Find all zeroes of the polynomial $(2 x^{4} - 9 x^{3} + 5 x^{2} + 3 x + 1)$ if two of its zeroes are $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$ .

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Solution

Here, $p (x) = 2 x^{4} - 9 x^{3} + 5 x^{2} + 3 x - 1$ And two of its zeroes are $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$ .
Quadratic polynomial with zeros is given by,
${x - (2 + \sqrt{3})}$ . ${x - (2 - \sqrt{3})}$
$\Rightarrow (x - 2 - \sqrt{3}) (x - 2 + \sqrt{3})$
$\Rightarrow (x - 2)^{2} - (\sqrt{3})^{2}$
$\Rightarrow x^{2} - 4 x + 4 - 3$
$\Rightarrow x^{2} - 4 x + 1 = g (x)$ (say)

Now, g(x) will be a factor of p(x) so g(x) will be divisible by p(x)

For other zeroes,
$2 x^{2} - x - 1 = 0 2 x^{2} - 2 x + x - 1 = 0 o r 2 x (x - 1) + 1 (x - 1) = 0 (x - 1) (2 x + 1) = 0 x - 1 = 02 x + 1 = 0 x = 1, x = \frac{- 1}{2}$
Zeroes of p(x) are
1, $\frac{- 1}{2}, 2 + \sqrt{3}$ and $2 - \sqrt{3}$ .

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