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Simple Polynomial Factoring (page 2 of 3)

Sections: Simple factoring, Factoring "in pairs"

• 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)

Remember: When "nothing" is left after factoring, a "1" is left behind in the parentheses.

• 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)

You can use the Mathway widget below to practice "Factoring Polynomials", subtopic "Factoring out GCF". Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)

To activate, click in the "Enter Problem" box below:

(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free seven-day trial of the software.)

• Factor 2(x – y) – b(x – y). Copyright © Elizabeth Stapel 2002-2011 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", the 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"....

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