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One such decomposition is: $$ \dfrac{A}{x+1} + \dfrac{B}{x-5} + \frac{C}{\left(x-5\right)^2}. $$

Why can't we decompose like below?

$$ \dfrac{A}{x+1} + \dfrac{B}{(x-5)^2} $$

TShiong
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Haaarish
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    Welcome to Mathematics SE. Please try to use MathJax syntax to typeset equations inside your questions. This makes them easier to read for everyone. – cs89 Apr 28 '23 at 20:21
  • the correct two piece version is $$ \frac{A}{x+1} + \frac{Bx +D}{(x-5)^2} $$ – Will Jagy Apr 28 '23 at 20:31
  • Ok. But my question was why does the above decomposition fails. ? – Haaarish Apr 28 '23 at 20:33
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    Try it! That is, $$\frac{3x^2+1}{(x+1)(x-5)^2}=\frac{A}{x+1} + \frac{B}{(x-5)^2}.$$ Get common denominator. You get a system of linear equations to solve for $A,B$ (three equations in two unknowns). There is no solution! So that decomposition fails. – GEdgar Apr 28 '23 at 20:57
  • A(x-5)^2 + B(x+1) = 3x^2 + 1 . We can plug in x = 5 and -1 and get A, B right ? Please correct me if i am wrong – Haaarish Apr 28 '23 at 21:22

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Suppose we rewrite the putative decomposition $$\frac{A}{x + 1} + \frac{B}{(x - 5)^2}$$ in standard form: $$\frac{A x^2 + (-10 A + B) x + (25 A + B)}{(x + 1) (x - 5)^2} .$$ Comparing coefficients of the numerator shows that any rational function $$\frac{a x^2 + b x + c}{(x + 1) (x - 5)^2}$$ that can be written in the form $\frac{A}{x + 1} + \frac{B}{(x - 5)^2}$ must satisfy $35 a + b - c = 0$, which not all of them do.

Heuristically, for a fixed denominator $q(x)$, a proper rational expression $\frac{p(x)}{q(x)}$ is specified by $n := \deg q$ parameters, namely, the coefficients $p_0, \ldots, p_{n - 1}$ of $p(x)$. So, we expect that generically a partial fraction decomposition will need to be specified by $n$ parameters. In our case, $\deg ((x + 1) (x - 5)^2) = 3$, so we expect to need $3$ parameters rather than $2$. One way to rectify this situation is to write $$\frac{ax^2 + b x + c}{(x + 1)(x - 5)^2} = \frac{A}{x + 1} + \frac{B}{x - 5} + \frac{C}{(x - 5)^2} ,$$ which turns out to determine $A, B, C$ uniquely.

Travis Willse
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