<s>[INST] How do I demonstrate that a polynomial of degree $2$ divides one of degree $n$? Let $f$ and $g$ the polynomials
$$f(x) = (x+1)^{2n-1}+(-1)^n(x+2)^{n+1}\qquad\text{and}\qquad g(x) = x^2 + 3x + 3$$
How do I demonstrate that $g$ divides $f$? I tried finding the roots of $g$ then testing them for $f$, it didn't work ...
I also wrote g as $(x+1)^2 + x + 2$ but I couldn't get anywhere.
 [/INST] Let $r$ be any of the two roots of $g$; since $g(r) = (r+1) (r+2) + 1$, obviously $r+1 \ne 0$ and we get $r+2 = -\frac 1 {r+1}$. To have $g \mid f$ we should have $f(r) = 0$, so let us compute $f(r)$.
Since
$$f(r) = (r+1)^{2n-1} + (-1)^n (r+2)^{n+1} = (r+1)^{2n-1} + (-1)^n \left(- \frac 1 {r+1} \right)^{n+1} = \\
(r+1)^{2n-1} - \frac 1 {(r+1)^{n+1}} = \frac {(r+1)^{3n} - 1} {(r+1)^{n+1}} ,$$
let us finally show that $(r+1)^{3n} - 1 = 0$. To do this, notice that
$$(r+1)^3 = r^3 + 3r^2 + 3r + 1 = r g(r) + 1 = 1 ,$$
whence it follows that $(r+1)^{3n} - 1= \left((r+1)^3 \right)^n - 1 = 1^n - 1 = 0$, as desired.
This shows that indeed $f(r) = 0$ for each of the two roots $r$ of $g$, so indeed $g \mid f$.
</s>