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What is the reasoning or intuition that leads to the assumption that $$r(x) =\frac{x^2 + 2}{ (x+2)(x-1)^2}$$ can be expressed as $$r(x) = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$$

(For the sake of context, this problem arose when trying to solve an integral by the method of partial fractions.)

Moreover, how can the correct form of the partial fractions decomposition be found for other rational functions?

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    $\begingroup$ @ParasKhosla That doesn't answer the question at all! How do you know that there exist $A,B,C$ so that the numerator is the same as the numerator in the original? $\endgroup$ – David C. Ullrich Mar 2 at 17:15
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    $\begingroup$ @ParasKhosla Yes, but if you would add $\frac{B}{(x-1)^2} + \frac{C}{x+2}$ you would also get the same denominator as in the former. $\endgroup$ – K. Claesson Mar 2 at 17:16
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    $\begingroup$ @ParasKhosla ??? The replies to the first version of your comment apply equally to the second version - you haven't addressed the objections... $\endgroup$ – David C. Ullrich Mar 2 at 17:23
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    $\begingroup$ And I know my previous comment doesn't answer the question of why we can decompose. This comment should explain: $$$$To answer the "why decomposition is possible", work backwards. Create two fractions with variables, add them together, and see how many different numerators are possible. For example: $$\frac{a}{x+A} + \frac{b}{x+B} = \frac{(a+b)x+(aB+bA)}{(x+A)(x+B)}$$ There are many combinations of $a$ and $b$ that could form a coefficient, we just need to find two such numbers so that $aB + bA$ equal a specific, nonchangeable number, because $A$ and $B$ are non-changeable. $\endgroup$ – Christopher Marley Mar 2 at 17:37
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    $\begingroup$ See this article $\endgroup$ – steven gregory Mar 2 at 20:59
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Start with the rational function $\frac{N(x)}{D(x)},$ where $N(x)$ and $D(x)$ are polynomials over whichever field we happen to be working in (such as the real numbers or complex numbers) and the degree of $N(x)$ is less than the degree of $D(x).$

Consequences of Bezout's Identity

Suppose we can factor $D(x) = P_1(x)P_2(x),$ where $P_1(x)$ and $P_2(x)$ have no common factor (and therefore no common root). Then by Bezout's theorem for polynomials (aka Bezout's identity for polynomials), there are polynomials $F_1(x)$ and $F_2(x)$ such that $F_1(x)P_1(x) + F_2(x)P_2(x) = 1.$

(I believe the application of Bezout's identity here is why calculus books give the technique of partial fraction decomposition without proof. Bezout's identity comes from abstract algebra, which normally is not taught until you've had at least a couple of years of calculus; you wouldn't normally see it before university, and even then you'd probably only see it if you majored in math. Personally, I think it's a shame the curriculum is sequenced this way--I spent most of my time in first-year calculus griping about why we needed to memorize all that ugly ____, and only a few years later, when I finally got to the upper level courses, realized it was actually beautiful and made perfect sense--but that's enough ranting for one answer.)

By polynomial division we also have \begin{align} N(x)F_1(x) &= Q_2(x)P_2(x) + R_2(x), \\ N(x)F_2(x) &= Q_1(x)P_1(x) + R_1(x) \end{align} where the degree of $R_i(x)$ is less than the degree of $P_i(x).$ Therefore \begin{align} N(x) &= N(x)(F_1(x)P_1(x) + F_2(x)P_2(x)) \\ &= N(x)F_1(x)P_1(x) + N(x)F_2(x)P_2(x) \\ &= Q_2(x)P_1(x)P_2(x) + P_1(x)R_2(x) + Q_1(x)P_1(x)P_2(x) + P_2(x)R_1(x) \\ &= (Q_1(x) + Q_2(x))P_1(x)P_2(x) + P_1(x)R_2(x) + P_2(x)R_1(x). \end{align}

Since $\deg(R_1(x)) < \deg(P_1(x))$ and $\deg(R_2(x)) < \deg(P_2(x)),$ it follows that $\deg(P_2(x)R_1(x)) < \deg(P_1(x)P_2(x))$ and $\deg(P_1(x)R_2(x)) < \deg(P_1(x)P_2(x)).$ Since we must also have $\deg(N(x)) < \deg(P_1(x)P_2(x)),$ we must have $\deg((Q_1(x) + Q_2(x))P_1(x)P_2(x)) < \deg(P_1(x)P_2(x)),$ which is possible only if $Q_1(x) + Q_2(x) = 0.$ Therefore we can more simply write $$ N(x) = P_1(x)R_2(x) + P_2(x)R_1(x).$$

Therefore \begin{align} \frac{N(x)}{D(x)} &= \frac{P_1(x)R_2(x) + P_2(x)R_1(x)}{P_1(x)P_2(x)} \\ &= \frac{R_2(x)}{P_2(x)} + \frac{R_1(x)}{P_1(x)}. \tag1 \end{align}

Taking out a first-degree factor

To apply this to partial fraction decomposition, if $x - a$ divides $D(x)$ we find the greatest power of $x - a$ that divides $D(x).$ Suppose that this is the $n$th power. Set $P_1(x) = (x - a)^n$ and $P_2(x) = \frac{D(x)}{(x - a)^n}.$ Then $P_1(x)$ and $P_2(x)$ have no common factor, and the result $(1)$ above says that $$ \frac{N(x)}{D(x)} = \frac{R_2(x)}{P_2(x)} + \frac{R_1(x)}{(x - a)^n} $$ where $\deg(R_2(x)) < \deg(P_2(x))$ and $\deg(R_1(x)) < n = \deg((x - a)^n).$

Taking out an irreducible quadratic factor

If we are doing real analysis and don't allow polynomials to have complex coefficients, then $D(x)$ might have a factor of the form $x^2 + bx + c$ that cannot be factored into first-degree polynomials (that is, it is irreducible). In that case, if the highest power of $x^2 + bx + c$ that divides $D(x)$ is the $m$th power, then we can write $P_1(x) = (x^2 + bx + c)^m$ and $P_2(x) = \frac{D(x)}{(x^2 + bx + c)^m}.$ It follows that $P_1(x)$ and $P_2(x)$ have no common factor, and therefore (according to $(1)$ again) $$ \frac{N(x)}{D(x)} = \frac{R_2(x)}{P_2(x)} + \frac{R_1(x)}{(x^2 + bx + c)^m} $$ where $\deg(R_2(x)) < \deg(P_2(x))$ and $\deg(R_1(x)) < 2m = \deg((x^2 + bx + c)^m).$

Completing the decomposition

Provided that we are able to find all the first- and second-degree factors of the polynomial $D(x),$ we can repeatedly take out either first-degree factors or irreducible quadratic factors from $D(x)$ and then from the polynomial $P_2(x)$ that we get after taking out the previous factor, until we end up with a $P_2$ that is itself a first-degree polynomial or an irreducible quadratic. We end up with something that looks like this: $$ \frac{N(x)}{D(x)} = \frac{S_1(x)}{(x - a_1)^{n_1}} + \cdots + \frac{S_h(x)}{(x - a_h)^{n_h}} + \frac{T_1(x)}{(x^2 + b_1x + c_1)^{m_1}} + \cdots + \frac{T_1(x)}{(x^2 + b_kx + c_k)^{m_k}}. $$

The final step of the proof is to show that if the degree of $U(x)$ is less than the degree of $(V(x))^p,$ then $$ \frac{U(x)}{(V(x))^p} = \frac{U_1(x)}{V(x)} + \frac{U_2(x)}{(V(x))^2} + \cdots + \frac{U_p(x)}{(V(x))^p} $$ where the degree of each $U_i(x)$ is less than the degree of $V(x).$ We can get this result by dividing $U(x)$ by $V(x)$ (the remainder is $U_p(x)$), then dividing the quotient of that division by $V(x)$ again (the remainder is $U_{p-1}(x)$), and so on repeatedly until we get a quotient whose degree is less than that of $V(x),$ which will happen after at most $p-1$ divisions. That's why, when you have a factor that occurs more than once in the factorization of $D(x)$, you get a term for each power of that factor up to the highest power that divides $D(x).$

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  • $\begingroup$ This is a very nice explanation. $\endgroup$ – John Wayland Bales Mar 3 at 16:38
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It's not an assumption. What you're trying to do is rewrite the fraction on the left as a sum of fractions on the right. Since the antiderivatives of the fractions on the right are easier to compute, the internal is easier to take.

For every factor, one gets a fraction. If the factor is linear, the numerator is a constant. If the factor is an irreducible quadratic, then the numerator is a linear term. If the factor has multiplicity $m$, then we get m fractions, one for each power.

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    $\begingroup$ This doesn't answer the question at all. How do you know that the decomposition is possible? $\endgroup$ – David C. Ullrich Mar 2 at 17:26
  • $\begingroup$ I know the decomposition is possible because 1)the fundamental theorem of algebra says that every polynomial can be decomposed into linear and irreducible quadratic factors and 2) addition of fractions. $\endgroup$ – Joel Pereira Mar 3 at 16:01
  • $\begingroup$ That's nonsense. Consider the specific example in the OP. We're given a factorization of the denominator. "addition of fractions" shows that the sum on the RHS exists. So what? The question is how do you know that there exist $A,B,C$ such that the RHS is actually equall to the LHS... $\endgroup$ – David C. Ullrich Mar 3 at 16:24
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To answer your second question: You can find the proper form for the partial-fractions decomposition for an arbitrary rational function in more or less any calculus book.

I don't think your main question really has a good answer - I don't believe that there is a proof that ""partial fractions works" that's going to make sense to a typical calculus student.

I know of two proofs. One uses "abstract algebra", in particular "Bezout's identity" for polynomials. I haven't read it carefully, but since it uses Bezout's identity I imagine the other answer, the one that uses this identity, is correct.

As a matter of possible interest to other readers, one can prove that partial fractions works if one knows a little "complex analysis". Sketch of how thhat goes for the example you ask about:

Define a rational function $q(z)$ in the plane by $$q(z)=\frac{z^2 + 2}{ (z+2)(z-1)^2} -\left(\frac{A}{z-1} + \frac{B}{(z-1)^2} + \frac{C}{z+2}\right).$$It's not hard to see that there exist $A,B,C$ so that the "principal part" of $q$ at every pole vanishes. (Note for example that the original fraction $r(z)$ has a pole of order $2$ at $z=1$, hence the principal part there has the form $A/(z-1)+B/(z-1)^2$.) So then $q$ is a rational function with no poles, which says $q$ is a polynomial. It's clear that $q$ tends to $0$ at infinity, hence $q=0$.

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