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Factoring: Sections: Differences of squares, Sums and differences of cubes, Recognizing patterns When you learn to factor quadratics, there are three other formulas that they usually introduce at the same time. The first is the "difference of squares" formula. Remember from your translation skills that "difference" means "subtraction". So a difference of squares is something that looks like x^{2} – 4. That's because 4 = 2^{2}, so you really have x^{2} – 2^{2}, a difference of squares. To factor this, do your parentheses, same as usual: x^{2} – 4 = (x )(x ) You need factors of –4 that add up to zero, so use –2 and +2: x^{2} – 4 = (x – 2)(x + 2) (Review Factoring Quadratics, if this example didn't make sense to you.) Note that we had x^{2} – 2^{2}, and ended up with (x – 2)(x + 2). Differences of squares (something squared minus something else squared) always work this way:
For a^{2} – b^{2}, do the parentheses: ( )( ) ...put the first squared thing in front: (a )(a ) ...put the second squared thing in back: (a b)(a b) ...and alternate the signs in the middles: (a – b)(a + b) Memorize this formula! It will come in handy later, especially when you get to rational expressions (polynomial fractions), and you'll probably be expected to know the formula for your next test. Here are examples of some typical homework problems:
This is x^{2} – 4^{2}, so I get: x^{2} – 16 = x^{2} – 4^{2} = (x – 4)(x + 4)
This is (2x)^{2} – 5^{2}, so I get: 4x^{2} – 25 = (2x)^{2} – 5^{2} = (2x – 5)(2x + 5)
This is (3x^{3})^{2} – (y^{4})^{2}, so I get: 9x^{6} – y^{8} = (3x^{3})^{2} – (y^{4})^{2} = (3x^{3} – y^{4})(3x^{3} + y^{4})
This is (x^{2})^{2} – 1^{2}, so I get: x^{4} – 1 = (x^{2})^{2} – 1^{2} = (x^{2} – 1)(x^{2} + 1) Note that I'm not done yet, because x^{2} – 1 is itself a difference of squares, so I need to apply the formula again to get the fullyfactored form. Since x^{2} – 1 = (x – 1)(x + 1), then: x^{4} – 1 = (x^{2})^{2} – 1^{2} = (x^{2} – 1)(x^{2} + 1) = ((x)^{2} – (1)^{2})(x^{2} + 1) = (x – 1)(x + 1)(x^{2} + 1) The answer to this last exercise depended on the fact that 1, to any power at all, is still just 1. Warning: Never forget that this formula is for the difference of squares (with variables); the polynomial sum of squares is always prime (that is, it can't be factored with whole numbers or fractions). You can use the Mathway widget below to practice factoring a difference of squares. 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.)
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.) Top  1  2  3  Return to Index Next >>


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