Question

If $a{x}^2+bx+c$ and $b{x}^2+ax+c$ have a common factor $x+1$ then show that $c=0$ and $a=b$

Answer 1

The given polynomials are $P\left(x\right)=a{x}^2+bx+c$ and $Q\left(x\right)=b{x}^2+ax+c$

It is also given that $(x+1)$ is the common factor of $P\left(x\right)$ and$Q\left(x\right)$ which means that $P(-1)=0$ and $Q(-1)=0$.

Let us first substitute $P(-1)=0$ in $P\left(x\right)=a{x}^2+bx+c$ as shown below:

$P\left(x\right)=a{x}^2+bx+c\Rightarrow P(-1)=a(-1{)}^2+(b\times -1)+c\Rightarrow 0=(a\times 1)-b+c\Rightarrow a-b+c=0.........(1)$

Now, substitute $Q(-1)=0$ in $Q\left(x\right)=b{x}^2+ax+c$ as shown below:

$Q\left(x\right)=b{x}^2+ax+c\Rightarrow Q(-1)=b(-1{)}^2+(a\times -1)+c\Rightarrow 0=(b\times 1)-a+c\Rightarrow -a+b+c=0.........(2)$

Now subtracting the equations 1 and 2, we get

$(a-(-a\left)\right)-b-b+c-c=0\Rightarrow a+a-2b=0\Rightarrow 2a-2b=0\Rightarrow 2a=2b\Rightarrow a=b$

Now substitute $a=b$ in equation 1:

$a-a+c=0$

$\Rightarrow c=0$

Hence $a=b$ and $c=0$.