Skepticism concerning Heaviside's "Cover-up Method" for partial fraction decomposition

Question

I was reading this paper from MIT and it introduces Heaviside’s Cover-up Method for partial fraction decomposition. In that paper in Example $1$ it solves a problem using that method and just when explaining why it works (on the same page-1) it says-

Why does the method work? The reason is simple. The “right” way to determine $A$ from equation $(1)$ would be to multiply both sides by $(x −1)$ ; this would give $$\frac{x − 7}{ ~~~~~~~~~(x + 2)} = A + \frac{B}{ x + 2} (x − 1) ~~~~~~~~\qquad(4)$$

Now if we substitute $x = 1$, what we get is exactly equation $(2)$, since the term on the right disappears.

Which seems absurd to me since multiplying both sides by $x-1$ should render that $x \neq 1$ otherwise it would mean $\frac{0}{0}$ is equal to $1$ because we could've written $A$ as such $\frac{A\cdot(x-1)}{x-1}$ and substituting by $x = 1$ would give us $\frac{A\cdot 0}{0}$. I looked over other places too where this method is used but those more or less follows the same way.

Note that I read few questions about it on this site eg,. this answer.

Can someone please help me make sense of it? Any help is genuinely appreciated.

Answer 1

I've at last found what I was looking for and contributed the answer.

$A(x+2)+B(x-1)$ is a polynomial in $x$, as is $x-7$. I wanted values of $A$ and $B$ that make these two polynomials equal for all real numbers except $1$ and $-2$. But polynomials are continuous and so two polynomials that agree at infinitely many real numbers are necessarily identically equal and will therefore agree at all real numbers. In particular, $A(x+2)+B(x-1)$ will hold at $x=1$ and $x=-2$ if it holds at all other integers, so we can substitute $x=1$ and $x=-2$ as shortcuts to finding the correct values of $A$ and $B$.