Question 916690
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A 4th degree polynomial must have 4 zeroes.  If any of them are complex numbers, then an even number of them must be complex numbers.  But we are given three real zeroes; hence the 4th zero must be real and therefore one of the zeros 1, 4, or 10 must have a multiplicity of 2.  Since the polynomial is only positive on (4,10), the graph must cross (not be tangent to) the x-axis at 4 and 10.  That leaves 1 as the zero that must have the multiplicity of 2.  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ a(x\ -\ 1)^2(x\ -\ 4)(x\ -\ 10)]


Multiply this out, then choose *[tex \Large a] such that *[tex \Large p(0)\ =\ -2].  Hint:  Choose *[tex \Large a] such that the constant term equals -2.


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