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Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies the given condition. -4i, 4i, 2; f(4) = -256
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yeh simple
f(x) =A (x+4i)(x-4i)(x-2)
okay and if x=4 than it must = -256, right?
and that's how you find a?
(x+4i)(x-4i) = x^2 - (-16) = x^2 +16
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f(x) = A (x^2 +16)(x-2)
when x=4, f= -256
-256 = 32(2) A
A= -4 f(x) = -4 ( x^2+16)(x-2)
Oh I see what I did wrong, I added a negative sign by accident, thanks once again!
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