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

(i) $p(x)=3 x+1, x=-\frac{1}{3}$

(ii) $p(x)=5 x-\pi, x=\frac{4}{5}$

(iii) $p(x)=x^{2}-1, x=1,-1$

(iv) $p(x)=(x+1)(x-2), x=-1,2$

(v) $p(x)=x^{2}, x=0$

(vi) $p(x)=l x+m$

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

(viii) $p(x)=2 x+1, x=\frac{1}{2}$
Solution:

(i) If $x=\frac{-1}{3}$ is a zero of given polynomial $p(x)=3 x+1$, then $p\left(-\frac{1}{3}\right)$ should be 0 .

Here, $p\left(\frac{-1}{3}\right)=3\left(\frac{-1}{3}\right)+1=-1+1=0$

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

(ii) If $x=\frac{4}{5}$ is a zero of polynomial $p(x)=5 x-\pi$, then $p\left(\frac{4}{5}\right)$ should be 0 .

Here, $p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi=4-\pi$

$A s p\left(\frac{4}{5}\right) \neq 0$

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

(iii) If $x=1$ and $x=-1$ are zeroes of polynomial $p(x)=x^{2}-1$, then $p(1)$ and $p(-1)$ should be 0 .

Here, $p(1)=(1)^{2}-1=0$, and

$p(-1)=(-1)^{2}-1=0$

Hence, $x=1$ and $-1$ are zeroes of the given polynomial.

(iv) If $x=-1$ and $x=2$ are zeroes of polynomial $p(x)=(x+1)(x-2)$, then $p(-1)$ and $p(2)$ should be 0 .

Here, $p(-1)=(-1+1)(-1-2)=0(-3)=0$, and

$p(2)=(2+1)(2-2)=3(0)=0$

Therefore, $x=-1$ and $x=2$ are zeroes of the given polynomial.

(v) If $x=0$ is a zero of polynomial $p(x)=x^{2}$, then $p(0)$ should be zero.

Here, $p(0)=(0)^{2}=0$

Hence, $x=0$ is a zero of the given polynomial.

(vi) If $x=\frac{-m}{l}$ is a zero of polynomial $p(x)=l x+m$, then $p\left(\frac{-m}{l}\right)$ should be 0 .

Here, $p\left(\frac{-m}{l}\right)=l\left(\frac{-m}{l}\right)+m=-m+m=0$

Therefore, $x=-\frac{m}{l}$ is a zero of the given polynomial.

(vii) If $x=\frac{-1}{\sqrt{3}}$ and $x=\frac{2}{\sqrt{3}}$ are zeroes of polynomial $p(x)=3 x^{2}-1$, then

$p\left(\frac{-1}{\sqrt{3}}\right)$ and $p\left(\frac{2}{\sqrt{3}}\right)$ should be 0

Here, $p\left(\frac{-1}{\sqrt{3}}\right)=3\left(\frac{-1}{\sqrt{3}}\right)^{2}-1=3\left(\frac{1}{3}\right)-1=1-1=0$, and

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

Hence, $x=\frac{-1}{\sqrt{3}}$ is a zero of the given polynomial. However, $x=\frac{2}{\sqrt{3}}$ is not a zero of the given polynomial.

(viii) If $x=\frac{1}{2}$ is a zero of polynomial $p(x)=2 x+1$, then $p\left(\frac{1}{2}\right)$ should be 0 .

Here, $p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1=1+1=2$

$\operatorname{As} p\left(\frac{1}{2}\right) \neq 0$

Therefore, $x=\frac{1}{2}$ is not a zero of the given polynomial.
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