Question

If $x+a$ is the HCF of $x^2+px+q$ and $x^2+1x+m$, then find the value of '$a$'.

Answer 1

Determine the value of $a$.

If $(x–a)$ is a factor of $p\left(x\right)$, then $p\left(a\right)=0$.

Let $f\left(x\right)=x^2+px+q$ and $g\left(x\right)=x^2+1x+m$.

Here, $x+a=x-\left(-a\right)$ is a factor of $f\left(x\right)$, then $f(-a)=0$.

$$\begin{gathered} \begin{array}{rcl}f\left(-a\right)& =& 0\\ & \Rightarrow & {\left(-a\right)}^2+p\left(-a\right)+q=0\\ & \Rightarrow & a^2-pa+q=0\end{array} \\ \begin{array}{rcl}& \Rightarrow & a^2=pa-q\dots \left(1\right)\end{array} \end{gathered}$$

Here, $x+a=x-\left(-a\right)$ is a factor of $g\left(x\right)$, then $g(-a)=0$.

$$\begin{gathered} g\left(-a\right)=0 \\ \Rightarrow {\left(-a\right)}^2+1\left(-a\right)+m=0 \\ \Rightarrow a^2=a-m\dots \left(2\right) \end{gathered}$$

Form equation $\left(1\right)$ and $\left(2\right)$, we get

$\begin{array}{rcl}pa-q& =& a-m\\ pa-a& =& q-m\\ a& =& \frac{q-m}{p-1}\end{array}$

Hence, the value of $a$ is $\frac{q-m}{p-1}$.