Question

Obtained all other zeroes of the polynomial $x^4-17x^2-36x-20$, if two of its zeros are $5$ and $-2$

Answer 1

$p\left(x\right)=x^4-17x^2-36x-20$

Let its roots be $\alpha ,\beta ,5$ and $-2$.

comparing $p\left(x\right)$ with $a{x}^4+b{x}^3+c{x}^2+dx+e,$

we get,

$a=1,b=0,c=-17,d=-36,e=-20$

Now, $\alpha +\beta +5+(-2)=\frac{-b}{a}=0$

$\Rightarrow \alpha +\beta =-3$-------$\left(A\right)$

And product of roots, $\alpha \beta \times 5\times (-2)=\frac{e}{a}=-20$

$\Rightarrow \alpha \beta =2$--------$\left(B\right)$

From $\left(A\right)$ and $\left(B\right)$, we get that $\alpha$ and $\beta$ are

$-1$ and $-2$.

Hence, zeroes of given polynomial are,

$5,-1,-2,-2$