College Algebra
posted by Patrick .
Find a polynomial function f(x), with real coefficients, that has 1 and 3+2i as zeros, and such that f(1)=2 (Multiply out and simplify your answer)

f(x) = a(x1)(x(3+2i))(x(32i))
= a(x1)(x^26x+13)
= a(x^37x^2+19x13
f(1) = a(171913) = 40a = 2
so, a = 1/20
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