Question 716369
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If you factor something and then multiply the factors you should get back to where you started.  So let's see what happens.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ -\ 3)(2x^2\ +\ 3)\ =\ 4x^3\ -\ 6x^2\ +\ 6x\ -\ 9\ \neq\ 4x^3\ +\ 12x^2\ +\ 4x\ +\ 12]


Hmmm.  Back to square 1.


First grouping:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^3\ +\ 4x\ +\ 12x^2\ +\ 12]


Factor *[tex \LARGE 4x] out of the first two terms, and *[tex \LARGE 12] out of the second two terms.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x(x^2\ +\ 1)\ +\ 12(x^2\ +\ 1)]


Now factor *[tex \LARGE (x^2\ +\ 1)] out:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\ +\ 12)(x^2\ +\ 1)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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