<s>[INST] Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $\mathbb{Q}$
What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$. 
Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that 
$5A-B=(-1)^n5^{n-1}$
But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?
 [/INST] Hint $\ $ Subtracting $\, C = 5\,$ then cancelling $x$ yields
$\qquad  f(x) := x^{\large n-1} = B + Ax + (x+5)^{\large 2} p(x)  $
Thus $\, B + Ax = f(-5) + f'(-5)(x+5)\,$ by Taylor expansion at $\,x = -5\,$
</s>