Question

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

Answer 1

The given polynomial is $Q\left(x\right)=p{x}^2+5x+r$

It is also given that $(x-2)$ and $(x-\frac{1}{2})$ are the factors of $Q\left(x\right)$ which means that $Q\left(2\right)=0$ and $Q\left(\frac{1}{2}\right)=0$.

Let us first substitute $Q\left(2\right)=0$ in $Q\left(x\right)=p{x}^2+5x+r$ as shown below:

$Q\left(x\right)=p{x}^2+5x+r\Rightarrow Q\left(2\right)=p(2{)}^2+(5\times 2)+r\Rightarrow 0=(p\times 4)+10+r\Rightarrow 0=4p+10+r$

$\Rightarrow 4p+r=-10.........\left(1\right)$

Now, substitute $Q\left(\frac{1}{2}\right)=0$ in $Q\left(x\right)=p{x}^2+5x+r$ as shown below:

$Q\left(x\right)=p{x}^2+5x+r\Rightarrow Q\left(\frac{1}{2}\right)=p{\left(\frac{1}{2}\right)}^2+(5\times \frac{1}{2})+r\Rightarrow 0=\frac{p}{4}+\frac{5}{2}+r\Rightarrow 0=\frac{p+(5\times 2)+(r\times 4)}{4}$

$\Rightarrow 0=\frac{p+10+4r}{4}\Rightarrow 0\times 4=p+10+4r\Rightarrow 0=p+10+4r$

$\Rightarrow p+4r=-10.........\left(2\right)$

Now subtracting the equations $\left(1\right)$ and $\left(2\right)$, we get

$(4p-p)+(r-4r)=-10-(-10)\Rightarrow 3p-3r=-10+10\Rightarrow 3p-3r=0\Rightarrow 3p=3r\Rightarrow p=r$

Hence, $p=r$ is proved.