How do you factor $49 a^{2} - 81 b^{2}$ $49a^2 - 81b^2$ ?

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How do you factor $49 a^{2} - 81 b^{2}$ $49a^2 - 81b^2$ ?

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Answer 1 · 2015-07-09T13:57:53

You may notice that $49 = 7^{2} and 81 = 9^{2}$ $49=7^2and81=9^2$ are both squares.

Explanation:

So $49 a^{2} = {(7 a)}^{2} and 81 b^{2} = {(9 b)}^{2}$ $49a^2=(7a)^2and81b^2=(9b)^2$ are also squares, and then we can use the special product of the form: $x^{2} - y^{2} = (x + y) (x - y)$ $x^2-y^2=(x+y)(x-y)$

$= (7 a + 9 b) (7 a - 9 b)$ $=(7a+9b)(7a-9b)$

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How do you factor $49 a^{2} - 81 b^{2}$ $49a^2 - 81b^2$ ?

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