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)

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This may look different from what I've done above, but really it's not. The two terms, 2(xy) and b(xy), do indeed have a common factor; namely, the parenthetical factor xy. 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(xy) – b(xy) = (xy)(           )

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

2(xy) – b(xy) = (xy)(2         )

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

2(xy) – b(xy) = (xy)(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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 Cite this article as: Stapel, Elizabeth. "Simple Polynomial Factoring." Purplemath. Available from     http://www.purplemath.com/modules/simpfact2.htm. Accessed [Date] [Month] 2016

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