Question

If $x+a$ is a factor of $x^2+px+q$ and $x^2+mx+n$, show that $a=\frac{n-q}{m-p}$.

Answer 1

Given :

$(x+a)$ is a factor of $x^2+px+q$ and $x^2+mx+n$

then using the factor theorem which says that the polynomial $f(x0$ has a factor $(x-k)$ if and only if $f\left(k\right)=0$

We have

${(-a)}^2+p(-a)+q=0⟶\left(1\right)$

$\Rightarrow a^2-ap+q=0⟶\left(2\right)$

and

${(-a)}^2+m(-a)+n=0⟶\left(3\right)$

$\Rightarrow a^2-ma+n=0⟶\left(4\right)$

Subtracting $\left(2\right)$ & $\left(4\right)$ we get

$-ap+am+q-n=0$

$\Rightarrow +a(m-p)=n-q$

$\Rightarrow a=\frac{n-q}{m-p}$

Hence, proved