How do you find all the zeros of $x^{3} + x^{2} + 9 x + 9$ $x^3 + x^2 + 9x + 9$ given zero 3i?

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How do you find all the zeros of $x^{3} + x^{2} + 9 x + 9$ $x^3 + x^2 + 9x + 9$ given zero 3i?

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Answer 1 · 2016-05-21T19:30:04

The zeros are $- 1$ $-1$ , $3 i$ $3i$ and $- 3 i$ $-3i$ .

Explanation:

We don't really need to be told that $3 i$ $3i$ is a zero, except that it does inform us that we should include Complex zeros in the answer.

This cubic factors by grouping:

$x^{3} + x^{2} + 9 x + 9$ $x^3+x^2+9x+9$

$= (x^{3} + x^{2}) + (9 x + 9)$ $=(x^3+x^2)+(9x+9)$

$= x^{2} (x + 1) + 9 (x + 1)$ $=x^2(x+1)+9(x+1)$

$= (x^{2} + 9) (x + 1)$ $=(x^2+9)(x+1)$

$= (x - 3 i) (x + 3 i) (x + 1)$ $=(x-3i)(x+3i)(x+1)$

So the zeros are $- 1$ $-1$ , $3 i$ $3i$ and $- 3 i$ $-3i$ .

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How do you find all the zeros of $x^{3} + x^{2} + 9 x + 9$ $x^3 + x^2 + 9x + 9$ given zero 3i?

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Explanation:

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