Partial fractions and using values not in domain I'm studying partial fraction decomposition of rational expression. In this video the guy decompose this rational expression:
$$ \frac{3x-8}{x^2-4x-5}$$
this becomes:
$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$
$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1}\times[(x-5)(x+1)]$$
$$3x-8 = A(x+1) + B(x-5)$$
then he substitutes $x$ with $-1$ this is when I feel something is wrong.
because $x=-1$ is not in the domain of the original function and all the statement below the original function hold true only if they comply within the domain of the original function isn't it? so at the time this guy substitute $x=-1$ the statement become false?
I'm newbie in math, please explain to me with easy understanding and step by step.

Observe that the equation
\begin{eqnarray}
\frac{3x-8}{(x+1)(x-5)}=\frac{A}{x-5}+\frac{B}{x+1}
\end{eqnarray}
is valid only for $x\in \mathbb{R}\backslash \{-1,5\}$, as you have rightly pointed out. However, the equation
\begin{eqnarray}
3x-8=A(x+1)+B(x-5)
\end{eqnarray}
is valid for all $x\in \mathbb{R}$.
