How do you write a polynomial with zeros 8, -i, i and leading coefficient 1?

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How do you write a polynomial with zeros 8, -i, i and leading coefficient 1?

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Answer 1 · 2017-03-02T14:15:40

$x^{3} - 8 x^{2} + x - 8$ $color(green)(x^3-8x^2+x-8)$

Explanation:

If the polynomial has zeros $8, - i, and i$ $8, -i, and i$
then it has factors $c \times (x - 8) (x + i) (x - i)$ $cxx(x-8)(x+i)(x-i)$ where $c$ $c$ is a constant.

Since we are asked for a polynomial with a leading coefficient of $1$ $1$ ,
$c = 1$ $c=1$ and we can ignore it from here on.

$(x + i) (x - i) = x^{2} - i^{2} = x^{2} - (- 1) = x^{2} + 1$ $color(blue)((x+i)(x-i))=x^2-i^2=x^2-(-1)=color(blue)(x^2+1)$
and
$(x - 8) (x + i) (x - i) = (x - 8) \times (x^{2} + 1) = x^{3} - 8 x^{2} + x - 8$ $(x-8)color(blue)((x+i)(x-i))=(x-8)xxcolor(blue)(""(x^2+1))=x^3-8x^2+x-8$

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How do you write a polynomial with zeros 8, -i, i and leading coefficient 1?

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