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2 July, 08:07

Write a polynomial in standard form that is a degree of 3, has zeros - 1, 2i, and P (0) = 8

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  1. 2 July, 09:34
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    p (x) = 2 (x³ + x² + 4x + 4)

    Step-by-step explanation:

    An imaginary root such as 2i must be joined by its complex conjugate, which here is - 2i.

    Thus, the zeros are - 1, 2i and - 2i.

    The polynomial is p (x) = a (x + 1) (x - 2i) (x + 2i). This is of degree 3.

    In expanded form, we have p (x) = a (x + 1) (x² + 4), or

    = a (x³ + 4x + x² + 4), or, in standard form,

    = a (x³ + x² + 4x + 4).

    Since this must equal 8 when x = 0, 8 = a (4). Thus, a = 2.

    The polynomial is thus p (x) = 2 (x³ + x² + 4x + 4)
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