Although it this is basic it has been a stumbling block for me.

Factor

$\displaystyle 2x^2-32$

The answer is $\displaystyle 2(x-4)(x+4)$ apparently. Would someone please show me how to solve this? Thanks

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- Jul 8th 2009, 10:45 AMallyourbass2212Basic Factoring Question
Although it this is basic it has been a stumbling block for me.

Factor

$\displaystyle 2x^2-32$

The answer is $\displaystyle 2(x-4)(x+4)$ apparently. Would someone please show me how to solve this? Thanks - Jul 8th 2009, 10:50 AMSENTINEL4
$\displaystyle 2x^2-32 = 2(x^2-16) = 2(x^2-4^2) = 2(x-4)(x+4)$

- Jul 8th 2009, 10:51 AMred_dog
$\displaystyle 2x^2-32=2(x^2-16)=2(x^2-4^2)=2(x-4)(x+4)$

Use the formula $\displaystyle a^2-b^2=(a-b)(a+b)$