = xy(           )

= xy(xy²        )

= xy(xy² + 1)

• Factor 3x³ + 6x² – 15x.

I can factor a "3" and an "x" out of each term: 3x³ = 3x(x²), 6x² = 3x(2x), and
–15x = 3x(–5). Being careful of my signs, I get:

3x³ + 6x² – 15x = 3x(                    )

= 3x(x²               )

= 3x(x² + 2x      )

= 3x(x² + 2x – 5)

• Factor 2(x – y) – b(x – y). Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

This may look different from what I've done above, but really it's not. The two terms, 2(x – y) and – b(x – y), do indeed have a common factor; namely, the parenthetical factor x – y. This may be different from what you're used to seeing referred to as being a "factor", but the factorization process works just the same as before.

First, I'll take the common factor out front:

2(x – y) – b(x – y) = (x – y)(           )

From the first term, I have a "2" left over:

2(x – y) – b(x – y) = (x – y)(2         )

From the second term, I have a "–b" left over:

2(x – y) – b(x – y) = (x – y)(2 – b)

• Factor x(x – 2) + 3(2 – x).

This is almost the same as the previous case, but not quite, because "x – 2" is not quite the same as "2 – x". If I'd had "x + 2" and "2 + x", they factors would have been the same, because order doesn't matter for addition. But order does matter for subtraction, so I don't actually have a common factor here.

But I would have a common factor if I could just flip (or "reverse the order of") that subtraction. What would happen if I did that? Take a look at the following numerical subtraction:

5 – 3 = 2

3 – 5 = –2

When I flipped the subtraction in the second line, I got the same answer except that the sign had changed. This is always true: When you flip a subtraction, you also change the sign out front. In our case, this means:

x(x – 2) + 3(2 – x) = x(x – 2) – 3(x – 2)

By reversing the subtraction in the second parenthetical, I have created a common factor, so I can now proceed as I had in the previous example:

x(x – 2) + 3(2 – x) = x(x – 2) – 3(x – 2)

= (x – 2)(x – 3)

These examples lead us to the next topic: factoring "in pairs"....

Top  |  1 | 2  |  Return to Index  Next >>

  Copyright © 2002-2008  Elizabeth Stapel   |   About   |   Terms of Use

 Feedback   |   Error?