How do you factor $9x^2 - 8y^2$ ?

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Answer 1 · 2015-07-13T00:21:33

$9x^2-8y^2 = (3x-2sqrt(2)y)(3x+2sqrt(2)y)$

This is a difference of squares, but only with the help of irrational coefficients.

$9x^2-8y^2 = (3x)^2 - (sqrt(8)y)^2 = (3x-sqrt(8)y)(3x+sqrt(8)y)$

using the difference of squares identity:

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

with $a=3x$ and $b=sqrt(8)y$

Note further that $sqrt(8) = sqrt(4*2) = sqrt(4)sqrt(2) = 2sqrt(2)$

So we can express the factorisation as

$9x^2-8y^2 = (3x-2sqrt(2)y)(3x+2sqrt(2)y)$

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