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The given function is\[p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2\] . The roots of the given function is \[\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)=\left( {{x}^{2}}-2 \right)\]

First arrange the term of dividend and the divisor in the decreasing order of their degrees.

To obtain the first term of the quotient divide the highest degree term of the dividend by the highest degree term of the divisor.

To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.

Continue this process till the degree of remainder is less than the degree of divisor.

\[\begin{align}

& {{x}^{2}}-2\overset{\,\,2{{x}^{2}}-3x+1}{\overline{\left){2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2}\right.}} \\

& \,\,\,\,\,\,\,\,\,\,-\underline{\left( 2{{x}^{4}} \right)\,\,\,\,\,\,\,+\left( -4{{x}^{2}} \right)} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{3}}+\,\,\,\,{{x}^{2}}+6x \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\underline{\left( -\,3{{x}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6x \right)} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2 \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\underline{\left( {{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2 \right)} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\

\end{align}\]

Finding the roots of \[2{{x}^{2}}-3x+1\] as,

\[\begin{align}

& 2{{x}^{2}}-3x+1=2{{x}^{2}}-2x-x+1 \\

& =2x\left( x-1 \right)-1\left( x-1 \right) \\

& =\left( 2x-1 \right)\left( x-1 \right)

\end{align}\]

Put the above expression equal to 0 as,

\[\begin{align}

& \left( 2x-1 \right)=0 \\

& 2x=1 \\

& x=\dfrac{1}{2}

\end{align}\]

\[\begin{align}

& \left( x-1 \right)=0 \\

& x=1

\end{align}\]