Solve the equation $6 x^{4} - 13 x^{3} - 35 x^{2} - x + 3 = 0$ , having given that one roots is $2 - \sqrt{3}$ .

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Solution

Since $2 - \sqrt{3}$ is a root, we know that $2 + \sqrt{3}$ is also a root, and corresponding to this pair of roots we have the quadratic factor $x^{2} - 4 x + 1$ .
Also $6 x^{4} - 13 x^{3} - 35 x^{2} - x + 3 = (x^{2} - 4 x + 1) (6 x^{2} + 11 x + 3)$ ;
Hence the other roots are obtained from
$6 x^{2} + 11 x + 3 = 0$ , or $(3 x + 1) (2 x + 3) = 0$ ;
Thus, the roots are $- \frac{1}{3}, - \frac{3}{2}, 2 + \sqrt{3}, 2 - \sqrt{3}$ .

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