Question

If $(x+a)$ is a factor of $x^2+px+q$ and $x^2+mx+n$, then the value of $a$ is

• A. $\frac{m-p}{n-q}$
• B. $\frac{m+p}{n+q}$
• C. $\frac{n-q}{m-p}$
• D. $\frac{n+q}{m+p}$

Answer 1

The correct option is C

$\frac{n-q}{m-p}$

Let $f\left(x\right)$ = $x^2+px+q$
And $q\left(x\right)$ = $x^2+mx+n$
$(x+a)$ is a factor of $f\left(x\right)$ and $g\left(x\right)$
By remainder theorem,
$f(-a)={(-a)}^2+p(-a)+q=0$
$\Rightarrow a^2-ap+q=0$ ...(i)
Also,
$g(-a)={(-a)}^2+m(-a)+n=0$
$\Rightarrow a^2-am+n=0$ ...(ii)
Subtracting (ii) from (i)
$-a(p-m)+q-n=0$
$\Rightarrow -a(p-m)=-(q-n)\Rightarrow a(p-m)=q-n\Rightarrow a=\frac{n-q}{m-p}$