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

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Answer 1 · 2016-04-26T21:37:55

$2 x^{2} - 32 = 2 (x - 4) (x + 4)$ $2x^2-32=2(x-4)(x+4)$

Separate out the common scalar factor $2$ $2$ then use the difference of squares identity:

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

with $a = x$ $a=x$ and $b = 4$ $b=4$ as follows:

$2 x^{2} - 32 = 2 (x^{2} - 16) = 2 (x^{2} - 4^{2}) = 2 (x - 4) (x + 4)$ $2x^2-32=2(x^2-16)=2(x^2-4^2)=2(x-4)(x+4)$

How do you factor completely 2x2−322x^2 - 32?