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How do you write a polynomial in standard form given the zeros x=5, 8i, and -8i?

1 Answer

Nov 2, 2016

$f (x) = x^{3} - 5 x^{2} + 64 x - 320$ $f(x) = x^3-5x^2+64x-320$

Explanation:

We find the simplest such polynomial by multiplying the corresponding linear factors:

$f (x) = (x - 5) (x - 8 i) (x + 8 i)$ $f(x) = (x-5)(x-8i)(x+8i)$

$f (x) = (x - 5) (x^{2} - {(8 i)}^{2})$ $color(white)(f(x)) = (x-5)(x^2-(8i)^2)$

$f (x) = (x - 5) (x^{2} + 64)$ $color(white)(f(x)) = (x-5)(x^2+64)$

$f (x) = x^{3} - 5 x^{2} + 64 x - 320$ $color(white)(f(x)) = x^3-5x^2+64x-320$

Any polynomial in $x$ $x$ with these zeros will be a multiple (scalar or polynomial) of this $f (x)$ $f(x)$ .

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