Find the zero of the polynomial in each of the following cases :
$p (x) = c x + d, c \neq 0, c, d$ are real numbers.

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Solution

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

To find zero of a polynomial we put $p (x) = 0$ and find $x$

$\Rightarrow c x + d = 0$

$\Rightarrow c x = - d$

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

Therefore $x = - \frac{d}{c}$ is the zero of the given polynomial.

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Q. Find the zero of the polynomial in each of the following cases:
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(i) $p (x) = x + 5$
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Q. Question 4
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) = 2x + 5
(iv) p(x) = 3x - 2
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Find the zero of the polynomial in each of the following cases :p(x)=cx+d,c≠0,c,d are real numbers.

Given p(x)=cx+d,c≠0,c,d are real numbers.To find zero of a polynomial we put p(x)=0 and find x⇒cx+d=0⇒cx=−d⇒x=−dcTherefore x=−dc is the zero of the given polynomial.

Find the zero of the polynomial in each of the following cases :
$p (x) = c x + d, c \neq 0, c, d$ are real numbers.

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

To find zero of a polynomial we put $p (x) = 0$ and find $x$

$\Rightarrow c x + d = 0$

$\Rightarrow c x = - d$

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

Therefore $x = - \frac{d}{c}$ is the zero of the given polynomial.