How do you find all the zeros of $3 x^{3} - 4 x^{2} + 8 x + 8$ $3x^3-4x^2+8x+8$ given the zero: $1 - \sqrt{3} i$ $1-sqrt 3i$ ?

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Answer 1 · 2016-04-19T17:09:17

$x = - \frac{2}{3} and (1 \pm i \sqrt{3})$ $x = -2/3 and (1+-i sqrt 3)$ .

Complex roots occur in conjugate pairs.
So, two of the roots are $1 \pm i \sqrt{3}$ $1+-i sqrt 3$ .
The sum of these roots = 2..

The sum of the three roots is
$-$ $-$ (coefficient of $x^{2}$ $x^2$ )/(coefficient of $x^{3})$ $x^3)$ = $\frac{4}{3}$ $4/3$ .

So, the third root = $\frac{4}{3} - 2 = - \frac{2}{3}$ $4/3-2=-2/3$ ..

How do you find all the zeros of 3x3−4x2+8x+83x^3-4x^2+8x+8 given the zero: 1−√3i1-sqrt 3i?