Partial fraction decomposition

There exist two polynomials E and F₁ such that

$${\displaystyle {\frac {F}{G}}=E+{\frac {F_{1}}{G}},}$$

and

$${\displaystyle \deg F_{1}<\deg G,}$$

where ${\displaystyle \deg P}$ 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 ${\displaystyle F=EG+F_{1}}$ and ${\displaystyle \deg F_{1}<\deg G.}$

This allows supposing in the next steps that ${\displaystyle \deg F<\deg G.}$

If ${\displaystyle \deg F<\deg G,}$ and

$${\displaystyle G=G_{1}G_{2},}$$

where G₁ and G₂ are coprime polynomials, then there exist polynomials ${\displaystyle F_{1}}$ and ${\displaystyle F_{2}}$ such that

$${\displaystyle {\frac {F}{G}}={\frac {F_{1}}{G_{1}}}+{\frac {F_{2}}{G_{2}}},}$$

and

$${\displaystyle \deg F_{1}<\deg G_{1}\quad {\text{and}}\quad \deg F_{2}<\deg G_{2}.}$$

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

$${\displaystyle CG_{1}+DG_{2}=1}$$

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

Let ${\displaystyle DF=G_{1}Q+F_{1}}$ with ${\displaystyle \deg F_{1}<\deg G_{1}}$ be the Euclidean division of DF by ${\displaystyle G_{1}.}$ Setting ${\displaystyle F_{2}=CF+QG_{2},}$ one gets

$${\displaystyle {\begin{aligned}{\frac {F}{G}}&={\frac {F(CG_{1}+DG_{2})}{G_{1}G_{2}}}={\frac {DF}{G_{1}}}+{\frac {CF}{G_{2}}}\\&={\frac {F_{1}+G_{1}Q}{G_{1}}}+{\frac {F_{2}-G_{2}Q}{G_{2}}}\\&={\frac {F_{1}}{G_{1}}}+{\frac {F_{2}}{G_{2}}}.\end{aligned}}}$$

It remains to show that ${\displaystyle \deg F_{2}<\deg G_{2}.}$ By reducing the last sum of fractions to a common denominator, one gets ${\displaystyle F=F_{2}G_{1}+F_{1}G_{2},}$ and thus

$${\displaystyle {\begin{aligned}\deg F_{2}&=\deg(F-F_{1}G_{2})-\deg G_{1}\leq \max(\deg F,\deg(F_{1}G_{2}))-\deg G_{1}\\&<\max(\deg G,\deg(G_{1}G_{2}))-\deg G_{1}=\deg G_{2}\end{aligned}}}$$

Using the preceding decomposition inductively one gets fractions of the form ${\displaystyle {\frac {F}{G^{k}}},}$ with ${\displaystyle \deg F<\deg G^{k}=k\deg G,}$ 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, ${\displaystyle 1}$ is a greatest common divisor of the polynomial and its derivative. If ${\displaystyle G'}$ is the derivative of G, Bézout's identity provides polynomials C and D such that ${\displaystyle CG+DG'=1}$ and thus ${\displaystyle F=FCG+FDG'.}$ Euclidean division of ${\displaystyle FDG'}$ by ${\displaystyle G}$ gives polynomials ${\displaystyle H_{k}}$ and ${\displaystyle Q}$ such that ${\displaystyle FDG'=QG+H_{k}}$ and ${\displaystyle \deg H_{k}<\deg G.}$ Setting ${\displaystyle F_{k-1}=FC+Q,}$ one gets

$${\displaystyle {\frac {F}{G^{k}}}={\frac {H_{k}}{G^{k}}}+{\frac {F_{k-1}}{G^{k-1}}},}$$

with ${\displaystyle \deg H_{k}<\deg G.}$

Iterating this process with ${\displaystyle {\frac {F_{k-1}}{G^{k-1}}}}$ in place of ${\displaystyle {\frac {F}{G^{k}}}}$ 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 :

$${\displaystyle g=\prod _{i=1}^{k}p_{i}^{n_{i}}.}$$

There are (unique) polynomials b and a_ij with deg a_ij < deg p_i such that

$${\displaystyle {\frac {f}{g}}=b+\sum _{i=1}^{k}\sum _{j=1}^{n_{i}}{\frac {a_{ij}}{p_{i}^{j}}}.}$$

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 a_ij 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 p_i have degree one, and all numerators ${\displaystyle a_{ij}}$ are constants. When K is the field of real numbers, some of the p_i 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 p_i 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.

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

$${\displaystyle {\frac {3x+5}{(1-2x)^{2}}}={\frac {A}{(1-2x)^{2}}}+{\frac {B}{(1-2x)}}.}$$

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

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

$${\displaystyle {\frac {3x+5}{(1-2x)^{2}}}={\frac {13/2}{(1-2x)^{2}}}+{\frac {-3/2}{(1-2x)}}.}$$

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

$${\displaystyle f(x)=\sum _{i}\left({\frac {a_{i1}}{x-x_{i}}}+{\frac {a_{i2}}{(x-x_{i})^{2}}}+\cdots +{\frac {a_{ik_{i}}}{(x-x_{i})^{k_{i}}}}\right).}$$

Let

$${\displaystyle g_{ij}(x)=(x-x_{i})^{j-1}f(x),}$$

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

$${\displaystyle a_{ij}=\operatorname {Res} (g_{ij},x_{i}).}$$

This is given directly by the formula

$${\displaystyle a_{ij}={\frac {1}{(k_{i}-j)!}}\lim _{x\to x_{i}}{\frac {d^{k_{i}-j}}{dx^{k_{i}-j}}}\left((x-x_{i})^{k_{i}}f(x)\right),}$$

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

$${\displaystyle a_{i1}={\frac {P(x_{i})}{Q'(x_{i})}},}$$

when

$${\displaystyle f(x)={\frac {P(x)}{Q(x)}}.}$$

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

$${\displaystyle f(x)={\frac {p(x)}{q(x)}}}$$

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

$${\displaystyle q(x)=(x-a_{1})^{j_{1}}\cdots (x-a_{m})^{j_{m}}(x^{2}+b_{1}x+c_{1})^{k_{1}}\cdots (x^{2}+b_{n}x+c_{n})^{k_{n}}}$$

where a₁,..., a_m, b₁,..., b_n, c₁,..., c_n are real numbers with b_i² − 4c_i < 0, and j₁,..., j_m, k₁,..., k_n are positive integers. The terms (x − a_i) are the linear factors of q(x) which correspond to real roots of q(x), and the terms (x_i² + b_ix + c_i) 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:

$${\displaystyle f(x)={\frac {p(x)}{q(x)}}=P(x)+\sum _{i=1}^{m}\sum _{r=1}^{j_{i}}{\frac {A_{ir}}{(x-a_{i})^{r}}}+\sum _{i=1}^{n}\sum _{r=1}^{k_{i}}{\frac {B_{ir}x+C_{ir}}{(x^{2}+b_{i}x+c_{i})^{r}}}}$$

Here, P(x) is a (possibly zero) polynomial, and the A_ir, B_ir, and C_ir 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 A_ir, B_ir, and C_ir. 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).

$${\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}}$$

Here, the denominator splits into two distinct linear factors:

$${\displaystyle q(x)=x^{2}+2x-3=(x+3)(x-1)}$$

so we have the partial fraction decomposition

$${\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}={\frac {A}{x+3}}+{\frac {B}{x-1}}}$$

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

$${\displaystyle 1=A(x-1)+B(x+3)}$$

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

$${\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}={\frac {1}{4}}\left({\frac {-1}{x+3}}+{\frac {1}{x-1}}\right)}$$

$${\displaystyle f(x)={\frac {x^{3}+16}{x^{3}-4x^{2}+8x}}}$$

After long division, we have

$${\displaystyle f(x)=1+{\frac {4x^{2}-8x+16}{x^{3}-4x^{2}+8x}}=1+{\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}}$$

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

$${\displaystyle {\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}={\frac {A}{x}}+{\frac {Bx+C}{x^{2}-4x+8}}}$$

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

$${\displaystyle 4x^{2}-8x+16=A\left(x^{2}-4x+8\right)+\left(Bx+C\right)x}$$

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,

$${\displaystyle f(x)=1+2\left({\frac {1}{x}}+{\frac {x}{x^{2}-4x+8}}\right)}$$

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:

$${\displaystyle {\frac {x}{x^{2}-4x+8}}={\frac {D}{x-(2+2i)}}+{\frac {E}{x-(2-2i)}}}$$

Multiplying through by the denominator gives:

$${\displaystyle x=D(x-(2-2i))+E(x-(2+2i))}$$

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

$${\displaystyle D={\frac {1+i}{2i}}={\frac {1-i}{2}},\qquad E={\frac {1-i}{-2i}}={\frac {1+i}{2}}.}$$

Thus we have a complete decomposition:

$${\displaystyle f(x)={\frac {x^{3}+16}{x^{3}-4x^{2}+8x}}=1+{\frac {2}{x}}+{\frac {1-i}{x-(2+2i)}}+{\frac {1+i}{x-(2-2i)}}}$$

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

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

$${\displaystyle f(x)={\frac {x^{9}-2x^{6}+2x^{5}-7x^{4}+13x^{3}-11x^{2}+12x-4}{x^{7}-3x^{6}+5x^{5}-7x^{4}+7x^{3}-5x^{2}+3x-1}}}$$

After long division and factoring the denominator, we have

$${\displaystyle f(x)=x^{2}+3x+4+{\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}}$$

The partial fraction decomposition takes the form

$${\displaystyle {\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}={\frac {A}{x-1}}+{\frac {B}{(x-1)^{2}}}+{\frac {C}{(x-1)^{3}}}+{\frac {Dx+E}{x^{2}+1}}+{\frac {Fx+G}{(x^{2}+1)^{2}}}.}$$

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

$${\displaystyle {\begin{aligned}&2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\[4pt]={}&A\left(x-1\right)^{2}\left(x^{2}+1\right)^{2}+B\left(x-1\right)\left(x^{2}+1\right)^{2}+C\left(x^{2}+1\right)^{2}+\left(Dx+E\right)\left(x-1\right)^{3}\left(x^{2}+1\right)+\left(Fx+G\right)\left(x-1\right)^{3}\end{aligned}}}$$

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

$${\displaystyle {\begin{cases}4=4C&x=1\\2+2i=(Fi+G)(2+2i)&x=i\\0=A-B+C-E-G&x=0\end{cases}}}$$

Solving this we have:

$${\displaystyle {\begin{cases}C=1\\F=0,G=1\\E=A-B\end{cases}}}$$

Using these values we can write:

$${\displaystyle {\begin{aligned}&2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\[4pt]={}&A\left(x-1\right)^{2}\left(x^{2}+1\right)^{2}+B\left(x-1\right)\left(x^{2}+1\right)^{2}+\left(x^{2}+1\right)^{2}+\left(Dx+\left(A-B\right)\right)\left(x-1\right)^{3}\left(x^{2}+1\right)+\left(x-1\right)^{3}\\[4pt]={}&\left(A+D\right)x^{6}+\left(-A-3D\right)x^{5}+\left(2B+4D+1\right)x^{4}+\left(-2B-4D+1\right)x^{3}+\left(-A+2B+3D-1\right)x^{2}+\left(A-2B-D+3\right)x\end{aligned}}}$$

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

$${\displaystyle {\begin{cases}A+D=2\\-A-3D=-4\end{cases}}\quad \Rightarrow \quad A=D=1.}$$

Therefore:

$${\displaystyle 2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x=2x^{6}-4x^{5}+(2B+5)x^{4}+(-2B-3)x^{3}+(2B+1)x^{2}+(-2B+3)x}$$

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

$${\displaystyle f(x)=x^{2}+3x+4+{\frac {1}{(x-1)}}+{\frac {1}{(x-1)^{3}}}+{\frac {x+1}{x^{2}+1}}+{\frac {1}{(x^{2}+1)^{2}}}.}$$

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

$${\displaystyle 2\cdot 6-4\cdot 5+5\cdot 4-3\cdot 3+2+3=A\cdot (0+0)+B\cdot (4+0)+8+D\cdot 0}$$

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

$${\displaystyle f(z)={\frac {z^{2}-5}{(z^{2}-1)(z^{2}+1)}}={\frac {z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}}}$$

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

$${\displaystyle {\frac {P(z_{i})}{Q'(z_{i})}}={\frac {z_{i}^{2}-5}{4z_{i}^{3}}},}$$

are

$${\displaystyle 1,-1,{\tfrac {3i}{2}},-{\tfrac {3i}{2}},}$$

respectively, and

$${\displaystyle f(z)={\frac {1}{z+1}}-{\frac {1}{z-1}}+{\frac {3i}{2}}{\frac {1}{z+i}}-{\frac {3i}{2}}{\frac {1}{z-i}}.}$$

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

$${\displaystyle {\frac {1}{x^{3}-1}}}$$

First, factor the denominator which determines the decomposition:

$${\displaystyle {\frac {1}{x^{3}-1}}={\frac {1}{(x-1)(x^{2}+x+1)}}={\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}.}$$

Multiplying everything by ${\displaystyle x-1}$, and taking the limit when ${\displaystyle x\to 1}$, we get

$${\displaystyle \lim _{x\to 1}\left((x-1)\left({\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}\right)\right)=\lim _{x\to 1}A+\lim _{x\to 1}{\frac {(x-1)(Bx+C)}{x^{2}+x+1}}=A.}$$

On the other hand,

$${\displaystyle \lim _{x\to 1}{\frac {(x-1)}{(x-1)(x^{2}+x+1)}}=\lim _{x\to 1}{\frac {1}{x^{2}+x+1}}={\frac {1}{3}},}$$

and thus:

$${\displaystyle A={\frac {1}{3}}.}$$

Multiplying by x and taking the limit when ${\displaystyle x\to \infty }$, we have

$${\displaystyle \lim _{x\to \infty }x\left({\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}\right)=\lim _{x\to \infty }{\frac {Ax}{x-1}}+\lim _{x\to \infty }{\frac {Bx^{2}+Cx}{x^{2}+x+1}}=A+B,}$$

and

$${\displaystyle \lim _{x\to \infty }{\frac {x}{(x-1)(x^{2}+x+1)}}=0.}$$

This implies A + B = 0 and so ${\displaystyle B=-{\frac {1}{3}}}$.

For x = 0, we get ${\displaystyle -1=-A+C,}$ and thus ${\displaystyle C=-{\tfrac {2}{3}}}$.

Putting everything together, we get the decomposition

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

Suppose we have the indefinite integral:

$${\displaystyle \int {\frac {x^{4}+x^{3}+x^{2}+1}{x^{2}+x-2}}\,dx}$$

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

$${\displaystyle \int \left(x^{2}+3+{\frac {-3x+7}{(x+2)(x-1)}}\right)dx}$$

Upon this, we may now perform partial fraction decomposition.

$${\displaystyle \int \left(x^{2}+3+{\frac {-3x+7}{(x+2)(x-1)}}\right)dx=\int \left(x^{2}+3+{\frac {A}{(x+2)}}+{\frac {B}{(x-1)}}\right)dx}$$

so:

$${\displaystyle A(x-1)+B(x+2)=-3x+7}$$

.

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:

$${\displaystyle A={\frac {-13}{3}}\ ,B={\frac {4}{3}}}$$

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

$${\displaystyle \int \left(x^{2}+3+{\frac {-13/3}{(x+2)}}+{\frac {4/3}{(x-1)}}\right)\,dx={\frac {x^{3}}{3}}\ +3x-{\frac {13}{3}}\ln(|x+2|)+{\frac {4}{3}}\ln(|x-1|)+C}$$

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

$${\displaystyle {\frac {P}{Q_{i}}}=A_{i}+O((x-\lambda _{i})^{\nu _{i}}),\qquad {\text{for }}x\to \lambda _{i},}$$

so ${\displaystyle A_{i}}$ is the Taylor polynomial of ${\displaystyle {\tfrac {P}{Q_{i}}}}$, because of the unicity of the polynomial expansion of order ${\displaystyle \nu _{i}-1}$, and by assumption ${\displaystyle \deg A_{i}<\nu _{i}}$.

Conversely, if the ${\displaystyle A_{i}}$ are the Taylor polynomials, the above expansions at each ${\displaystyle \lambda _{i}}$ hold, therefore we also have

$${\displaystyle P-Q_{i}A_{i}=O((x-\lambda _{i})^{\nu _{i}}),\qquad {\text{for }}x\to \lambda _{i},}$$

which implies that the polynomial ${\displaystyle P-Q_{i}A_{i}}$ is divisible by ${\displaystyle (x-\lambda _{i})^{\nu _{i}}.}$

For ${\displaystyle j\neq i,Q_{j}A_{j}}$ is also divisible by ${\displaystyle (x-\lambda _{i})^{\nu _{i}}}$, so

$${\displaystyle P-\sum _{j=1}^{r}Q_{j}A_{j}}$$

is divisible by ${\displaystyle Q}$. Since

$${\displaystyle \deg \left(P-\sum _{j=1}^{r}Q_{j}A_{j}\right)<\deg(Q)}$$

we then have

$${\displaystyle P-\sum _{j=1}^{r}Q_{j}A_{j}=0,}$$

and we find the partial fraction decomposition dividing by ${\displaystyle Q}$.