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Pre-Calc

A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point.

Degree: 3
Zeros: -2,2+2√2i
Solution Point: f(−1) = −68

(a) Write the function in completely factored form.

(b) Write the function in polynomial form.

Help Please my teacher doesn't really teach!!!

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  1. you have one complex root, but they always come as conjugate pairs
    The other must be 2-2√2i
    You also know that one factor is (x+2)

    the sum of the complex roots
    = 2+2√2i + 2-2√2i = 4
    product of the complex roots
    = (2+2√2i)(2-2√2i) = 4 + 8 = 12

    So the quadratic factor is x^2 - 4x + 12

    f(x) = a(x+2)(x^2 - 4x + 12)
    we also have a point (-1,-68)

    -68 = a(1)(17)
    a = -4

    f(x) = -4(x+2)(x^2 - 4x + 12)

    I will leave it up to you to expand it.

    Check of my answer:
    https://www.wolframalpha.com/input/?i=-4(x%2B2)(x%5E2+-+4x+%2B+12)%3D0

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  2. write a polynomial function of minimum degree with real coefficients whose zeros include those listed. write the polynomial in standard form calculator

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