Finding the Remainder Given the polynomials $$P(x) = nx^n+(n-1)x^{n-1}+(n-2)x^{n-2}+\cdots+x+1$$ and $$Q(x)=x(x-1)^2$$ find the remainder of the division $\dfrac{P (x)}{Q (x)}$.
 A: Hint $P(x)=Q(x)q(x)+r(x)$ where $r(x)=ax^2+bx+c$ is a quadratic, and $Q(0)=Q(1)=Q'(1)=0$
Set $x=0$ to find $r(0)=1$
Set $x=1$ to find $r(1)=n+(n-1)+\dots + 1+1$
Differentiate with respect to $x$ and set $x=1$ to obtain $2a+b=n^2+(n-1)^2+\dots +1$
This should give you three equations in three unknowns (the coefficients of $r(x)$).
Since you didn't show much work or motivation, I've left some gaps for you to fill in.
A: Hints:
Partial fractions:
$$\frac{P(x)}{x(x-1)^2}=\frac Ax+\frac B{x-1}+\frac C{(x-1)^2}\implies$$
$$P(x)=A(x-1)^2+Bx(x-1)+Cx$$
But
$$1\stackrel{\text{why?}}=P(0)=A$$
$$\frac{n(n+1)}2\stackrel{\text{why?}}=P(1)=C$$
and etc.
A: HINT:
Clearly, $(x,(x-1)^2)=1$
Now, $\displaystyle P(x)\equiv1\pmod x\ \ \ \ (1)$
Again, $\displaystyle P(x)=1+\sum_{r=1}^n rx^r=1+\sum_{r=1}^nr(1+x-1)^r$
Using Binomial Expansion,
$\displaystyle P(x)\equiv 1+\sum_{r=1}^nr\{1+r(x-1)\}\pmod{(x-1)^2}$
$\displaystyle\implies  P(x)\equiv1+\sum_{r=1}^nr+(x-1)\sum_{r=1}^nr^2\ \ \ \ (2)$
Now, set $\displaystyle P(x)=A(x) x(x-1)^2+Bx^2+Cx+D$ where $B,C,D$ are constants and $A(x)$ is  a polynomial 
A: Let 

$$ q(x) = 1+x+x^2+\dots+x^{n-1}+x^{n}= \frac{1-x^{n+1}}{1-x}, $$

then the polynomial 

$$P(x) = 1+xq'(x)$$

Can you finish the problem? 
