# Find all the zeroes of  $2x^4 - 13x^3 + 19x^2 + 7x - 3$, if you know that two of its zeroes are $2+\sqrt{3}$  and $2-\sqrt{3}$ .

Given that  $2x^4 - 13x^3 + 19x^2 + 7x - 3$

Two zeroes are  $2 + \sqrt{3}$  and  $2 - \sqrt{3}$.

The factor of the given polynomial :

$\Rightarrow (x-(2+\sqrt{3}))(x-(x-\sqrt{3}))$

$\Rightarrow x^2-4x+1$

Now divide the polynomial with their factor

$\Rightarrow \frac{2x^4 - 13 x^2 + 19 x^2 + 7x -3}{x^2-4x+1}$

$\Rightarrow 2x^2-5x-3$

Now factor the above equation :

$\Rightarrow 2x^2-6x+x-3$

$\Rightarrow 2x(x-3)+1(x-3)$

$\Rightarrow (x-3)(2x+1)$

The zeroes are

$x-3=0\; \text{and} \; 2x+1=0$

$x=3,\frac{-1}{2}$ , $2+\sqrt{3}$  and $2-\sqrt{3}$ .

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