Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$p{x}^2+5x+r=0---\left(1\right)$
$(x-2)(x-\frac{1}{2})=0$ [since it is a factor]
$x^2-\frac{x}{2}-2x+1=0$
$x^2-\left(\frac{x+4x}{2}\right)+1=0$
$2x^2-5x+2=0---\left(2\right)$
Comparing equation(1) and equation(2), we get,
$p=r$