Find all zeros of the polynomial f(x) = 2x4 − 2x3 − 7x2 + 3x + 6,

Question:

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 $\sqrt{\frac{3}{2}}$.

Solution:

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, $f(x)=g(x) \times q(x)+r(x)$

$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=\left(2 x^{2}-3\right)\left(x^{2}-x-2\right)+0$

$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=(\sqrt{2} x+\sqrt{3})(\sqrt{2 x}-\sqrt{3})\left(x^{2}+1 x-2 x-2\right)$

$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=(\sqrt{2} x+\sqrt{3})(\sqrt{2} x-\sqrt{3})[x(x+1)-2(x+1)]$

$2 x^{4}-2 x^{3}-7 x^{2}+3 x+6=(\sqrt{2} x+\sqrt{3})(\sqrt{2} x-\sqrt{3})(x-2)(x+1)$