Question 265636
<font face="Garamond" size="+2">


The factored form of all polynomial functions that have *[tex \Large x]-intercepts at 1, -2, and 10 is:


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


However, we want the particular function such that *[tex \Large p(x)\ =\ -32] when *[tex \Large x\ =\ 2], so substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\left((2)\,-\,1\right)\left((2)\,+\,2\right)\left((2)\,-\,10\right)\ =\ -32]


and solve for *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\left(1\right)\left(4\right)\left(-8\right)\ =\ -32]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -32a\ =\ -32]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ (x\,-\,1)(x\,+\,2)(x\,-\,10)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>