Question 907297
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If the zeros are 5, 5, -3, and -3, then the polynomial function has the form *[tex \Large f(x)\ =\ \alpha(x\ -\ 5)^2(x\ +\ 3)^2]


Since *[tex \Large f(0)\ =\ -450] then it must be true that *[tex \Large \alpha(-5)^2(3)^2\ =\ -450], which is to say *[tex \Large \alpha\ =\ -2]


Hence the desired polynomial has the factored form *[tex \Large f(x)\ =\ -2(x\ -\ 5)^2(x\ +\ 3)^2].  All that is left is for you to expand the binomials, multiply the two resulting trinomials, collect like terms, and then multiply through by the lead constant term.


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