Solve the equations:
$3 x^{4} - 10 x^{3} + 4 x^{2} - x - 6 = 0$ , one root being $\frac{1 + \sqrt{- 3}}{2}$ .

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Solution

$p (x) = 3 x^{4} - 10 x^{3} + 4 x^{2} - x - 6$

One of the root is $\frac{1 + \sqrt{- 3}}{2}$ then one other root is $\frac{1 - \sqrt{- 3}}{2}$ as imaginary roots occur in pair

$\Rightarrow (x - \frac{1 + \sqrt{- 3}}{2}) (x - \frac{1 - \sqrt{- 3}}{2})$ is a factor of $p (x)$

$\Rightarrow (x^{2} - x + 1)$ is a factor of $p (x)$

Dividing $p (x)$ by $(x^{2} - x - 1)$

$\Rightarrow p (x) = (3 x^{2} - 7 x - 6) (x^{2} - x + 1) \Rightarrow (3 x^{2} - 7 x - 6) (x^{2} - x + 1) = 0 \Rightarrow 3 x^{2} - 7 x - 6 = 0 \Rightarrow 3 x^{2} - 9 x + 2 x - 6 = 0 \Rightarrow 3 x (x - 3) - 2 (x - 3) = 0 \Rightarrow (3 x - 2) (x - 3) = 0 \Rightarrow x = 3, \frac{3}{2}$

So the other two roots are $3, \frac{3}{2}$

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