Question

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

Answer 1

$itx-2\&x-\frac{1}{2}\text{are facters of}p{x}^2+5x+r\left(1\right)$

$thenx=2\&x=\frac{1}{2}\text{are --- of --}$

$\therefore 4p+10tr=0\Rightarrow 4p+r=-10$

$Also\frac{1}{4}p+\frac{5}{2}+v=0\Rightarrow p+4r=-10$

$\Rightarrow 4p+r=p+4r$

$Also\Rightarrow p=r$

$\text{By solving, we have}4p+r=-10$

$7p+16r=-40$

$-15r=30$

$\Rightarrow r=-2$

$p=-10-4r$

$=10+8=-2$

$\Rightarrow p=r$