Question

Find the zero of the polynomial in each of the following cases:
(i) $p\left(x\right)=x+5$
(ii) $p\left(x\right)=x-5$
(iii) $p\left(x\right)=2x+5$
(iv) $p\left(x\right)=3x-2$
(v) $p\left(x\right)=3x$
(vi) $p\left(x\right)=ax,a\ne 0$
(vii) $p\left(x\right)=cx+d,c\ne 0$ and $c,d$ are real numbers.

Answer 1

To get the zero of polynomial $p\left(x\right)$, take $p\left(x\right)=0$ and hence solve the equation.

(i) $p\left(x\right)=x+5$

$x+5=0$

$\Rightarrow x=-5$

$\Rightarrow -5$ is the zero of $p\left(x\right)=x+5$.

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

$x+5=0$

$\Rightarrow x=5$

$\Rightarrow 5$ is the zero of $p\left(x\right)=x-5$.

(iii) $p\left(x\right)=2x+5$

$2x+5=0$

$\Rightarrow 2x=-5$

$\Rightarrow x=\frac{-5}{2}$

$\Rightarrow -\frac{5}{2}$ is the zero of $p\left(x\right)=2x+5$

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

$3x-2=0$

$\Rightarrow 3x=2$

$\Rightarrow x=\frac{2}{3}$

So $\frac{2}{3}$ is the zero of $p\left(x\right)=3x-2$.

(v) $p\left(x\right)=3x$

$3x=0$

$\Rightarrow x=0$

So $0$ is the zero of $p\left(x\right)=3x$.

(vi) $p\left(x\right)=ax,a\ne 0$

$ax=0,a\ne 0$

$\Rightarrow x=\frac{0}{a}=0$

So, $0$ is the zero of .$p\left(x\right)=ax,a\ne 0$

(vii) $p\left(x\right)=cx+d,c\ne 0$ and $c,d$ are real numbers.

$cx+d=0,(c\ne 0)$

$\Rightarrow cx=-d$

$\Rightarrow x=\frac{-d}{c}$

So, $\frac{-d}{c}$ is the zero of $p\left(x\right)=cx+d$.