Question 275631
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x^2\ -\ 2x]


Factor out *[tex \Large 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\left(4x^2\ -\ 1\right)]


Recall the difference of two squares factorization, namely:  *[tex \LARGE a^2\ -\ b^2\ =\ \left(a\ +\ b\right)\left(a\ -\ b\right)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\left(2x\ +\ 1\right)\left(2x\ -\ 1\right)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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