Question

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p\left(x\right)=3x+1,x=-\frac{1}{3}$
(ii) $p\left(x\right)=5x-\pi ,x=\frac{4}{5}$
(iii) $p\left(x\right)=x^2-1,x=1,-1$
(iv) $p\left(x\right)=(x+1)(x-2),x=-1,2$
(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) $p\left(x\right)=2x+1,x=\frac{1}{2}$

Answer 1

In order to verify the values are zeros of polynomial $p\left(x\right)$, we must replace the variable $x$ with the given values.

If $p\left(x\right)=0$, then that given value is zero of polynomial $p\left(x\right)$.

(i) $p\left(x\right)=3x+1$:

Put $x=-\frac{1}{3}$, we get,

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

So, $x=-\frac{1}{3}$ is the zero of the given polynomial $p\left(x\right)$.

(ii) $p\left(x\right)=5x-\pi$:

Put $x=\frac{4}{5}$, we get,

$p\left(x\right)=p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi =4-\pi \ne 0$.

So, $x=\frac{4}{5}$ is not the zero of the given polynomial $p\left(x\right)$.

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

Put $x=1$, we get,

$p\left(x\right)=p\left(1\right)=(1{)}^2-1=1-1=0$.

So, $x=1$ is the zero of the given polynomial $p\left(x\right)$.

Now put $x=-1$, we get,

$p\left(x\right)p(-1)=(-1{)}^2-1=1-1=0$.

So, $x=-1$ is the zero of the given polynomial $p\left(x\right)$.

(iv) $p\left(x\right)=(x+1)(x-2)$:

Put $x=-1$, we get,

$p\left(x\right)=(-1+1)(-1-2)=0(-3)=0$.

So, $x=-1$ is the zero of the given polynomial $p\left(x\right)$.

Now put $x=2$, we get,

$p\left(x\right)=(2+1)(2-2)=3\left(0\right)=0$.

So, $x=2$ is the zero of the given polynomial $p\left(x\right)$.

(v) $p\left(x\right)=x^2$:

Put $x=0$, we get,

$p\left(x\right)=p\left(0\right)=(0{)}^2=0$.

So, $x=0$ is the zero of the given polynomial $p\left(x\right)$.

(vi) $p\left(x\right)=lx+m$:

Put $x=-\frac{m}{l}$, we get,

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

So, $x=-\frac{m}{l}$ is the zero of the given polynomial $p\left(x\right)$.

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

Put $x=-\frac{1}{\sqrt{3}}$, we get,

$p\left(x\right)=p(-\frac{1}{\sqrt{3}})=3{(-\frac{1}{\sqrt{3}})}^2-1=(3\times \frac{1}{3})-1=1-1=0$.

So, $x=-\frac{1}{\sqrt{3}}$ is the zero of the given polynomial $p\left(x\right)$.

Now put $x=\frac{2}{\sqrt{3}}$, we get,

$p\left(x\right)=p\left(\frac{2}{\sqrt{3}}\right)=3{\left(\frac{2}{\sqrt{3}}\right)}^2-1=(3\times \frac{4}{3})-1=4-1=3\ne 0$.

So, $x=\frac{2}{\sqrt{3}}$ is not the zero of the given polynomial $p\left(x\right)$.

(viii) $p\left(x\right)=2x+1$:

Put $x=\frac{1}{2}$, we get,

$p\left(x\right)=p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1=1+1=2\ne 0$.

So, $x=\frac{1}{2}$ is not the zero of the given polynomial $p\left(x\right)$.