How do you write a polynomial function of least degree given the zeros 2i, -2i, 2+2i?

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How do you write a polynomial function of least degree given the zeros 2i, -2i, 2+2i?

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Answer 1 · 2016-10-14T19:45:19

Please see the explanation for the process.

$y = k (x^{4} - 4 x^{3} + 12 x^{2} - 16 x + 32)$ $y = k(x^4 - 4x^3 + 12x^2 -16x + 32)$

Explanation:

You cannot have an odd number of complex or imaginary roots, because they always exist is conjugate pairs, therefore, another root must be 2 - 2i and the polynomial is of the form:

$y = k (x - 2 i) (x + 2 i) (x - 2 - 2 i) (x - 2 + 2 i)$ $y = k(x - 2i)(x + 2i)(x - 2 -2i)(x - 2 + 2i)$

$y = k (x^{2} - 4 i^{2}) (x - 2 - 2 i) (x - 2 + 2 i)$ $y = k(x^2 - 4i^2)(x - 2 -2i)(x - 2 + 2i)$

$y = k (x^{2} + 4) (x - 2 - 2 i) (x - 2 + 2 i)$ $y = k(x^2 + 4)(x - 2 -2i)(x - 2 + 2i)$

$y = k (x^{2} + 4) (x^{2} - 2 x + 2 i x - 2 x + 4 - 4 i - 2 i x + 4 i - 4 i^{2})$ $y = k(x^2 + 4)(x^2 - 2x + 2ix -2x + 4 - 4i - 2ix +4i - 4i^2)$

$y = k (x^{2} + 4) (x^{2} - 4 x + 4 - 4 i^{2})$ $y = k(x^2 + 4)(x^2 - 4x + 4 - 4i^2)$

$y = k (x^{2} + 4) (x^{2} - 4 x + 4 + 4)$ $y = k(x^2 + 4)(x^2 - 4x + 4 + 4)$

$y = k (x^{2} + 4) (x^{2} - 4 x + 8)$ $y = k(x^2 + 4)(x^2 - 4x + 8)$

$y = k (x^{4} - 4 x^{3} + 8 x^{2} + 4 x^{2} - 16 x + 32)$ $y = k(x^4 - 4x^3 + 8x^2 + 4x^2 -16x + 32)$

$y = k (x^{4} - 4 x^{3} + 12 x^{2} - 16 x + 32)$ $y = k(x^4 - 4x^3 + 12x^2 -16x + 32)$

k exists to allow the polynomial to pass through a specified point.

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How do you write a polynomial function of least degree given the zeros 2i, -2i, 2+2i?

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