How do you factor $12y^2 x^6 - 48k^4 m^8$ ?

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Answer 1 · 2015-05-05T12:50:44

First extract the (obvious) common factor
$12y^2x^6-48k^4m^8 = 12(y^2x^6-4k^4m^8)$

Restate the second term as the difference of squares
$= 12((yx^3)^2- (2k^2m^4)^2)$

Use the Difference of Squares Factoring Formula
$=12(yx^3+2k^2m^4)(yx^3-2k^2m^4)$

How do you factor 12y^2 x^6 - 48k^4 m^812y^2 x^6 - 48k^4 m^8?