Question

If both $x-2$ and $x-\frac{1}{2}$ are factors of $p{x}^2+5x+r$. Which of the conditions hold true?

• A. $p=-r$
• B. $p=r=0$
• C. $p=r=-1$
• D. $p=r$

Answer 1

The correct option is D $p=r$

Given polynomial is $q\left(x\right)=p{x}^2+5x+r$ ...... $\left(i\right)$
If $x-2$ is the factor of $q\left(x\right)$ then $q\left(2\right)=0$
$p(2{)}^2+5\times 2+r=0$
$4p+r+10=0$ ...... $\left(ii\right)$
If $x-\frac{1}{2}$ is the factor $q\left(x\right)$, then $q\left(\frac{1}{2}\right)=0$

$\Rightarrow p{\left(\frac{1}{2}\right)}^2+5\times \frac{1}{2}+r=0$
$\Rightarrow \frac{p}{4}+\frac{5}{2}+r=0$
$\Rightarrow p+4r+10=0$ ..... $\left(iii\right)$

Subtracting $\left(iii\right)$ from $\left(ii\right)$
$4p+r+10-(p+4r+10)=0$
$\Rightarrow 3p-3r=0$
$\Rightarrow 3p=3r$
$\Rightarrow p=r$