Verify whether the following are zeroes of the polynomial, indicated against them. 

(i) $p\left(x\right)=3x+1, x=-\frac{1}{3}$ (ii) $p\left(x\right)=5x-\mathrm{\pi }, x=\frac{4}{5}$

(iii) $p\left(x\right)=x^2-1, x=1,-1$ (iv) $p\left(x\right)=\left(x+1\right)\left(x-2\right), x=-1,2$

(v) $p\left(x\right)=x^2, x=0$ (vi) $p\left(x\right)=lx+m, x=-\frac{m}{l}$

(vii) $p\left(x\right)=3x^2-1, x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$ (viii) $p\left(x\right)=2x+1, x=\frac{1}{2}$

(i) To verify the zeroes of the polynomial, we have:

$\begin{array}{rcl}p\left(x\right)& =& 3x+1\\ p\left(-\frac{1}{3}\right)& =& 3\left(-\frac{1}{3}\right)+1\\ & =& -1+1\\ & =& 0\end{array}$

So, after putting the value of $x$, the polynomial becomes zero.

Hence, $x=-\frac{1}{3}$is a zero of the given polynomial.

(ii) To verify the zeroes of the polynomial, we have:

$\begin{array}{rcl}p\left(x\right)& =& 5x-\mathrm{\pi }\\ p\left(\frac{4}{5}\right)& =& 5\left(\frac{4}{5}\right)-\mathrm{\pi }\\ & =& 4-\mathrm{\pi }\end{array}$

So, after putting the value of $x$, the polynomial does not become zero.

Hence, $x=\frac{4}{5}$is not a zero of the given polynomial.