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How does partial fraction decomposition avoid division by zero?
Say you have the rational function:
$\frac{x^2 + 1}{(x-1)(x-2)(x-3)}$
This means that the function is undefined when x is equal to 1, 2, or 3.
Then to decompose it, you can equate that function to: $\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$
If you clear the fraction then you will get: x^2 + 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
Then if you let x = 1,2,3 you can find A,B, and C but why are you even allowed to do that? From the beginning don't we define the domain of the function to be all real numbers besides 1,2, and 3? So why can we can go against the domain of the function to solve for the coefficients ?