Question

Find all zeroes of the polynomial $3x^4+6x^3-2x^2-10x-5$ if two of its zeroes are $\frac{\sqrt{5}}{3}$ and $-\frac{\sqrt{5}}{3}$

Answer 1

Let other two roots be $\alpha ,\beta$

Sum of roots $=\frac{-6}{3}\left(\frac{-b}{a}\right)$

$=\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{3}+\alpha +\beta$

$\Rightarrow \alpha +\beta =-2$

Product of roots $(=\frac{e}{a})=\frac{-5}{3}$

$\Rightarrow \frac{-5}{9}\alpha \beta =\frac{-5}{3}$

$\Rightarrow \alpha \beta =3$

$\Rightarrow \alpha (-2-\alpha )=3$

$\Rightarrow -{\alpha }^2-2\alpha =3$

$\Rightarrow {\alpha }^2+2\alpha +3=0$

But discriminant of this $=b^2-4ac=2^2-4\left(3\right)=4-12<0$

$\Rightarrow$ Roots are not real

$\Rightarrow \alpha =\frac{-2\pm \sqrt{-8}}{2}=\frac{-2\pm 2\sqrt{2}i}{2}$

$\Rightarrow \alpha ,\beta =-1\pm \sqrt{2}i$.