Question

Verify that $3,-1,\frac{-1}{3}$are the zeros of cubic polynomial $p\left(x\right)=3x^3-5x^2-11x-3$.

Answer 1

Put the values of $x$ in the polynomial to verify the zeroes of the polynomial:

The zeroes of a polynomial are the values of $x$ that satisfies the equation and makes the polynomial equal to zero as a whole.

Here, the polynomial $p\left(x\right)=3x^3-5x^2-11x-3$

$\begin{array}{rcl}& \Rightarrow & p\left(3\right)=3{\left(3\right)}^3-5{\left(3\right)}^2-11\left(3\right)-3\\ & =& 81-45-33-3\\ & =& 81-81\\ & =& 0\end{array}$

$\begin{array}{rcl}& \Rightarrow & p(-1)=3{(-1)}^3-5{(-1)}^2-11(-1)-3\\ & =& -3-5+11-3\\ & =& -11+11\\ & =& 0\end{array}$

$\begin{array}{rcl}p\left(\frac{-1}{3}\right)& =& 3{\left(\frac{-1}{3}\right)}^3-5{\left(\frac{-1}{3}\right)}^2-11\left(\frac{-1}{3}\right)-3\\ & =& 3\left(\frac{-1}{27}\right)-\frac{5}{9}+\frac{11}{3}-3\\ & =& \frac{-1-5+33-27}{9}\\ & =& \frac{33-33}{9}\\ & =& 0\end{array}$

Hence, proved that $3,-1,\frac{-1}{3}$are the zeros of cubic polynomial $p\left(x\right)=3x^3-5x^2-11x-3$.