# Why can't partial fractions expansion be “normally” done is this case?

I've learned partial fractions but I couldn't really understand one thing. When we have a case when one of the factors has multiplicity $> 1$, we got to make a kind of "stairs". e.g.

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

But I thought $why$ couldn't it be written like this:

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

After some pencil and paper working, I realized it doesn't seem to work (and I don't know why!) I've been trying this for weeks, but I've got nowhere. So, if you could help me visualize a reason, I'd be grateful.

Thank you.

• Essentially, you are trying to write a "proper" fraction (denominator of greater degree than the numerator) as a sum of "proper" fractions. But a proper fraction with a quadratic denominator may have numerator of degree up to $1$, and there are situations where this occurs, so you must include that possibility. – Arturo Magidin Feb 14 '12 at 16:37
• I changed the TeX displays from "inline" to "displayed" form (since each had a line all to itself). What was done was that a fraction was inside of TeX and a sum of fractions was inside of TeX, but the "$=$" between them was not! That is not proper use of TeX, and in some cases it results in mismatches of size or alignment. – Michael Hardy Feb 14 '12 at 19:29

The most simple way to think about it is as follows: When you have a sum of rational functions, where denominators are polynomials, you always use the "commmon factor rule". You put in the denominator all uncommon factors and all common factors with the greatest exponent. So if you had

$$\frac{1}{{x - 2}} + \frac{1}{{{{\left( {x - 2} \right)}^2}}} + \frac{1}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^3}}}$$

Upong merging the functions you'd get

$$\frac{{A\left( x \right)}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 1} \right)}^3}}}$$

Where $A$ is polynomial. But now you see you "discarded" the lower exponents. This is why you have to consider the "full" decomposition to get a satisfactory result. However, it is also useful to consider a decomposition as follows:

$$\frac{{A\left( x \right)}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 1} \right)}^3}}} = \frac{{Ax + B}}{{{{\left( {x - 2} \right)}^2}}} + \frac{{C{x^2} + Dx + E}}{{{{\left( {x + 1} \right)}^3}}}$$

i.e. you have to consider the polynomial of lower degree that you'd get when merging all the lower factors you have. For example in the first expression you'd get:

$$\frac{1}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^3}}} = \frac{{{x^2} + 2x + 2}}{{{{\left( {x + 1} \right)}^3}}}$$

Hope this helped.

You can do it almost like your second case, but since $(x+2)^2$ is quadratic (degree 2), you'd need $Bx+C$ in the numerator: $$\frac{A}{(x-1)} + \frac{Bx+C}{(x+2)^2}$$

Generally, it's easier to do it the first way than the second way.

• This is the kind of thing I was looking for. – Ian Mateus Feb 18 '12 at 0:08

Hint: every proper fraction of form $f(x)/x^n$ can be written as $(c_0 + c_1 x +\:\cdots\: c_{n-1} x^{n-1})/x^{n}$. Such proper fractions form an $n$ dimensional vector space over $\mathbb R$ with basis $x^{-1},\:x^{-2},\:\cdots\:x^{-n}\:$. Therefore if $n > 1$ then the single fraction $1/x^n$ does not span the whole space. The same holds true if one applies a shift automorphism $x\mapsto x+a$.

Assume you are given a rational function of the form $$R(x)={p(x)\over (x-a)^m (x-b)^n}\ .$$ Using long division you may write $$R(x)={q(x) \over(x-a)^m (x-b)^n} + p_1(x)\ ,$$ where now ${\rm deg}(q)< m+n$, and $p_1$ is another polynomial. This $q$ will have $m+n$ coefficients of arbitrary values; therefore any "Ansatz" with less than $m+n$ constants $A_k$, $B_k$ to be determined is doomed to be unsuccessful. But we are lucky: It is an algebraic theorem that the identity $${q(x) \over(x-a)^m (x-b)^n} \equiv {A_m\over (x-a)^m}+\ldots +{A_1\over x-a} +{B_m\over (x-b)^m}+\ldots+{B_1\over x-b}$$ can be satisfied by a suitable choice of the $m+n$ indeterminate coefficients $A_k$, $B_k$.