Question 1183867
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If *[tex \Large \alpha] is a zero of a polynomial function, then *[tex \Large (x\,-\,\alpha)] is a factor of the polynomial.  The number of factors of the polynomial is equal to the degree of the polynomial.  So your polynomial has the following factors:


*[tex \Large x]


*[tex \Large x]


*[tex \Large x\,-\,5]


*[tex \Large x\,-\,5]


*[tex \Large x\,-\,(-5)], which is to say *[tex \Large x\,+\,5]


The set of all 5th-degree polynomials with real coefficients that have this set of zeros can be described as:


*[tex \LARGE P(x)\ =\ a\(x^2\)\(x\,-\,5\)^2\(x\,+\,5\)]


Where *[tex \Large a] is the lead coefficient.  Your lead coefficient is given to be 1, so:


*[tex \LARGE P(x)\ =\ \(x^2\)\(x\,-\,5\)^2\(x\,+\,5\)]


is a valid possible formula for *[tex \LARGE P(x)].  You could expand that, but that is a lot of extra work that is not required by the question posed.

																
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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