Let $F=2X^3-X^2+5X-6$, $G \in \mathbb{R}[X]$, $\deg(G)=2$. If $F + G$ has roots $-1,0,1$, find lowest degree $H$ s.t. $G\mid H$, $\gcd(H,H')=X^2(X-6)$

Question

I began by noting that, if $X$ is a root of $F+G$, then $$(F+G)(X) = F(X) + G(X) = 0$$

Since $F(1)=0, F(0)=-6, F(-1)=-14$, we conclude that $G(1)=0,G(0)=6,G(-1)=14$. Moreover, given that $\deg(G)=2$, this is enough information to conclude that $G=X^2-7X+6=(X-1)(X-6)$

We know that $G \mid H$, so $H=GQ=(X-1)(X-6)Q$. But we also know that $\gcd(H,H')=X^2(X-6)$. This implies that two of $H$'s roots, $0$ and $6$, have multiplicity $3$ and $2$ respectively.

Hence, the lowest degree polynomial that verifies all the conditions is

$$H=X^3(X-6)^2(X-1)$$

Is this correct and properly justified?

Answer 1

As has been pointed out in the comments, I would maybe spend a line or two justifying why $\gcd(H, H') = X^2(X-6)$ implies what you say it implies about the degree of the roots of $H$.

Otherwise, this looks fine to me.