If both $x-2$ and $x-\frac{1}{2}$ are factors of $p x^{2}+5 x+r,$ show that $p=r$

Let $p(x)=p x^{2}+5 x+r$

As $(x-2)$ is a factor of $p(x)$

So, $p(2)=0 \Rightarrow P(2)^{2}+5(2)+r=0$

$\Rightarrow \quad 4 p+10+r=0.....(1)$

Again, $\left(x-\frac{1}{2}\right)$ is factor of $p(x)$

$\therefore \quad p\left(\frac{1}{2}\right)=0$

Now, $\quad p\left(\frac{1}{2}\right)=p\left(\frac{1}{2}\right)^{2}+5\left(\frac{1}{2}\right)+r$

$=\frac{1}{4} p+\frac{5}{2}+r$

$\therefore \quad p\left(\frac{1}{2}\right)=0 \Rightarrow \frac{1}{4} p+\frac{5}{2}+r=0....(2)$

From $(1),$ we have $4 p+r=-10$

From $(2),$ we have $p+10+4 r=0$

$\Rightarrow \quad p+4 r=-10$

$\therefore \quad 4 p+r=p+4 r$ $[\because$ Each $=-10]$

$\therefore \quad 3 p=3 r \Rightarrow p=r$

Hence, proved.