How do you factor $32 x^{2} y - 8 y$ $32x^2y - 8y$ ?

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Answer 1 · 2015-05-06T12:14:35

You can collect $8 y$ $8y$ between the two terms to get:
$8 y (4 x^{2} - 1) = 8 y (2 x + 1) (2 x - 1)$ $8y(4x^2-1)=8y(2x+1)(2x-1)$

Using the fact that $(a^{2} - b^{2}) = (a + b) (a - b)$ $(a^2-b^2)=(a+b)(a-b)$

How do you factor 32x2y−8y32x^2y - 8y?