Question 627126
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This is only possible if your polynomial has complex coefficients.  If that is not the case, then you have made a typo when you wrote out the roots.  Complex zeros always come in conjugate pairs when you have a polynomial with real coefficients.


Regardless, if a given number *[tex \LARGE \alpha] is a zero of a polynomial, then *[tex \LARGE x\ -\ \alpha] is a factor of the polynomial.  So if your four zeros are *[tex \LARGE \alpha_1], *[tex \LARGE \alpha_2], *[tex \LARGE \alpha_3], and *[tex \LARGE \alpha_4], then the factored form of the polynomial is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ \alpha_1)(x\ -\ \alpha_2)(x\ -\ \alpha_3)(x\ -\ \alpha_4)\ =\ 0].


Multiply the four factors to get the standard form.
 

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