finding an nth degree polynomial
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If 2i is a zero then so is -2i. n=3 so the degree of x is x³.

The equation for f(x) is a(x-2i)(x+2i)(x-4) where a is a constant.

So f(x)=a(x²+4)(x-4)=a(x³-4x²+4x-16).

f(1)=30 so plug in x=1: a(1-4+4-16)=-15a=30, a=-2.

f(x)=-2x³+8x²-8x+32.

by Top Rated User (825k points)

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