How do you write a polynomial in standard form given the zeros 1 and 4-4i?

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Answer 1 · 2017-03-10T07:25:12

$x^{3} - 9 x^{2} + 40 x - 32$ $x^3 -9 x^2 +40x - 32$

Complex zeros always come in pairs: $4 \pm 4 i$ $4 +-4i$

Always multiply the complex zeros first to get rid of the complex terms by remembering that $i^{2} = - 1$ $i^2 = -1$

Multiply the complex factors: $(x - 4 - 4 i) (x - 4 + 4 i)$ $(x-4-4i)(x-4+4i)$

$= x^{2} - 4 x + 4 i x - 4 x + 16 - 16 i - 4 i x + 16 i - 16 i^{2}$ $= x^2 -4x + 4ix -4x +16 -16i -4ix +16i -16i^2$

Simplify by adding like terms:
$= x^{2} - 8 x + 16 - 16 (- 1)$ $= x^2 - 8x + 16 -16(-1)$
$= x^{2} - 8 x + 32$ $= x^2 - 8x + 32$

Add the $(x - 1)$ $(x-1)$ , $(x = 1)$ $(x=1)$ factor to the equation:

$(x - 1) (x^{2} - 8 x + 32)$ $(x-1)(x^2 - 8x + 32)$

Multiply using distribution:
$x^{3} - 8 x^{2} + 32 x - x^{2} + 8 x - 32$ $x^3 - 8x^2 +32x -x^2 +8x - 32$

Simplify by adding like terms:
$x^{3} - 9 x^{2} + 40 x - 32$ $x^3 -9 x^2 +40x - 32$