Question

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

Answer 1

$f\left(x\right)=p{x}^2+5x+r$

$(x-2)$ and $(x-\frac{1}{2})$ are factors of $f\left(x\right)$

$(x-2)(x-\frac{1}{2})=p{x}^2+5x+r$ Writing factors theorem

$x^2-(2+\frac{1}{2})x+2\times \frac{1}{2}=p{x}^2+5x+r$

$x^2-\frac{5}{2}x+1=p{x}^2+5x+r$

Equating the coefficients of $x^2,x$ and $r$, we get

$2x^2-5x+2=p{x}^2+5x+r$

$-2x^2+5x-2=p{x}^2+5x+r$

$\therefore p=-2,r=-2$

Hence $p=r$