Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. n=​4; 2i and 3i are​ zeros;

Question

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.
n=​4;
2i and 3i are​ zeros;
f(−1)equals=100

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Eloise 2 weeks 2022-01-15T17:46:14+00:00 1 Answer 0 views 0

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    2022-01-15T17:47:44+00:00

    Answer:

    f(x) = 2x⁴ + 26x² + 72

    Step-by-step explanation:

    Imaginary and complex roots come in conjugate pairs.  So if 2i and 3i are zeros, then -2i and -3i are also zeros.

    f(x) = a (x − 2i) (x + 2i) (x − 3i) (x + 3i)

    f(x) = a (x² − 4i²) (x² − 9i²)

    f(x) = a (x² + 4) (x² + 9)

    f(x) = a (x⁴ + 13x² + 36)

    f(-1) = 100

    100 = a ((-1)⁴ + 13(-1)² + 36)

    100 = a (1 + 13 + 36)

    100 = 50a

    a = 2

    f(x) = 2 (x⁴ + 13x² + 36)

    f(x) = 2x⁴ + 26x² + 72

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45:7+7-4:2-5:5*4+35:2 =? ( )