How do you factor $20 r^{4} - 45 n^{4}$ $20r^4-45n^4$ ?

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How do you factor $20 r^{4} - 45 n^{4}$ $20r^4-45n^4$ ?

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Answer 1 · 2017-02-27T20:04:20

We can factor this using the following rule:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2 - b^2 = (a - b)(a + b)$

Let $a^{2} = 20 r^{4}$ $a^2 = 20r^4$ therefore $\sqrt{a^{2}} = a = \sqrt{20 r^{4}} = \sqrt{20} r^{2}$ $sqrt(a^2) = a = sqrt(20r^4) = sqrt(20)r^2$

Let $b^{2} = 45 n^{4}$ $b^2 = 45n^4$ therefore $\sqrt{b^{2}} = b = \sqrt{45 n^{4}} = \sqrt{45} n^{2}$ $sqrt(b^2) = b = sqrt(45n^4) = sqrt(45)n^2$

Substituting gives:

$20 r^{4} - 45 n^{4} = (\sqrt{20} r^{2} - \sqrt{45} n^{2}) (\sqrt{20} r^{2} + \sqrt{45} n^{2})$ $20r^4 - 45n^4 = (sqrt(20)r^2 - sqrt(45)n^2)(sqrt(20)r^2 + sqrt(45)n^2)$

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How do you factor $20 r^{4} - 45 n^{4}$ $20r^4-45n^4$ ?

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