Question

Find the conditions that $x^3+p{x}^2+qx+r$ may be divisible by $x^2+ax+b$.

Answer 1

$f\left(x\right)=x^3+p{x}^2+qx+r$

& $x^2+ax+b$

because $x^3+p{x}^2+qx+r$ is divisible by $x^2+ax+b$

so result will be in unit power of x

Let result is $x+n$

So

$\Rightarrow x(x^2+ax+b)+n(x^2+ax+b)=x^3+p{x}^2+qx=r$

$\Rightarrow x^3+a{x}^2+bx+n{x}^2+anx+bn=x^3+p{x}^2+qx+r$

$(a+n)=P...\left(i\right)$

$b+an=q...\left(ii\right)$

$bn=r...\left(iii\right)$

$b=r/n$

$a=p-n$

From (ii)

$\frac{r}{n}+n(p-n)=q$

$np-n^2+\frac{r}{n}=q$ where n is real number