Question

If $\left( x+a \right)$ is a factor of two polynomials ${{x}^{2}}+px+q$ and ${{x}^{2}}+mx+n$ the prove that: $a=\dfrac{n-q}{m-p}$.

Answer 1

Hint: It is given that $\left( x+a \right)$ is a factor of two polynomials ${{x}^{2}}+px+q$ and ${{x}^{2}}+mx+n$ so substituting $x=-a$ in the given polynomials and after putting this x value, equate that expression to 0. Now, we have two equations in $''a''$ so rearrange these equations to get the value of $''a''$.

Complete step by step answer:
In the above problem, we have given two polynomial expressions as follows:
${{x}^{2}}+px+q$
${{x}^{2}}+mx+n$
Also, it is given that $\left( x+a \right)$ is a factor of these polynomial expressions. So, we are going to substitute the value of $x=-a$ in these two polynomial expression and we get,
$\begin{align}
  & {{\left( -a \right)}^{2}}+p\left( -a \right)+q; \\
 & {{\left( -a \right)}^{2}}+m\left( -a \right)+n \\
\end{align}$
As $\left( x+a \right)$ is a factor of these polynomials so putting the value of $x=-a$ in these polynomials will give answer 0 so equating them to 0 we get,
$\begin{align}
  & {{\left( -a \right)}^{2}}+p\left( -a \right)+q=0 \\
 & \Rightarrow {{a}^{2}}-ap+q=0.......(1) \\
 & {{\left( -a \right)}^{2}}+m\left( -a \right)+n=0 \\
 & \Rightarrow {{a}^{2}}-ma+n=0......(2) \\
\end{align}$
Subtracting eq. (1) from eq. (2) we get,
${{a}^{2}}-ma+n-\left( {{a}^{2}}-ap+q \right)=0$
Now, opening the bracket with negative sign in front of it in the above equation and we get,
$\begin{align}
  & \Rightarrow {{a}^{2}}-ma+n-{{a}^{2}}+ap-q=0 \\
 & \Rightarrow -ma+ap+n-q=0 \\
\end{align}$
Taking $''a''$ as common in the first two terms of the L.H.S of the above expression and we get,
$a\left( p-m \right)+n-q=0$
Rearranging the above equation to the form given in the above problem we get,
$a\left( p-m \right)=q-n$
Dividing $\left( p-m \right)$ on both the sides of the above equation and we get,
$a=\dfrac{q-n}{p-m}$
Multiplying and dividing by -1 in the R.H.S of the above equation we get,
$\begin{align}
  & \Rightarrow a=\dfrac{-\left( q-n \right)}{-\left( p-m \right)} \\
 & \Rightarrow a=\dfrac{n-q}{m-p} \\
\end{align}$
Hence, we have proved the given condition $a=\dfrac{n-q}{m-p}$.

Note: The mistake which could be possible in the above problem is that instead of putting $x=-a$, you might put $x=a$ in the above problem to prove the given condition. To avoid such a mistake, put the factor given above to 0 and then find the value of $''x''$ from that.