Question

Verify whether the values of $x$ given in each case are the zeroes of the polynomial or not?
$f\left(x\right)=3x^2-1;x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Answer 1

Given polynomial $p\left(x\right)=3x^2-1$ and value $x=(-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}})$

$p\left(x\right)=3x^2-1$

Replace x by $-\frac{1}{\sqrt{3}}$

$p(-\frac{1}{\sqrt{3}})=3{(-\frac{1}{\sqrt{3}})}^2-1$

$\Rightarrow p(-\frac{1}{\sqrt{3}})=1-1=0$

Replace x by $\frac{2}{\sqrt{3}}$

$p\left(\frac{2}{\sqrt{3}}\right)=3{\left(\frac{2}{\sqrt{3}}\right)}^2-1$

$\Rightarrow p\left(\frac{2}{\sqrt{3}}\right)=4-1=3$

Then polynomial $p\left(x\right)=3x^2-1$ is zero for value $x=(-\frac{1}{\sqrt{3}})$ and non zero value for $x=\left(\frac{2}{\sqrt{3}}\right)$

So polynomial $p\left(x\right)=3x^2-1$ is zero polynomial for value $x=(-\frac{1}{\sqrt{3}})$ and not zero polynomial value for $x=\left(\frac{2}{\sqrt{3}}\right)$