Verity that 3, −2, 1 are the zeros of the cubic polynomial

The given polynomial is $p(x)=\left(x^{3}-2 x^{2}-5 x+6\right)$

$\therefore p(3)=\left(3^{3}-2 \times 3^{2}-5 \times 3+6\right)=(27-18-15+6)=0$

$p(-2)=\left[\left(-2^{3}\right)-2 \times\left(-2^{2}\right)-5 \times(-2)+6\right]=(-8-8+10+6)=0$

$p(1)=\left(1^{3}-2 \times 1^{2}-5 \times 1+6\right)=(1-2-5+6)=0$

$\therefore 3,-2$ and 1 are the zeroes of $\mathrm{p}(x)$

Let $\alpha=3, \beta=-2$ and $\gamma=1$. Then we have:

$(\alpha+\beta+\gamma)=(3-2+1)=2=\frac{-\left(\text { coefficient of } x^{2}\right)}{\left(\text { coefficient of } x^{3}\right)}$

$(\alpha \beta+\beta \gamma+\gamma \alpha)=(-6-2+3)=\frac{-5}{1}=\frac{\text { coefficient of } x}{\text { coefficient of } x^{3}}$

$\alpha \beta \gamma=\{3 \times(-2) \times 1\}=\frac{-6}{1}=\frac{-(\text { constant term })}{\left(\text { coefficient of } x^{3}\right)}$