How do you factor $169 a^{2} b^{2} - 16 c^{4}$ $169a^2b^2-16c^ 4$ ?

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Answer 1 · 2016-06-12T21:21:47

$(13 a b + 4 c^{2}) (13 a b - 4 c^{2})$ $(13ab+4c^2)(13ab-4c^2)$

To factor $169 a^{2} b^{2} - 16 c^{4}$ $169a^2b^2-16c^4$ we must first recognize that this is the difference (subtraction) of two perfect squares.

$169 a^{2} b^{2} = (13 a b) (13 a b)$ $169a^2b^2 = (13ab)(13ab)$
$16 c^{4} = (4 c^{2}) (4 c^{2})$ $16c^4=(4c^2)(4c^2)$

When we factors the difference of two squares the binomials always have one $+$ $+$ and one $-$ $-$ .

$(13 a b + 4 c^{2}) (13 a b - 4 c^{2})$ $(13ab+4c^2)(13ab-4c^2)$

How do you factor 169a2b2−16c4169a^2b^2-16c^ 4?