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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 similar. In both cases, you are finding things that divide out evenly. But in this case, you are dividing numbers and variables out of expressions, not just 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. Just don't make the mistake of thinking that "factoring" means "dividing off and making disappear". Instead, remember that "factoring" means "dividing out to in front of the parentheses". Nothing disappears when you factor; things merely get rearranged. Here is an example of how to factor:

• Factor 3x – 12.

The only thing common between the two terms (that is, the only thing that can be divided out front) is a "3". So factor this out front:

3x – 12 = 3(          )

Now, when you divided the "3" out of the "3x", this left only the "x" behind, so put that in for the first term inside the parentheses:

3x – 12 = 3(x         )

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

3x – 12 = 3(x – 4)

This is your final answer:  3(x – 4)

Make sure you are careful not to drop "minus" signs when you factor.

Some books teach this topic by using the Greatest Common Factor, or GCF. In that case, you would 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 factor this out front:

7x – 7 = 7(         )

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

7x – 7 = 7(x       )

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

7x – 7 = 7(x – 1)

Remember the point from that last paragraph: 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, but you can divide out a common factor of "y".

12y² – 5y = y(          )

In the first term, you'll have the "12" and the other "y" factor left over:

12y² – 5y = y(12y     )

In the second term, you'll have the "5" left over:

12y² – 5y = y(12y – 5)

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

• Factor x²y³ + xy

You can factor an "x" and a "y" out of each term:

x²y³ + xy = xy(           )

= xy(xy²        )

= xy(xy² + 1)

• Factor 3x³ + 6x² – 15x.

You can factor a "3" and an "x" out of each term. Being careful of 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: x – y. This is different from what you're used to seeing as a "factor", but the process works just the same as before.

Take the common factor out front:

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

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

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

From the second term, you 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 the same as "2 – x". If we'd had "x + 2" and "2 + x", they would have been the same, because order doesn't matter for addition. But order does matter for subtraction, so we don't actually have a common factor here. But we would have a common factor if we could just flip the subtraction. What would happen if we did this? Take a look at this:

5 – 3 = 2

3 – 5 = –2

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

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

Now I do have a common factor, and I can proceed as I did 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".

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