Question

When a polynomial $f\left(x\right)$ is divided by $x^2-5$ the quotient is $x^2-2x-3$ and remainder is zero. Find the polynomial and all its zeroes.

Answer 1

$\frac{f\left(x\right)}{x^2-5}=(x^2-2x-3)$

$f\left(x\right)=(x^2-5)(x^2-2x-3)$

$f\left(x\right)=x^4-2x^3-8x^2+10x+15$

Let us check whether $x=-1$ is a root.

$f\left(x\right)=x^4-2x^3-8x^2+10x+15$

$f(-1)=1+2-8-10+15$

$=18-18=0$

$\therefore x=-1$ is a root.

Let us check whether $x=3$ is a root.

$f\left(x\right)=f\left(3\right)=81-54-72+30+15=0$

$\therefore x=3$ is also a root.

$\therefore (x+1)(x-3)=x^2-3x+x-3$

$=x^2-2x-3$

Dividing $f\left(x\right)$ by $x^2-2x-3$

$\frac{x^4-2x^3-8x^2+10x+15}{x^2-2x-3}$ $=x^2-5$

$x^2-5=0$

$(x+\sqrt{5})(x-\sqrt{5})=0$

$\therefore x=\sqrt{5},-\sqrt{5}$

$\therefore$ Roots are $-1,3,\sqrt{5},-\sqrt{5}$.