Find a polynomial $p(x)$ such that $p(x)(2x+6x^2+\cdots + n(n+1)x^n + \cdots ) = \frac{2x(1+x)^2}{(1-x)}$
I believe I am supposed to use the generating function $\frac{x(x+1)}{(1-x)^3} = \sum_{n=0}^{\infty}n^2x^n$, but I am having trouble because my knowledge of geometric functions is limited, so I am not sure fully how to approach this.
Hint
It could be easier to work first $$A(x)=\sum_{n=1}^\infty n(n+1)x^n=\sum_{n=1}^\infty \left(n(n-1)+2n\right)x^n=\sum_{n=1}^\infty n(n-1)x^n+2\sum_{n=1}^\infty nx^n$$ which is $$A(x)=x^2\left(\sum_{n=1}^\infty x^n\right)''+2x\left(\sum_{n=1}^\infty x^n\right)'$$ Now, remember what is $\sum_{n=1}^\infty x^n$, compute the derivatives and solve for $P$ $$P(x) \times A(x)=\frac{2x(1+x)^2}{(1-x)}\implies P(x)=\frac{2x(1+x)^2}{(1-x)\,A(x)}$$ I am sure that you can take it from here.