How do you factor $2 x^{4} y^{2} - 32$ $2x^4y^2-32$ ?

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Answer 1 · 2018-07-01T19:08:12

See a solution process below:

First, factor out a common term from each term in the expression:

$(2 \times x^{4} y^{2}) - (2 \times 16) \Rightarrow 2 (x^{4} y^{2} - 16)$ $(2 xx x^4y^2) - (2 xx 16) => 2(x^4y^2 - 16)$

Now, use this rule for quadratics to factor the term on the right:

$x^{2} - y^{2} = (x + y) (x - y)$ $color(red)(x)^2 - color(blue)(y)^2 = (color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y))$

$2 (x^{4} y^{2} - 16) \Rightarrow 2 (x^{2} y - 4) (x^{2} y + 4)$ $2(x^4y^2 - 16) => 2(x^2y - 4)(x^2y + 4)$

How do you factor 2x^4y^2-322x4y2−322x^4y^2-32?