I need to prove or disprove this:
If $P\in \mathbb R[X]$ polynomial with $P(1)=0$ and $P(-2)=5$ then $\gcd(P,x^2+x-2)=x-1$.
We know that $(x-1)\mid P$ and $(x+2)\nmid P$.
Both $(x-1),(x+2)\mid x+2+x-2$ since $x+2+x-2=(x-1)(x+2)$.
Now if $M\mid S$ and $M\mid T$ then $M\mid\gcd(S,T)$ so $(x-1)\mid\gcd(P,x^2+x-2)$.
I don't really know how to go from here. I think the statement is incorrect but I can't find an example to support that.
Thanks!
One step is left. $(x-1)|\gcd(P,x^2+x-2)$, and $\gcd(P,x^2+x-2)|(x^2+x-2),$ so $\gcd(P,x^2+x-2)$ is either $x-1$, or $(x-1)(x+2).$ As you've noticed, $(x+2)\nmid P$, so $\gcd$ also can't be divisible by $x+2$, so it's $x-1$.
A general method: by the GCD Distributive Law we can factor out $\,\color{#c00}{x\!-\!1}\,$ from the gcd, i.e.
$$\begin{align} &\ \gcd(\overbrace{(\color{#c00}{x\!-\!1})Q}^{\large P},\,(\color{#c00}{x\!-\!1})(x\!+\!2))\\[.2em] = &\ \ (\color{#c00}{x\!-\!1})\gcd(Q,\,x\!+\!2) \end{align}\qquad$$
$x\!+\!2\,$ is irreducible, so $\,\gcd(Q,\,x\!+\!2)=1\iff x\!+\!2\nmid Q\iff Q(-2)\neq 0$,
true by $P(-2)=5\neq 0,\,$ i.e. $\,P= (x\!-\!1)Q\,$ so $\,P(-2) = 0\iff Q(-2)= 0$.
