Question

Verify whether $2,3$ and $\frac{1}{2}$ are the zeroes of the polynomial
$p\left(x\right)=2x^3-11x^2+17x-6$

Answer 1

A number $\alpha$ is root / zero of a polynomial $p\left(x\right)$, if $P\left(\alpha \right)=0$

$p\left(2\right)=2(2{)}^3-11(2{)}^2+17\left(2\right)-6$

$⟹p\left(2\right)=16-44+34-6⟹p\left(2\right)=0$

$p\left(3\right)=2(3{)}^3-11(3{)}^2+17\left(3\right)-6$

$⟹p\left(3\right)=54-99+51-6⟹p\left(3\right)=0$

$p\left(\frac{1}{2}\right)=2(\frac{1}{2}{)}^3-11(\frac{1}{2}{)}^2+17\left(\frac{1}{2}\right)-6$

$⟹p\left(\frac{1}{2}\right)=\frac{1-11+34-24}{4}⟹p\left(\frac{1}{2}\right)=0$

Hence we say that $2$, $3$ and $\frac{1}{2}$ are zeroes of the given polynomial