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

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Answer:

p(x) = 2(x³ + x² + 4x + 4)

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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