Solve the equations:
$6 x^{4} - 13 x^{3} - 35 x^{2} - x + 3 = 0$ , one root being $2 - \sqrt{3}$ .

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Solution

$p (x) = 6 x^{4} - 13 x^{3} - 35 x^{2} - x - 3 = 0$

If one root is $2 - \sqrt{3}$ then one other root is $2 + \sqrt{3}$

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

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

$\Rightarrow x^{2} - 4 x + 1$ is a factor of $p (x)$

Dividing $p (x)$ by $x^{2} - 4 x + 1$

$\Rightarrow p (x) = (6 x^{2} + 11 x + 3) (x^{2} - 4 x + 1) \Rightarrow (6 x^{2} + 11 x + 3) (x^{2} - 4 x + 1) = 0 \Rightarrow 6 x^{2} + 11 x + 3 = 0 \Rightarrow 6 x^{2} + 2 x + 9 x + 3 = 0 \Rightarrow 2 x (3 x + 1) + 3 (3 x + 1) = 0 \Rightarrow (2 x + 3) (3 x + 1) = 0 \Rightarrow x = - \frac{3}{2}, - \frac{1}{3}$

So the other two roots are $- \frac{3}{2}, - \frac{1}{3}$

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