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asked Jul 27, 2014 in ALGEBRA 2 by anonymous

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The zeros of the function are 2, - 4 and (1 + 3i).

Because complex zeros occur in conjugate pairs, we know that (1 - 3i) is also a zero of the function. So, from the Linear Factorization Theorem, f(x) can be written as f(x) = a [x - 2] · [x + 4] · [x - (1 + 3i)] · [x - (1 - 3i)].

For simplicity, let a = 1 to obtain

f(x) = [x - 2] · [x + 4] · [(x - 1) - 3i)] · [(x - 1) + 3i)]

f(x) = [x2 + 2x - 8] · [ (x - 1)2 - 9i2 ]

f(x) = [x2 + 2x - 8] · [x2 - 2x + 1 + 9]

f(x) = [x2 + 2x - 8] · [x2 - 2x + 10]

f(x) = x2(x2 - 2x + 10) + 2x(x2 - 2x + 10) - 8(x2 - 2x + 10)

f(x) = x4 - 2x3 + 10x2 + 2x3 - 4x2 + 20x - 8x2 + 16x - 80

f(x) = x4 - 2x2 + 36x - 80.

answered Jul 27, 2014 by casacop Expert

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