How do you find a polynomial function of degree 3 with 5, i, -i as zeros?

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How do you find a polynomial function of degree 3 with 5, i, -i as zeros?

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Answer 1 · 2015-05-19T21:26:37

The best way is to recognise that, if $x = 5$ $x=5$ is a root, then $x - 5 = 0$ $x-5=0$ , and ditto for the other two roots. So we have $x - 5, x - i, x + i$ $x-5, x-i, x+i$ all equalling zero. To find our polynomial, we just multiply the three terms together:

$(x - 5) (x - i) (x + i)$ $(x-5)(x-i)(x+i)$
$= (x^{2} - i x - 5 x + 5 i) (x + i)$ $=(x^2-ix-5x+5i)(x+i)$
$= x^{3} + i x^{2} - i x^{2} - (i^{2}) x - 5 x^{2} - 5 i x + 5 i x + 5 i^{2}$ $=x^3+ix^2-ix^2-(i^2)x-5x^2-5ix+5ix+5i^2$
which simplifies to
$x^{3} - 5 x^{2} + x - 5$ $x^3-5x^2+x-5$ .

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How do you find a polynomial function of degree 3 with 5, i, -i as zeros?

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