Question 245871
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<b><i>Terminology note:</i></b> *[tex \LARGE x^3\ +\ 8x^2\ +\ 2x\ +\ 16] is <b><i>NOT</i></b> an equation.  An equation must have an equal sign in it somewhere, and there is no equal sign to be seen here.  This is a polynomial expression, specifically a 3rd degree polynomial expression.  If this was English class and you claimed that this was a sentence, your teacher would ask "Where is your verb?"


The method illustrated here only works under certain conditions, and in this case those conditions have been contrived by the constructor of the problem.  Keep this in your toolbox, but it is not a magic bullet that is always appropriate to use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ +\ 8x^2\ +\ 2x\ +\ 16]


Notice that the coefficient on the *[tex \Large x] term is 2 times the coefficient on the *[tex \Large x^3] term and the constant term is 2 times the coefficient on the *[tex \Large x^2] term.


Rearrange the terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ +\ 2x\ +\ 8x^2\ +\ 16]


Group the two pairs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x^3\ +\ 2x\right)\ +\ \left(8x^2\ +\ 16\right)]


Factor *[tex \Large x] from the first group and 8 from the second group:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(x^2\ +\ 2\right)\ +\ 8\left(x^2\ +\ 2\right)]


Now note that the resulting two terms have a factor of *[tex \Large x^2\ +\ 2] in common.  Factor it out:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 8\right)\left(x^2\ +\ 2\right)]


Since *[tex \Large x^2\ +\ 2] cannot be factored over the real numbers, if factoring over the real numbers is your assignment, then you are done.  If you are supposed to consider the complex factors, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 8\right)\left(x\ +\ i\sqrt{2}\right)\left(x\ -\ i\sqrt{2}\right)]



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