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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$\displaystyle 2x^2-32 = 2(x^2-16) = 2(x^2-4^2) = 2(x-4)(x+4)$
$\displaystyle 2x^2-32=2(x^2-16)=2(x^2-4^2)=2(x-4)(x+4)$ Use the formula $\displaystyle a^2-b^2=(a-b)(a+b)$
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