Question

# If both (x-2) an are factors of px? + 5x +r, prove that p=r.

Solution

$\Rightarrow P{x}^{2}+5x+r\phantom{\rule{0ex}{0ex}}(x-2)and(x-1/2)arefactor\phantom{\rule{0ex}{0ex}}x=2,x=1/2\phantom{\rule{0ex}{0ex}}\Rightarrow letf\left(x\right)=p{x}^{2}+5x+r\phantom{\rule{0ex}{0ex}}f\left(2\right)=4p+10+r=0.....1\phantom{\rule{0ex}{0ex}}f(1/2)=p/4+5/2+r=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4p+10+r=9/4+5/2+r\phantom{\rule{0ex}{0ex}}\frac{15p}{4}+15/2+r=r\phantom{\rule{0ex}{0ex}}\frac{\overline{)15}p}{4}=\frac{-\overline{)15}}{2}\Rightarrow \overline{)p=-2}\phantom{\rule{0ex}{0ex}}\Rightarrow eqn1-8+10+r=0\phantom{\rule{0ex}{0ex}}r=-2\phantom{\rule{0ex}{0ex}}\mathit{p}\mathbf{=}\mathit{r}\phantom{\rule{0ex}{0ex}}$