Question 285331
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 98]


The lead coefficient is 2 and the constant term is an even number.  Take out a 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(x^2\ -\ 49\right)]


*[tex \Large x^2] is a perfect square, and *[tex \Large 49] is a perfect square, hence we have the difference of two squares.


The difference of two squares factorization is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2\ =\ (a\ +\ b)(a\ -\ b)]


Follow the pattern:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(x\ +\ 7\right)\left(x\ -\ 7\right)]



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