Question 645903
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Use the sum and difference of cubes factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^3\ \pm\ b^3\ =\ (a\ \pm\ b)\left(a^2\ \mp\ ab\ +\ b^2\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ +\ 5^3\ =\ (x\ +\ 5)\left(x^2\ -\ 5x\ +\ 25\right)]


The *[tex \LARGE x\ +\ 5] factor gives you the one real zero, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 5x\ +\ 25\ =\ 0]


is easily solved with the quadratic formula to get the conjugate pair of complex zeros.


Bonus:  Here's a mnemonic to remember the signs in the factors of the sum and difference of cubes factorization.  San Diego Padres :: Same Different Plus.  The sign in the linear binomial factor matches the sum or difference of cubes binomial.  The first sign in the quadratic factor is the opposite (different) of the sign in the original binomial, and the second sign in the quadratic is always positive.


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