Find all zeros of the polynomial $f(x)=2 x^{4}-2 x^{3}-7 x^{2}+3 x+6$, if its two zeros are $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$.
Since $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$ are two zeros of $f(x)$. Therefore,
$=\left(x-\sqrt{\frac{3}{2}}\right)\left(x+\sqrt{\frac{3}{2}}\right)$
$=\left(x^{2}-\frac{3}{2}\right)$
$=\frac{1}{2}\left(2 x^{2}-3\right)$ is a factor of $f(x)$.
Also $2 x^{2}-3$ is a factor of $f(x)$.
Let us now divide $f(x)$ by $2 x^{2}-3$. We have,
By using division algorithm we have, $f(x)=g(x) \times q(x)+r(x)$
$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=\left(2 x^{2}-3\right)\left(x^{2}-x-2\right)+0$
$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=(\sqrt{2} x+\sqrt{3})(\sqrt{2 x}-\sqrt{3})\left(x^{2}+1 x-2 x-2\right)$
$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=(\sqrt{2} x+\sqrt{3})(\sqrt{2} x-\sqrt{3})[x(x+1)-2(x+1)]$
$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=(\sqrt{2} x+\sqrt{3})(\sqrt{2} x-\sqrt{3})(x-2)(x+1)$
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