Partial fraction decomposition

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In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.[1]

The importance of the partial fraction decomposition lies in the fact that it provides algorithms for various computations with rational functions, including the explicit computation of antiderivatives,[2] Taylor series expansions, inverse Z-transforms, and inverse Laplace transforms. The concept was discovered independently in 1702 by both Johann Bernoulli and Gottfried Leibniz.[3]

In symbols, the partial fraction decomposition of a rational fraction of the form where f and g are polynomials, is its expression as

where p(x) is a polynomial, and, for each j, the denominator gj (x) is a power of an irreducible polynomial (that is not factorable into polynomials of positive degrees), and the numerator fj (x) is a polynomial of a smaller degree than the degree of this irreducible polynomial.

When explicit computation is involved, a coarser decomposition is often preferred, which consists of replacing "irreducible polynomial" by "square-free polynomial" in the description of the outcome. This allows replacing polynomial factorization by the much easier to compute square-free factorization. This is sufficient for most applications, and avoids introducing irrational coefficients when the coefficients of the input polynomials are integers or rational numbers.

Basic principlesEdit



be a rational fraction, where F and G are univariate polynomials in the indeterminate x. The existence of the partial fraction can be proved by applying inductively the following reduction steps.

Polynomial partEdit

There exist two polynomials E and F1 such that




where   denotes the degree of the polynomial P.

This results immediately from the Euclidean division of F by G, which asserts the existence of E and F1 such that   and  

This allows supposing in the next steps that  

Factors of the denominatorEdit

If   and


where G1 and G2 are coprime polynomials, then there exist polynomials   and   such that




This can be proved as follows. Bézout's identity asserts the existence of polynomials C and D such that


(by hypothesis, 1 is a greatest common divisor of G1 and G2).

Let   with   be the Euclidean division of DF by   Setting   one gets


It remains to show that   By reducing to the same denominator the last sum of fractions, one gets   and thus


Powers in the denominatorEdit

Using the preceding decomposition inductively one gets fractions of the form   with   where G is an irreducible polynomial. If k > 1, one can decompose further, by using that an irreducible polynomial is a square-free polynomial, that is,   is a greatest common divisor of the polynomial and its derivative. If   is the derivative of G, Bézout's identity provides polynomials C and D such that   and thus   The Euclidean division of  `by   gives polynomials   and   such that   and   Setting   one gets



Iterating this process with   in place of   leads eventually to the following theorem.


Theorem — Let f and g be nonzero polynomials over a field K. Write g as a product of powers of distinct irreducible polynomials :


There are (unique) polynomials b and aij with deg aij < deg pi such that


If deg f < deg g, then b = 0.

The uniqueness can be proved as follows. Let d = max(1 + deg f, deg g). All together, b and the aij have d coefficients. The shape of the decomposition defines a linear map from coefficient vectors to polynomials f of degree less than d. The existence proof means that this map is surjective. As the two vector spaces have the same dimension, the map is also injective, which means uniqueness of the decomposition. By the way, this proof induces an algorithm for computing the decomposition through linear algebra.

If K is field of complex numbers, the fundamental theorem of algebra implies that all pi have degree one, and all numerators   are constants. When K is the field of real numbers, some of the pi may be quadratic, so, in the partial fraction decomposition, quotients of linear polynomials by powers of quadratic polynomials may also occur.

In the preceding theorem, one may replace "distinct irreducible polynomials" by "pairwise coprime polynomials that are coprime with their derivative". For example, the pi may be the factors of the square-free factorization of g. When K is the field of rational numbers, as it is typically the case in computer algebra, this allows to replace factorization by greatest common divisor computation for computing a partial fraction decomposition.

Application to symbolic integrationEdit

For the purpose of symbolic integration, the preceding result may be refined into

Theorem — Let f and g be nonzero polynomials over a field K. Write g as a product of powers of pairwise coprime polynomials which have no multiple root in an algebraically closed field:


There are (unique) polynomials b and cij with deg cij < deg pi such that


where   denotes the derivative of  

This reduces the computation of the antiderivative of a rational function to the integration of the last sum, which is called the logarithmic part, because its antiderivative is a linear combination of logarithms. In fact, we have


There are various methods to compute above decomposition. The one that is the simplest to describe is probably the so-called Hermite's method. As the degree of cij is bounded by the degree of pi, and the degree of b is the difference of the degrees of f and g (if this difference is non negative; otherwise, b=0), one may write these unknowns polynomials as polynomials with unknown coefficients. Reducing the two members of above formula to the same denominator and writing that the coefficients of each power of x are the same in the two numerators, one gets a system of linear equations which can be solved to obtain the desired values for the unknowns coefficients.


Given two polynomials   and  , where the αi are distinct constants and deg P < n, partial fractions are generally obtained by supposing that


and solving for the ci constants, by substitution, by equating the coefficients of terms involving the powers of x, or otherwise. (This is a variant of the method of undetermined coefficients.)

A more direct computation, which is strongly related with Lagrange interpolation consists of writing


where   is the derivative of the polynomial  .

This approach does not account for several other cases, but can be modified accordingly:

  • If   then it is necessary to perform the Euclidean division of P by Q, using polynomial long division, giving P(x) = E(x) Q(x) + R(x) with deg R < n. Dividing by Q(x) this gives
and then seek partial fractions for the remainder fraction (which by definition satisfies deg R < deg Q).
  • If Q(x) contains factors which are irreducible over the given field, then the numerator N(x) of each partial fraction with such a factor F(x) in the denominator must be sought as a polynomial with deg N < deg F, rather than as a constant. For example, take the following decomposition over R:
  • Suppose Q(x) = (xα)rS(x) and S(α) ≠ 0. Then Q(x) has a zero α of multiplicity r, and in the partial fraction decomposition, r of the partial fractions will involve the powers of (xα). For illustration, take S(x) = 1 to get the following decomposition:


In an example application of this procedure, (3x + 5)/(1 – 2x)2 can be decomposed in the form


Clearing denominators shows that 3x + 5 = A + B(1 – 2x). Expanding and equating the coefficients of powers of x gives

5 = A + B and 3x = –2Bx

Solving this system of linear equations for A and B yields A = 13/2 and B = –3/2. Hence,


Residue methodEdit

Over the complex numbers, suppose f(x) is a rational proper fraction, and can be decomposed into




then according to the uniqueness of Laurent series, aij is the coefficient of the term (x − xi)−1 in the Laurent expansion of gij(x) about the point xi, i.e., its residue


This is given directly by the formula


or in the special case when xi is a simple root,




Over the realsEdit

Partial fractions are used in real-variable integral calculus to find real-valued antiderivatives of rational functions. Partial fraction decomposition of real rational functions is also used to find their Inverse Laplace transforms. For applications of partial fraction decomposition over the reals, see

General resultEdit

Let f(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials functions p(x) and q(x)≠ 0, such that


By dividing both the numerator and the denominator by the leading coefficient of q(x), we may assume without loss of generality that q(x) is monic. By the fundamental theorem of algebra, we can write


where a1,..., am, b1,..., bn, c1,..., cn are real numbers with bi2 − 4ci < 0, and j1,..., jm, k1,..., kn are positive integers. The terms (xai) are the linear factors of q(x) which correspond to real roots of q(x), and the terms (xi2 + bix + ci) are the irreducible quadratic factors of q(x) which correspond to pairs of complex conjugate roots of q(x).

Then the partial fraction decomposition of f(x) is the following:


Here, P(x) is a (possibly zero) polynomial, and the Air, Bir, and Cir are real constants. There are a number of ways the constants can be found.

The most straightforward method is to multiply through by the common denominator q(x). We then obtain an equation of polynomials whose left-hand side is simply p(x) and whose right-hand side has coefficients which are linear expressions of the constants Air, Bir, and Cir. Since two polynomials are equal if and only if their corresponding coefficients are equal, we can equate the coefficients of like terms. In this way, a system of linear equations is obtained which always has a unique solution. This solution can be found using any of the standard methods of linear algebra. It can also be found with limits (see Example 5).


Example 1Edit


Here, the denominator splits into two distinct linear factors:


so we have the partial fraction decomposition


Multiplying through by the denominator on the left-hand side gives us the polynomial identity


Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that


Example 2Edit


After long-division, we have


The factor x2 − 4x + 8 is irreducible over the reals, as its discriminant (−4)2 − 4×8 = − 16 is negative. Thus the partial fraction decomposition over the reals has the shape


Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity


Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that −8 = −4A + C = −8 + C, so C = 0. Altogether,


The fraction can be completely decomposed using complex numbers. According to the fundamental theorem of algebra every complex polynomial of degree n has n (complex) roots (some of which can be repeated). The second fraction can be decomposed to:


Multiplying through by the denominator gives:


Equating the coefficients of x and the constant (with respect to x) coefficients of both sides of this equation, one gets a system of two linear equations in D and E, whose solution is


Thus we have a complete decomposition:


One may also compute directly A, D and E with the residue method (see also example 4 below).

Example 3Edit

This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system.


After long-division and factoring the denominator, we have


The partial fraction decomposition takes the form


Multiplying through by the denominator on the left-hand side we have the polynomial identity


Now we use different values of x to compute the coefficients:


Solving this we have:


Using these values we can write:


We compare the coefficients of x6 and x5 on both side and we have:




which gives us B = 0. Thus the partial fraction decomposition is given by:


Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at   in the above polynomial identity. (To this end, recall that the derivative at x = a of (xa)mp(x) vanishes if m > 1 and is just p(a) for m = 1.) For instance the first derivative at x = 1 gives


that is 8 = 4B + 8 so B = 0.

Example 4 (residue method)Edit


Thus, f(z) can be decomposed into rational functions whose denominators are z+1, z−1, z+i, z−i. Since each term is of power one, −1, 1, −i and i are simple poles.

Hence, the residues associated with each pole, given by




respectively, and


Example 5 (limit method)Edit

Limits can be used to find a partial fraction decomposition.[4] Consider the following example:


First, factor the denominator which determines the decomposition:


Multiplying everything by  , and taking the limit when  , we get


On the other hand,


and thus:


Multiplying by x and taking the limit when  , we have




This implies A + B = 0 and so  .

For x = 0, we get   and thus  .

Putting everything together, we get the decomposition


Example 6 (integral)Edit

Suppose we have the indefinite integral:


Before performing decomposition, it is obvious we must perform polynomial long division and factor the denominator. Doing this would result in:


Upon this, we may now perform partial fraction decomposition.




Upon substituting our values, in this case, where x=1 to solve for B and x=-2 to solve for A, we will result in:


Plugging all of this back into our integral allows us to find the answer:


The role of the Taylor polynomialEdit

The partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let


be real or complex polynomials assume that




Also define


Then we have


if, and only if, each polynomial   is the Taylor polynomial of   of order   at the point  :


Taylor's theorem (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients.

Sketch of the proofEdit

The above partial fraction decomposition implies, for each 1 ≤ i ≤ r, a polynomial expansion


so   is the Taylor polynomial of  , because of the unicity of the polynomial expansion of order  , and by assumption  .

Conversely, if the   are the Taylor polynomials, the above expansions at each   hold, therefore we also have


which implies that the polynomial   is divisible by  

For   is also divisible by  , so


is divisible by  . Since


we then have


and we find the partial fraction decomposition dividing by  .

Fractions of integersEdit

The idea of partial fractions can be generalized to other integral domains, say the ring of integers where prime numbers take the role of irreducible denominators. For example:



  1. ^ Larson, Ron (2016). Algebra & Trigonometry. Cengage Learning. ISBN 9781337271172.
  2. ^ Horowitz, Ellis. "Algorithms for partial fraction decomposition and rational function integration." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971.
  3. ^ Grosholz, Emily (2000). The Growth of Mathematical Knowledge. Kluwer Academic Publilshers. p. 179. ISBN 978-90-481-5391-6.
  4. ^ Bluman, George W. (1984). Problem Book for First Year Calculus. New York: Springer-Verlag. pp. 250–251.


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