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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 ?

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marked as duplicate by David Mitra, J. M., Gerry Myerson, t.b., Zev Chonoles Jul 25 '12 at 6:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Sorry for the duplication, I am not sure how I missed that. –  user1086516 Jul 5 '12 at 22:39
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Once we are looking at the equation $x^2 + 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$ , we are no longer solving the initial problem -- rather, we are ascertaining that the polynomials on the left hand and right hand sides are equal for all values of $x$, and as such we can figure out what the values of $A$, $B$, and $C$ are.

Traditionally, to the best of my knowledge, this is done by simply expanding out the polynomial on the right hand side, then getting three linear equations in three unknowns between the coefficients of $x^2$, $x$, and the constants $(x^0)$. However, because polynomials are perfectly nice and continuous, etc., and because a degree $n$ polynomial can always be precisely specified by $n+1$ points with different $x$ values, then the values you find by constraining the above quadratic with 3 points will be the same as if you used the 'traditional' method, in every case. Is this sufficient?

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When you clear the denominator, you are left with an equality of polynomial functions, rather than simply rational functions. Even though the domain of the original rational function excludes $1,2,3,$ the polynomials are equal iff they are equal as functions on the entire domain.

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