Question 974906
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Complex zeros of polynomials always occur in conjugate pairs. Hence if *[tex \Large a\ +\ bi] is a zero of a polynomial then *[tex \Large a\ -\ bi] is a zero as well.


Your five zeros are *[tex \Large -5], *[tex \Large 0\ -\ i], *[tex \Large 0\ +\ i], *[tex \Large -9\ +\ i], and *[tex \Large -9\ -\ i].


If *[tex \Large \alpha] is a zero of a polynomial in *[tex \Large x], then *[tex \Large x\ -\ \alpha] is a factor of the polynomial.


Your five factors are therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 5)(x\ +\ i)(x\ -\ i)(x\ +\ 9\ -\ i)(x\ +\ 9\ +\ i)]


Just multiply the five factors together and collect like terms.  I'll get you started by multiplying the complex factors to get rid of all the imaginary parts.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 5)(x^2\ +\ 1)(x^2\ +\ 18x\ +\ 82)]


You can do the rest yourself.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \