Question

The product of two zeroes of the polynomial $p\left(x\right)=x^3-3x^2-6x+8$ is $(-2).$ Find all the zeroes of the polynomial.

Answer 1

Given, $p\left(x\right)=x^3-{3x}^2-6x+8$.

Let $\alpha ,\beta$ & $\gamma$ be zeroes of $p\left(x\right)$.

Then, $\alpha \beta \gamma =-8$, $\alpha +\beta +\gamma =3$, $\alpha \beta +\beta \gamma +\gamma \alpha =-6$.

According to question, $\alpha \beta =-2$.

$\therefore$ $\gamma =4$.

$\alpha +\beta =3-4=-1$.

$\Rightarrow \alpha \beta =-2$ & $\alpha +\beta =-1$

$\Rightarrow \alpha (-1-\alpha )=-2$

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

$\Rightarrow \alpha =1$ or $\alpha =-2$.

$\therefore$ Zeroes are $1$, $-2$, $4$.