Question

If $a{x}^3+b{x}^2+cx+d$ is divisible by $x^2+h^2$, prove that $ad=bc$.

Answer 1

Let the cubic equation $a{x}^3+b{x}^2+cx+d$ is divisible by $x^2+h^2$.

Then we can write ,

$a{x}^3+b{x}^2+cx+d=(x^2+h^2)(px+q)$

$a{x}^3+b{x}^2+cx+d=px(x^2+h^2)+q(x^2+h^2)$

$a{x}^3+b{x}^2+cx+d=p{x}^3+q{x}^2+p{h}^{2}x+q{h}^2$

Therefore ,

upon comparing the both eqution we get ,

$p=a$

$q=b$

$p{h}^2=c$

$q{h}^2=d$

Thus ,

$ad=pq{h}^2$

$bc=pq{h}^2$

Thus $ad=bc$