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 a "difference" means a "subtraction". So a difference of squares is something that looks like x^{2} – 4. That's because 4 = 2^{2}, so we really have x^{2} – 2^{2}, which is a difference of squares.
To factor this, I'll start by writing my parentheses, in the same way as usual for factoring:
x^{2} – 4 = (x )(x )
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For this quadratic factorization, I need factors of –4 that add up to zero, so I'll use –2 and +2:
x^{2} – 4 = (x – 2)(x + 2)
(Review Factoring Quadratics, if the steps in this example didn't make sense to you.)
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Note that we had x^{2} – 2^{2}, and ended up with (x – 2)(x + 2). Differences of squares (being something squared minus something else squared) always work this way:
For a^{2} – b^{2}, I start by doing the parentheses:
( )( )
Then I put the first squared thing in front:
(a )(a )
...and I put the second squared thing in back:
(a b)(a b)
...and then I alternate the signs in the middles:
(a – b)(a + b)
Because the factoring always works out exactly the same way, we can turn it into a formula:
Difference-of-Squares Formula:
For a difference of squares a^{2} – b^{2}, the factorization is:
(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 this formula for your next test.
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By the way, no, the order of the factors doesn't matter. Since multiplication is commutative (that is, since you can move the factors around without changing the value of the product), the difference of squares can also be stated as:
(a + b)(a – b)
Don't get hung up on the order of the factors. Either way is fine.
Here are examples of some typical homework problems:
This quadratic can be restated as x^{2} – 4^{2}, which is a difference of squares. Applying the formula, I get:
x^{2} – 16 = x^{2} – 4^{2}
= (x – 4)(x + 4)
This quadratic is (2x)^{2} – 5^{2} so, applying the formula, I get:
4x^{2} – 25 = (2x)^{2} – 5^{2}
= (2x – 5)(2x + 5)
This can be restated as (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, applying the formula, I get:
x^{4} – 1 = (x^{2})^{2} – 1^{2}
= (x^{2} – 1)(x^{2} + 1)
Note that I'm not done yet, because one of the factors I got — namely, the x^{2} – 1 factor — is itself a difference of squares, so I need to apply the formula again to get the fully-factored form. Since x^{2} – 1 = x^{2} – 1^{2} = (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.
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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).
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