Question

If two zeroes of polynomial $x^4+3x^3-20x^2-6x+36$ are $\sqrt{2}$ and $-\sqrt{2}$ find all other zeros of polynomial.

Answer 1

Given
$f\left(x\right)=2x^4-3x^3-3x^2+6x-2$
Two zeros of polynomial $f\left(x\right)$ are $\sqrt{2}$ and $-\sqrt{2}$
$(x-\sqrt{2})(x+\sqrt{2})=x^2-2$, will be a factor of $f\left(x\right)$
Dividing polynomial $f\left(x\right)$ by $x^2-2$
$2x^2-3x+1x^2-2)\overline{2x^4-3x^3-3x^2+6x-2}2x^4-4x^2-+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_-3x^3+x^2+6x-2-3x^3+6x+-\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_x^2-2x^2-2-+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_0$
$f\left(x\right)=(x^2-2)(2x^2-3x+1)$
$=(x^2-2)(2x^2-2x-x+1)$
$=(x^2-2)\left[2x\right(x-1)-1(x-1\left)\right]$
$=(x^2-2)(x-1)(2x-1)$
To find zero of polynomial, $f\left(x\right)=0$
$\therefore (x^2-2)(x-1)(2x-1)=0$
when $x^2-2=0$ then $x=\pm \sqrt{2}$
If $x-1=0$ then $x=1$
and $2x-1=0$ then $x=\frac{1}{2}$
$\therefore$ all the zeros of polynomial $f\left(x\right)$ are
$\sqrt{2},-\sqrt{2},1$ and $\frac{1}{2}$
$\sqrt{2},-\sqrt{2},1$ and $\frac{1}{2}$
Thus other two zero are $1$and $\frac{1}{2}$.