question_answer

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 \[\left( 2-\sqrt{3} \right)\]

Answer:

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 \[2-\sqrt{3}\].