Question 1177241
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If a polynomial function has a zero at some number *[tex \Large r], then *[tex \Large x\,-\,r] must be a factor of the polynomial.  Hence, the desired polynomial is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ a(x\,-\,0)(x\,+\,3)(x\,+\,2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ a\(x^3\ +\ 5x^2\ +\ 6x\)]


Then

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(-4)\ =\ a\((-4)^3\,+\,5(-4)^2\,+\,6(-4)\)]


Do the arithmetic, solve for *[tex \Large a], plug that value back into *[tex \Large f(x)\ =\ a\(x^3\ +\ 5x^2\ +\ 6x\)], and simplify.

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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