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Write $\dfrac{4x+1}{x^2 - x - 2}$ using partial fractions.

$$ \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$$

$$4x+1 = A(x-2)+B(x+1)$$

$$x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3$$

$$x = -1 \Rightarrow 4(-1) +1 = A(-3)+ B(0) \Rightarrow A = 1$$

Thus,

$$\frac{4x+1}{x^2-x-2} = \frac{1}{x+1} + \frac{3}{x-2}\textrm{.}$$

The substitution of $x$ ($x = 2, -1$) is a common method to find out the coefficient of the partial fractions. However, the equation on the third line is obtained by multiplying $(x+1)(x-2)$, which is assumed to be nonzero. Here we have a contradiction. Furthermore, the original function is not defined at $x=-1,2$.

How can we substitute these value for $x$? So is this method valid and rigorous? How to modify it so that it is rigorous?

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  • $\begingroup$ Where do you have the contradiction ? This is the rigorous method. Subsition method is just a "method". You can say a sort of shortcut $\endgroup$ – abkds Jun 5 '15 at 16:02
  • $\begingroup$ It looks valid to me. in terms of getting the $A,B$ coefficients. So what is the contradiction? you are meerly stating that $$\frac{4x+1}{x^2-x-2} = \frac{A(x-2)+B(x+1)}{x^2-x-2}$$ you are just equating the numerators. yes multiplying both sides by $0$ is not wise..but what you did seems correct to me. $\endgroup$ – Chinny84 Jun 5 '15 at 16:02
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    $\begingroup$ ah I see .. how can you equate two undefined functions. you dont need to use those points to be honest. you just need to have two points to solve the two unknowns. It is just easier to use the poles. $\endgroup$ – Chinny84 Jun 5 '15 at 16:06
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    $\begingroup$ For fun, do it for say $\frac{x^2}{x^2-x-2}$ by the substitution method, refusing to notice the top has degree $\ge$ the degree of the bottom. We will get values for $A$ and $B$ that don't work. The reason that it works wwhne used the right way is that there is a background theorem that says there is a unique pair $(A, B)$ for which the decomposition holds. $\endgroup$ – André Nicolas Jun 5 '15 at 16:09
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    $\begingroup$ Possible duplicate of How does partial fraction decomposition avoid division by zero? $\endgroup$ – Hans Lundmark Sep 10 '18 at 7:23
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This is a classical source of confusion. But the the problem is avoided completely if one takes the limit of both sides as $x \to 2$, instead of substituting $x=2$.

Clearly the result is the same, but with this formulation there is no evaluation of functions at points where they are undefined.

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$A(x-2)+B(x+1) = 4x+1$ gives us a system of two equations:

$$\begin{split} A+B &= 4 \\ -2A + B &= 1 \end{split}$$

But really the first equation as really $(A+B)x = 4x$, so the substitution $x=2$ simply gives us the system:

$$\begin{split} 2A+2B &= 8 \\ -2A + B &= 1 \\ \end{split}$$

Which conveniently sums to $3B = 9$, $B=3$. It's just a choice of substitution that makes it easy to solve the system. Any such substitution is valid, just that $x=2$ and $x=-1$ happen to be particularly good choices.

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