ACGNmath wrote: Real numbers $a \neq 0, b, c$ are given. Prove that there is a polynomial $P(x)$ with real coefficients such that the polynomial $x^2+1$ divides the polynomial $aP(x)^2+bP(x)+c$. Proposed by A. Golovanov Let $u+iv$ ($u,v\in\mathbb R$) any root of equation $az^2+bz+c=0$ Just choose $P(x)=u+vx$

What if $b^2 \geq 4ac$?

Mamat wrote: What if $b^2 \geq 4ac$? Polynomial $P(x)$ can be constant (take $P(x)\equiv u$). Also you can take $P(x)=x^2+1+u$, it's also works.

$P(x)=u+vx$ $a(u+vx)^2+b(u+vx)+c\equiv a(u^2-v^2)+bu+c+(2au+b)vx\pmod{x^2+1}$ set $a(u^2-v^2)+bu+c=(2au+b)v=0$ $u=\frac{-b}{2a},v=\pm\frac{\sqrt{4ac-b^2}}{2a}$ or $v=0,u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$