# Skepticism concerning Heaviside’s “Cover-up Method” for partial fraction decomposition

algebra-precalculusfractionspartial fractionspolynomials

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.

$$A(x+2)+B(x-1)$$ 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 $$B$$.