Clarification on partial fraction expansion I would like to use the cover up method for the following equation.
$$\frac{1}{x^2(x+1.79)}$$
and it breaks down into
$$\frac{A}{x}\quad\frac{B}{x^2}\quad\frac{C}{(x + 1.79)}$$
I realize that you cover up $(x+1.79)$ to get $x = -1.79$ but what do you do for the other variables?
Thanks for your help.

Here it is another way to approach for the sake of curiosity.
On one hand, we have that
\begin{align*}
\frac{1}{x^{2}(x+k)} & = \frac{1}{k}\left(\frac{k}{x^{2}(x+k)}\right)\\\\
& = \frac{1}{k}\left(\frac{(x + k) - x}{x^{2}(x+k)}\right)\\\\
& = \frac{1}{kx^{2}} - \frac{1}{kx(x+k)}
\end{align*}
On the other hand, we have that
\begin{align*}
\frac{1}{x(x+k)} & = \frac{1}{k}\left(\frac{k}{x(x+k)}\right)\\\\
& = \frac{1}{k}\left(\frac{(x + k) - x}{x(x+k)}\right)\\\\
& = \frac{1}{kx} - \frac{1}{k(x+k)}
\end{align*}
Gathering both results, it results that
\begin{align*}
\frac{1}{x^{2}(x+k)} = \frac{1}{kx^{2}} - \frac{1}{k^{2}x} + \frac{1}{k^{2}(x+k)}
\end{align*}
At your case, $k = 1.79$.
