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17 February, 19:22

Given a polynomial that has zeros of - 4, 9i, and - 9i and has a value of 492 when x=-1. Write the polynomial in standard form axn+bxn-1+ ... Answer using reduced fractions when necessary.

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  1. 17 February, 21:26
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    The polynomial that has zeros of - 4, 9i, and - 9i, and has a value 492 when x = - 1 is:

    2x³ + 8x² + 162x + 648 = 0

    Step-by-step explanation:

    Given that the polynomial that has zeros of - 4, 9i, and - 9i, we can say that

    x = - 4 = > x + 4 = 0

    x = 9i and - 9i = > x = ±9i = > x² = (9i) ²

    Because i² = - 1, x² = - 81

    => x² + 81 = 0

    So together, we can write the polynomial as

    (x + 4) (x² + 81) = 0

    Expanding the bracket, we have

    x³ + 81x + 4x² + 324 = 0

    x³ + 4x² + 81x + 324 = 0

    Again, we are told that the polynomial is 492 at x = - 1.

    Put x = - 1 in

    x³ + 4x² + 81x + 324 = 0

    (-1) ³ + 4 (-1) ² + 81 (-1) + 324

    -1 + 4 - 81 + 324

    = 246 ≠ 492

    But 492 = 2*246

    Since multiplying the polynomial by 2 doesn't change anything, our result is set.

    Now,

    2 (x³ + 4x² + 81x + 324) = 2 (0)

    2x³ + 8x² + 162x + 648 = 0

    When x = - 1

    2 (-1) ³ + 8 (-1) ² + 162 (-1) + 648

    -2 + 8 - 162 + 648

    = 492

    As required.
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