Question

Show that $3$ is the zero of the polynomial $p\left(x\right)=x^3-8x^2+8x+21$.

Answer 1

Verify whether $\mathbf{3}$ is a zero of the given polynomial:

Any real number $a$ is called the zero of the polynomial $p\left(x\right)$ if and only if $p\left(a\right)=0$.

Substitute $x=3$ into the polynomial $p\left(x\right)$.

$\begin{array}{rcl}& \Rightarrow & p\left(3\right)={\left(3\right)}^3-8{\left(3\right)}^2+8\left(3\right)+21\\ & =& 27-72+24+21\\ & =& 0\end{array}$

Thus, $p\left(3\right)=0$.

Therefore, $3$ is the zero of the polynomial $p\left(x\right)=x^3-8x^2+8x+21$.

Hence, proved.