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

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Solution

Given polynomial $f (x) = p x + q$
we know that for zero polynomial f(x)=0
$p x + q = 0$
$p x + q - q = - q$
$\Rightarrow p x = - q$
$x = - \frac{q}{p}$ (because $p \neq 0$ ,pq are real number )
Then for value of x is $- \frac{q}{p}$ the polynomial is zero polynomial

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Find the zero of the polynomial in each of the following cases.f(x)=px+q,p≠0,pq are real numbers

Given polynomial f(x)=px+qwe know that for zero polynomial f(x)=0px+q=0px+q−q=−q⇒px=−qx=−qp (because p≠0,pq are real number )Then for value of x is −qp the polynomial is zero polynomial

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

Given polynomial $f (x) = p x + q$
we know that for zero polynomial f(x)=0
$p x + q = 0$
$p x + q - q = - q$
$\Rightarrow p x = - q$
$x = - \frac{q}{p}$ (because $p \neq 0$ ,pq are real number )
Then for value of x is $- \frac{q}{p}$ the polynomial is zero polynomial