Question

Question 3
if both x – 2 and $x-\frac{1}{2}$ are factors of $p{x}^2+5x+r$, then show that p = r.

Answer 1

Let $f\left(x\right)=p{x}^2+5x+r$
Since x – 2 is a factor of f(x), f(2) = 0.
$\therefore p(2{)}^2+5(2)+r=0$
$\to 4p+10+r=0$ …(i)
Since $x-\frac{1}{2}$ is a factor of f(x), f(2) = 0.
$\therefore p{\left(\frac{1}{2}\right)}^2+5\left(\frac{1}{2}\right)+r=0$
$\Rightarrow p\times \frac{1}{4}+\frac{5}{2}+r=0$
$\Rightarrow p+10+4r=0$ ...(ii)
Since $x-2$ and $x-\frac{1}{2}$ are factors of $f\left(x\right)=p{x}^2+5x+r$,
from equations (i) and (ii),
$4p+10+r=p+10+4r\Rightarrow 3p=3r$
$p=r$