
Simple Polynomial Factoring (page 2 of 3) Sections: Simple factoring, Factoring "in pairs"
I can factor an "x" and a "y" out of each term: x^{2}y^{3 }= xy(x^{1}y^{2}) = xy(xy^{2}) and xy = xy(1). x^{2}y^{3} + xy = xy( ) = xy(xy^{2} ) = xy(xy^{2} + 1) Remember: When "nothing" is left after factoring, a "1" is left behind in the parentheses.
I can factor a "3"
and an "x"
out of each term: 3x^{3} = 3x(x^{2}),
6x^{2} = 3x(2x),
and 3x^{3} + 6x^{2} – 15x = 3x( ) = 3x(x^{2} ) = 3x(x^{2} + 2x ) = 3x(x^{2} + 2x – 5) You can use the Mathway widget below to practice finding the GCF of the terms of a polynomial. Try the entered exercise, or type in your own exercise. Then click the "paperairplane" button to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)
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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)
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".... << Previous Top  1  2  3  Return to Index Next >>


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