• # question_answer 14) Find all zeroes of the polynomial $(2{{x}^{4}}-9{{x}^{3}}+5{{x}^{2}}+3x-1)$  if two of its zeroes are $\left( 2+\sqrt{3} \right)$ and $\left( 2-\sqrt{3} \right)$

 Here, $p\left( x \right)=2{{x}^{4}}-9{{x}^{3}}+5{{x}^{2}}+3x-1$ And two of its zeroes are $\left( 2+\sqrt{3} \right)$ and $\left( 2-\sqrt{3} \right)$ Quadratic polynomial with zeroes is given by, $\left\{ x-\left( 2+\sqrt{3} \right) \right\}.\left\{ x-\left( 2-\sqrt{3} \right) \right\}$ $\Rightarrow$   $\left( x-2-\sqrt{3} \right)\left( x-2+\sqrt{3} \right)$ $\Rightarrow$   ${{(x-2)}^{2}}-{{(\sqrt{3})}^{2}}$ $\Rightarrow$   ${{x}^{2}}-4x+4-3$ $\Rightarrow$   ${{x}^{2}}-4x+1=g(x)$ (say) Now, g (x) will be a factor of p(x) so g (x) will be divisible by p (x) For other zeroes, $2{{x}^{2}}-x-1=0$ $2{{x}^{2}}-2x+x-1=0$ or          $2x(x-1)+1\text{ (}x-1)=0$ $(x-1)(2x+1)=0$ $x-1=0\,\,\,\,2x+1=0$ $x=1,x=\frac{-1}{2}$ Zeroes of p(x) are $1,\frac{-1}{2},2+\sqrt{3}$ and $2-\sqrt{3}$.