Question

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

• A. $p-r=0$
• B. $p+r=0$
• C. $2p+r=0$
• D. $p+2r=0$

Answer 1

The correct option is A $p-r=0$

Given $(x-2)\text{is a factor of}q\left(x\right)=p{x}^2+5x+r.$

$\text{By factor theorem, if}(x-a)\text{is a factor}\text{of}p\left(x\right),\text{then}p\left(a\right)=0.$

$\text{Using factor theorem,}q\left(2\right)=0p\times 2^2+5\times 2+r=04p+r=-10...\left(i\right)$

$(x-\frac{1}{2})\text{is also a factor of}q\left(x\right).\text{Using factor thorem,}q\left(\frac{1}{2}\right)=0p\times (\frac{1}{2}{)}^2+5\times (\frac{1}{2})+r=0\frac{p}{4}+\frac{5}{2}+r=0p+4r=-10...(ii)$

$\text{Subtracting (ii) from (i), we get:}(4p+r)-(p+4r)=-10-(-10)3p-3r=0p=r\Rightarrow p-r=0$