Question 908699
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Let *[tex \Large u\ =\ x^2] and substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 53u\ +\ 196]


Note:  *[tex \Large -4\ \times\ -49\ =\ 196] and *[tex \Large -4\ +\ (-49)\ =\ -53], hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 53u\ +\ 196\ =\ (u\ -\ 4)(u\ -\ 49)]


Substitute back


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x^2\ -\ 4)(x^2\ -\ 49)]


Each of the two binomial factors is the difference of two squares which you should be able to factor for yourself.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \