Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$
Why is the denominator under $C$ not simply $x+1$? It is $x$ times $(x+1)^2$ and not $(x+1)^3$

You need an $(x+1)^2$ in the denominator of at least one of the partial fractions, or when you sum them they would not have an $(x+1)^2$ term in the denominator of the sum.
But you might find it easier to solve:
$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{Bx + C}{(x+1)^2}$
And that is completely valid.
