Question

If $a{x}^3+b{x}^2+cx+d$ is divisible by $a{x}^2+c$, then show that $a,b,c,d$ are not nessarily in $G.P$.

Answer 1

If $a{x}^3+b{x}^2+cx+d$ is divisible by $a{x}^2+x$ then

$\therefore a{x}^3+b{x}^2+cx+d=(a{x}^2+x)(mx+n)$

$\therefore a{x}^3+b{x}^2+cx+d=am{x}^3+an{x}^2+cmx+cn$

$\therefore$ Matching coefficients:-

$m=1$

$n=\frac{d}{c}$

$\therefore a{x}^3+b{x}^2+cx+d=a{x}^3+\left(ad/c\right)x^2+cx+d$

$\therefore b=ad/c$

$\therefore b/a=d/c$

To be in G.P, we need to show that:

$b/a=c/b=d/c$

We have already shown that $b/a=d/c$ but $c/b$ is not necessarily equal to these.

Hence proved.