Find the zero of the polynomial in each of the following cases:

(i) $p(x)=x+5$

(ii) $p(x)=x-5$

(iii) $p(x)=2 x+5$

(iv) $p(x)=3 x-2$

(v) $p(x)=3 x$

(vi) $p(x)=a x, a \neq 0$

(vii) $p(x)=c x+d, c \neq 0, c$, are real numbers.
Solution:

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0 .

(i) $p(x)=x+5$

$p(x)=0$

$x+5=0$

$x=-5$

Therefore, for $x=-5$, the value of the polynomial is 0 and hence, $x=-5$ is a zero of the given polynomial.

(ii) $p(x)=x-5$

$p(x)=0$

$x-5=0$

$x=5$

Therefore, for $x=5$, the value of the polynomial is 0 and hence, $x=5$ is a zero of the given polynomial.

(iii) $p(x)=2 x+5$

$p(x)=0$

$2 x+5=0$

$2 x=-5$

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

Therefore, for $x=-\frac{5}{2}$, the value of the polynomial is 0 and hence, $x=\frac{-5}{2}$ is a zero of the given polynomial.

(iv) $p(x)=3 x-2$

$p(x)=0$

$3 x-2=0$

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

Therefore, for $x=\frac{2}{3}$, the value of the polynomial is 0 and hence, $x=\frac{2}{3}$ is a zero of the given polynomial,

(v) $p(x)=3 x$

$p(x)=0$

$3 x=0$

$x=0$

Therefore, for $x=0$, the value of the polynomial is 0 and hence, $x=0$ is a zero of the given polynomial.

(vi) $p(x)=a x$

$p(x)=0$

$a x=0$

$x=0$

Therefore, for $x=0$, the value of the polynomial is 0 and hence, $x=0$ is a zero of the given polynomial.

(vii) $p(x)=c x+d$

$p(x)=0$

$C X+d=0$

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

Therefore, for $x=\frac{-d}{c}$, the value of the polynomial is 0 and hence, $x=\frac{-d}{c}$ is a zero of the given polynomial.