Techniques of Partial Fraction Decomposition

Partial fraction decomposition is very often used for many applications, such as integration of rational fractions, inverse Laplace transforms, and Taylor series expansions. In this article, we skip discussions on the basics of partial fraction decomposition in view of many other resources online, however we do highlight techniques to efficiently decompose a fraction into partial fractions.

Before we start, we emphasize not to forget the following steps before decomposing the fraction:
  1. Ensure the numerator and denominator are both polynomials.
  2. Ensure the numerator of the fraction has an order less than the denominator.
  3. Ensure the denominator is completely reduced to irreducible factors.
We direct the reader to a document on how to divide polynomials. A table of the decomposition of a fraction of rational functions is provided below.
Irreducible factor in denominator Term in decomposition
$ax+b$ $\frac{C_1}{ax+b}$
$(ax+b)^n$ $\frac{C_1}{ax+b}+\frac{C_2}{(ax+b)^2}+...+\frac{C_n}{(ax+b)^n}$
$ax^2+bx+c$ $\frac{C_1x+D_1}{ax^2+bx+c}$
$(ax^2+bx+c)^n$ $\frac{C_1x+D_1}{ax^2+bx+c}+\frac{C_2x+D_2}{(ax^2+bx+c)^2}+...+\frac{C_nx+D_n}{(ax^2+bx+c)^n}$
The table above does not include factors of higher order since all higher polynomials must be converted to a multiple of irreducible factors located in the table. This procedure is described below.

Factorization of Polynomials to Linear/Quadratic Factors

Any polynomial can be represented by the product of irreducible, real linear and quadratic factors. This is the result of the fundamental theorem of algebra. The factorization of the rational denominator is required before partial fraction decomposition can commence. There are a few methods to do so, involving the rational root theorem to find linear factors with rational roots. However, this is somewhat restrictive. A more general method for factorizing any polynomial is to first solve (often graphically) for the roots to find the linear factors. After dividing the polynomial by all determined linear factors, the remaining function should be a product of solely irreducible quadratics. The coefficients for the quadratics are found by solving a system of nonlinear equations. This method is best illustrated with a comprehensive example.

Example: Reduce the polynomial into irreducible linear and quadratic factors with real coefficients. $$f(x)=2x^7-9x^6+31x^4-21x^3+24x^2-9x+54$$ At first, this looks extremely intimidating, but initial factorization is simple. Graph the polynomial and it is immediately obvious that two roots exist: $x=-2$ with a multiplicity of 1 and $x=3$ with a multiplicity of 2. Therefore, the linear factors, which are comprised of the real roots, are $x+2$ and $(x-3)^2$ (multiplicity is incorporated as a power). Long division of $f(x)$ with the product of the linear factors produces $g(x)=2x^4-x^3+2x^2+3$. With an order of 4, this function is therefore a product of two quadratics. We denote these two quadratics as: \begin{align*} ax^2+bx+c\\ ux^2+vx+w\end{align*} Multiplying the two polynomials and setting it equal to $g(x)$ results in: $$(au)x^4+(av+bu)x^3+(aw+bv+cu)x^2+(bw+cv)x+cw=2x^4-x^3+2x^2+3$$ Comparison of coefficients yields: \begin{align*} au&=2 \\ av+bu&=-1 \\ aw+bv+cu&=2 \\ bw+cv&=0 \\ cw&=3\end{align*} There are five equations and six unknowns yielding one degree of freedom. We arbitratily set $a=1$ making $u=2$ through the first equation. With some algebra, it can be shown that there are two sets of polynomials that satisfy the equations above. One solution produces the two polynomials $x^2+x+1$ and $2x^2-3x+3$. So, the final answer is: $$f(x)=(x-3)^2(x+2)(x^2+x+1)(2x^2-3x+3)$$ It is significantly easier to do this factorization through a mathematical software. MATLAB provides a function, factor, that can be used for this purpose.
syms x
F = factor(2*x^7-9*x^6+31*x^4-21*x^3+24*x^2-9*x+54)
Output: F = [ x + 2, x^2 + x + 1, 2*x^2 - 3*x + 3, x - 3, x - 3]

Method of Undetermined Coefficients

This method is often used as a fail-safe method for decomposing fractions. The unknown coefficients of the decomposed fractions are solved by forcing the partial fraction decomposition into a form synonymous with the original function. This function will have the same denominator as the original function, with unknown variables all in the numerator. The unknown coefficients are subsequently solved through comparing the two functions' numerators. Consider the function: $$f(x)=\frac{4x^2+6x+6}{x^3-7x-6}$$ By factorizing the denominator through the method described in the previous section, we end up with the following: $$f(x)=\frac{4x^2+6x+6}{(x-3)(x+1)(x+2)}$$ The partial fraction decomposition of the function, let's call $g(x)$, is: $$g(x)=\frac{A}{x-3}+\frac{B}{x+1}+\frac{C}{x+2}$$ Note that we are solving for the unknown parameters $A,B,C$ to make $f(x)$ and $g(x)$ equal. First, combine the partial fractions into one fraction. This is done by first making the denominators of the individual partial fractions the same as the original function's denominator. $$g(x)=\frac{A}{x-3} \times \frac{(x+1)(x+2)}{(x+1)(x+2)}+\frac{B}{x+1} \times \frac{(x-3)(x+2)}{(x-3)(x+2)}+\frac{C}{x+2} \times \frac{(x-3)(x+1)}{(x-3)(x+1)}$$ We've merely multiplied all the fractions by one, since the multiplying factors all have the same numerator and denominator. $$g(x)=\frac{A(x+1)(x+2)}{(x-3)(x+1)(x+2)}+\frac{B(x-3)(x+2)}{(x-3)(x+1)(x+2)}+\frac{C(x-3)(x+1)}{(x-3)(x+1)(x+2)}$$ $$=\frac{A(x+1)(x+2)+B(x-3)(x+2)+C(x-3)(x+1)}{(x-3)(x+1)(x+2)}$$ At this point, comparing the numerator with the original function's numerator is not useful. We need to force it into the same rational form as the original function's numerator. Thus, we need to simplify the numerator into the form $c_1x^2+c_2x+c_3$. Fortunately this is relatively straightforward but requires some algebra: $$g(x)=\frac{A(x^2+3x+2)+B(x^2-x-6)+C(x^2-2x-3)}{(x-3)(x+1)(x+2)}$$ $$=\frac{Ax^2+3Ax+2A+Bx^2-Bx-6B+Cx^2-2Cx-3C}{(x-3)(x+1)(x+2)}$$ $$=\frac{(A+B+C)x^2+(3A-B-2C)x+(2A-6B-3C)}{(x-3)(x+1)(x+2)}$$ The denominator of both functions $f(x)$ and $g(x)$ are the same, meaning that if the numerators are the same, then the two functions are equal! $$(A+B+C)x^2+(3A-B-2C)x+(2A-6B-3C)=4x^2+6x+6$$ Now that both sides are in the same form, that is, a polynomial of order 2, all that is needed for the numerators to be the same is for the coefficient on the left side to match up with the coefficients on the right hand side. Let us represent this equality slightly differently: \begin{alignat*}{2}(A+B+C)x^2&{}+{}&(3A-B-2C)x&{}+{}&(2A-6B-3C)\\=4x^2&{}+{}&6x&{}+{}&6\end{alignat*} It is now evident that for the coefficients to match up: \begin{alignat*}{2}&A&{}+{}&&B&{}+{}&&C&{}={}&4 \label{eq:1} \tag{1}\\ 3&A&{}-{}&&B&{}-{}&2&C&{}={}&6 \label{eq:2} \tag{2}\\ 2&A&{}-{}&6&B&{}-{}&3&C&{}={}&6 \label{eq:3} \tag{3}\end{alignat*} The first step in solving this system is to subtract equation \ref{eq:2} by equation \ref{eq:3} then by \ref{eq:1} to get $B=-1$. Then multiply equation \ref{eq:1} by 2 and subtract \ref{eq:3} by \ref{eq:1}, yielding $C=2$. Plug in $B$ and $C$ into equation \ref{eq:1} to get $A=3$. Therefore: $$f(x)=\frac{3}{x-3}-\frac{1}{x+1}+\frac{2}{x+2}$$ This method of solving for the unknowns in the partial fraction decomposition works all the time for any sort of partial fraction decomposition. However it is clear that it requires a lot of algebra and thus there is a greater chance calculation errors may arise.

Substitution Method

Consider the previous function again and its partial fraction expansion: $$\frac{4x^2+6x+6}{(x-3)(x+1)(x+2)}=\frac{A}{x-3}+\frac{B}{x+1}+\frac{C}{x+2}$$ It is important to realize that the equality above must hold for all values of $x$ in the original function's domain. Thus, by simply substituting three arbitrarily different values of $x$ into the equation above, we can derive a system of three equations of $A,B,C$ and solve accordingly. This results in a system of equations that is determined quicker than the previous Method of Undetermined Coefficients. However, a system of equations is still needing to be solved, and for large system of equations, this is very tedious. This method is rarely used to do a complete partial fraction decomposition, however it is used in conjunction with other methods and so the concept is important to understand.

Multiplication Method

This method solves for parameters without necessarily needing to solve systems. Consider the previous function again and its partial fraction expansion: $$\frac{4x^2+6x+6}{(x-3)(x+1)(x+2)}=\frac{A}{x-3}+\frac{B}{x+1}+\frac{C}{x+2}$$ To solve for $A$, multiply both sides of the equation above by the denominator $x-3$ associated with $A$. Take note that when $x=3, x-3$ is equal to zero. $$\frac{4x^2+6x+6}{(x+1)(x+2)}=A+\frac{B(x-3)}{x+1}+\frac{C(x-3)}{x+2}$$ Now, set $x=3$ as previously found. Parameters $B$ and $C$ are now cancelled with the substitution as they are both multiplied by zero. While using this method, make sure there are no divisions by zero! $$\frac{4(3)^2+6(3)+6}{(3+1)(3+2)}=A+\frac{B(3-3)}{3+1}+\frac{C(3-3)}{3+2}$$ $$\frac{60}{20}=A+0+0$$ $$\therefore A=3$$ Similarly, $B$ and $C$ are found the same way.
Now, let's consider another more complicated partial fraction. Consider the function and its partial fraction expansion: $$\frac{6x^2+9x+3}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2} \label{eq:4} \tag{4}$$ $A$ can be found through this method. The steps for showing $A=2$ is omitted. However, note we cannot solve for $B$ using this method. If we proceed to solve for $B$ using this method, we first need to multiply both sides by $x+2$. $$\frac{6x^2+9x+3}{(x-1)(x+2)}=\frac{A(x+2)}{x-1}+B+\frac{C}{x+2}$$ If we substitute $x=-2$, we get a division by zero on the left and right-hand side. However, if we instead solve for $C$, multiplying both sides by $(x+2)^2$ yields: $$\frac{6x^2+9x+3}{x-1}=\frac{A(x+2)^2}{x-1}+B(x+2)+C$$ Substituting $x=-2$ now works: $$\frac{6(-2)^2+9(-2)+3}{-2-1}=\frac{A(-2+2)^2}{-2-1}+B(-2+2)+C$$ $$\therefore C=\frac{9}{-3}=-3$$ To solve for $B$, one may use the Method of Undetermined Coefficients, however it is easier in this case just to use the Substitution method and substitute one number into equation \ref{eq:4}. Any number that does not result in a division by zero can be substituted, so we substitute $x=-1$, which is the root of the numerator in the original function. Note this value is in the domain of the original function. There is no division by zero as a result and the left-hand side of equation \ref{eq:4} goes to zero: $$0=\frac{A}{-1-1}+\frac{B}{-1+2}+\frac{C}{(-1+2)^2}$$ Values of $A$ and $C$ have already been solved using the Multiplication method. $$0=\frac{2}{-1-1}+\frac{B}{-1+2}-\frac{3}{(-1+2)^2}$$ $$\therefore B=1+3=4$$ $$\frac{6x^2+9x+3}{(x-1)(x+2)^2}=\frac{2}{x-1}+\frac{4}{x+2}-\frac{3}{(x+2)^2}$$ Therefore, for the partial fractions that result from repeated linear factors, the Multiplication Method should be used on the associated partial fraction with the denominator raised to the highest power. This method cannot be solely used to solve for the other factors raised to a lower power.
For fractions where the denominator has quadratics or quadratic terms raised to a power, all unknown parameters can be calculated with this method since a division by zero is mitigated. However, the use of complex numbers is required. For example, observe the function and its decomposition below: $$\frac{-x^2 + 3 x - 11}{(x - 2) (x^2 - 2 x + 3)}=\frac{A}{x-2}+\frac{Bx+C}{x^2-2x+3}$$ $A$ is solved easily to yield $A=-3$. To solve for $B$ and $C$, multiply both sides by $x^2-2x+3$. $$\frac{-x^2 + 3 x - 11}{x - 2}=\frac{A(x^2-2x+3)}{x-2}+Bx+C$$ The roots to $x^2-2x+3$ are $1+i\sqrt{2}$ and $1-i\sqrt{2}$. For the first root: $$\frac{-(1+i\sqrt{2})^2+3(1+i\sqrt{2})-11}{1+i\sqrt{2}-2}=\frac{A(0)}{x-2}+B(1+i\sqrt{2})+C$$ $$\frac{-7+i\sqrt{2}}{-1+i\sqrt{2}}=B(1+i\sqrt{2})+C$$ $$3 + 2 i \sqrt{2}=B(1+i\sqrt{2})+C \label{eq:5} \tag{5}$$ At this point, we know $B$ and $C$ must be real numbers, so that automatically forces $B$ to equal 2. Then it follows that $C=1$. A review of complex number operations is located here.
Complex number algebra may be a bit tedious so in this case, the Method of Undetermined Coefficients to solve for $B$ and $C$ may have produced a quicker answer. However, if a fraction has many linear factors in its denominator and only one or two quadratic terms, this method for partial fraction decomposition may be faster compared to the Method of Undetermined Coefficients since this method does not require the tedious step of combining all partial fractions into one fraction.
With the same analysis, it can be shown that for partial fractions that result from repeated quadratic factors, the Multiplication Method should be used on the associated partial fraction with the denominator raised to the highest power. This method cannot be solely used to solve for the other factors raised to a lower power. This is the same conclusion as the case for repeated linear factors.

Heaviside Cover-up Method

The Heaviside Cover-up Method uses the exact method as the Multiplication Method, but removes the intermediate algebra steps to drastically improve the time in solving for the unknown coefficients. It is especially useful for a partial fraction decomposition of many terms. However, it is important that the Heaviside Method has the same requirements presented in the multiplication method; this method works all the time for a decomposition with only linear factors but special exceptions apply for decompositions with other types of factors. Using the same example as before, $$\frac{4x^2+6x+6}{(x-3)(x+1)(x+2)}=\frac{A}{x-3}+\frac{B}{x+1}+\frac{C}{x+2}$$ The decomposition only involved linear factors, and so the Heaviside Cover-up method will work to solve for all parameters. Now, to solve for $A$, first notice that the denominator associated with $A$ in the partial fraction equals zero when $x=3$. Next, we simply 'cover up' or remove the factor $x-3$ associated with $A$ in the original function. $$\frac{4x^2+6x+6}{\hphantom{(x-3)}(x+1)(x+2)}$$ Lastly, we plug in the $x$ value of 3 that we previously solved for and plug it into the remaining function. $$A=\frac{4(3)^2+6(3)+6}{\hphantom{(x-3)}(3+1)(3+2)}=3$$ Solving for $B$ this time, $x=-1$ to solve for the root of the denominator $x+1$ associated with $B$. Next, 'cover up' $x+1$ from the original function. $$\frac{4x^2+6x+6}{(x-3)\hphantom{(x+1)}(x+2)}$$ Substitute $x=-1$ into the remaining function. $$B=\frac{4(-1)^2+6(-1)+6}{(-1-3)\hphantom{(x+1)}(-1+2)}=-1$$ Solve $C$ through the same way. $$C=\frac{4(-2)^2+6(-2)+6}{(-2-3)(-2+1)\hphantom{(x+2)}}=2$$ Now, for the case of non-linear factors in the denominator: $$\frac{4x^2-x-2}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$$ Heaviside method can be used for the fractions with linear denominators as was the case in the multiplication method. $$C=\frac{4(1)^2-1-2}{(1)^2\hphantom{(x-1)}}=1$$ The Heaviside method can be used for the partial fraction with the highest order denominator that resulted from the $x^2$ term: $$B=\frac{4(0)^2-0-2}{\hphantom{x^2}(0-1)}=2$$ The substitution method / method of undetermined coefficients is used to solve for the last unknown parameter. Selecting an arbitrary value to substitute $x=2$, $$\frac{4(2)^2-2-2}{2^2(2-1)}=\frac{A}{2}+\frac{B}{2^2}+\frac{C}{2-1}$$ $$3=\frac{A}{2}+\frac{2}{2^2}+\frac{1}{1}$$ which quickly shows that $A=3$. A table is conveniently provided below to list out restrictions when using the Heaviside method.
Irreducible factor in denominator Heaviside method restriction
$ax+b$ None.
$(ax+b)^n$ May only be used on associated partial fraction with denominator raised to highest power.
$ax^2+bx+c$ None, complex numbers required. May be quicker to use Method of Undetermined Coefficients.
$(ax^2+bx+c)^n$ May only be used on associated partial fraction with denominator raised to highest power. Complex numbers required. May be quicker to use Method of Undetermined Coefficients.

Partial Fraction Decomposition through Mathematical Software

Given the opportunity, this method is the easiest. MATLAB provides a function, partfrac, that can be used. The function accepts the function, and optionally the independent variable. For the same function given first in the Methods of Undetermined Coefficient method $\frac{4x^2+6x+6}{x^3-7x-6}$, the partial fraction is calculated through the code:
syms x
Output: f = 2/(x + 2) - 1/(x + 1) + 3/(x - 3)
Use the optional second parameter to specify the independent variable. In the case of $\frac{a}{(x+a)(x+2)}$, $x$ is the independent variable and $a$ is a constant:
syms x a
Output: f = a/((a - 2)*(x + 2)) - a/((a + x)*(a - 2))


  1. Decompose $f(x)=\frac{4x^2+6x+8}{x^2(x+4)}$ into partial fractions using all the methods shown.
  2. Decompose $f(x)=\frac{a}{(x-b)(x-c)}$ into partial fractions, where $a,b,c$ are real numbers and $b \neq c$.

Is BaSO4 soluble or insoluble?

Answer: BaSO4 would typically be considered insoluble due to the presence of a sulfate ion. This usually precludes an ion from disassociating easily in an aqueous solution; this being water. 

The idea of BaSO4 being considered insoluble is further borne out in the Ksp values. While these cannot be directly related to molar solubility, they provide a strong hint that Barium sulfate is more stringent than other comparable molecules. For example Ba(OH)2's ksp is approximately seven orders of magnitude less which means that it is much more likely to ionize. You can tell that the molar solubility of barium hydroxide is orders of magnitude greater without performing the necessary calculations. As a reminder a compound is considered insoluble if less than 1% of the compound ionizes. 

Barium Products Ksp Values
ksp values of some common Barium products (source)

What is the molar solubility of BaSO4?

The calculation is based on the chemical equation for the disassociation of BaSO4 and the formula for ksp (equilibrium constants only include aqueous and gaseous species). 

BaSO4 (s) => Ba(2+)(aq) + SO4(2-)(aq)

ksp = [Ba(2+)][SO4(2-)] 

Since when we ionize we create the two ions in equal molar quantity we can state the following

[Ba(2+)] = [SO4(2-)] = x

$1.1*10^-10 = x^2$

x = $1.05*10^-5$ M

This concentration indicates that less than 1% of the solution is composed of ionized species. 

How does BaSO4 theoretically ionize? 

As the equation (BaSO4 (s) => Ba(2+)(aq) + SO4(2-)(aq)) indicates, barium sulfate would disassociate and produce the barium and sulfate ion. In the real world; however, the low molar solubility indicates that very little barium sulfate actually ionizes. 

Is H2S Polar or Nonpolar?

Answer: H2S is a polar molecule due to the presence of lone pair electrons at the top of the molecule causing a region of partial negative charge due to electron-electron repulsion. 

H2S has a very similar structure as H2O (see the lewis dot structure for H2O and the polar/nonpolar explanation at the linked addresses). However, due to the larger size of the sulfur atom compared to oxygen, the bond angle (i.e. the smaller angle between the two hydrogen atoms) is only 92˚ compared to 107.5˚ for H2O. Sulfur contains many more electrons which ultimately due to electron-electron repulsion require a lot more space. Nevertheless the decreased electronegativity of sulfur when compared to hydrogen (2.58 vs. 2.20, respectively) means that the molecule is much less polar overall when compared to H2O. This means that it has a much lower melting and boiling point at -82˚C and -60˚C, respectively. Like SO2 the presence of sulfur means that this molecule has a pungent odor in gaseous form although it is colorless. 

H2S Ball and Stick Diagram
H2S Ball and Stick Diagram. Created with MolView.

How is H2S utilized in the real world?

Hydrogen sulfide appears in many different ways within the natural world. For starters it is an important constituent of the sulfur cycle. Bacteria oftentimes convert the sulfur from organic elements to inorganic molecules such as H2S. The main use of hydrogen sulfide is as a storage compound which can be converted to pure sulfur during reactions to form all kinds of sulfur-containing compounds. Hydrogen sulfide may have also caused a mass extinction due to its buildup within the atmosphere. Based on this fact it is not difficult to imagine the toxicity of hydrogen sulfide towards life forms such as human beings. It negatively impacts proper nervous system functioning primarily although it will affect other body systems as well. Nevertheless there are certain organisms adapted to live in high-H2S conditions due to those environments existing in deep underwater volcanic sea vents.  

The Lewis Dot Structure for Acetic Acid (CH3COOH)

Lewis Dot Structure for Acetic Acid
Created by MakeTheBrainHappy.

Above is the Lewis Dot Structure for Acetic Acid (CH3COOH). You could alternatively also draw the structure by including two dots for every bond. As you can see every single element has a filled valence shell with the two oxygen's each containing two lone pairs of electrons, the only instance of this phenomena within the Lewis Structure. In a sense this is a modified structure of methane (CH4) with the replacement of one hydrogen with the replacement of a carboxylic group (-COOH). As we will see the properties we observe within the Lewis structure have a significant impact on the properties of acetic acid.

The Greek Philosopher Theophrastus. Source

What is the history of acetic acid?

Due to the importance of different alcohols such as beer and wine in early civilizations, vinegar became one of the earliest chemical substances that was familiar to ancient peoples. Vinegar is formed by natural fermentation processes and contains approximately 5% acetic acid. One of the earliest mentions of acetic acid was by the Greek Philosopher Theophrastus who explained how to form different pigments, including those for white and green colors, with vinegar as an important constituent ingredient. 

Is Acetic Acid (CH3COOH) Polar or Nonpolar?

Acetic Acid (CH3COOH) is a polar molecule due to the presence of the functional group COOH, a carboxylic acid. It is also an acid in solution, releasing a small number of protons into solution which form H3O+. Due to its polar qualities it is found as a liquid at standard temperature and pressure in relation to its rather light molar mass. It has a melting point is between 16˚C to 17˚C while the boiling point is between 118˚C and 119˚C. These are again elevated due to the polarity of acetic acid. 

How is Acetic Acid (CH3COOH) utilized in the real world?

Acetic acid is useful due to its properties as a polar solvent and building block for other molecules, containing both a methyl (CH3) and COOH functional group. Nearly one third of produced acetic acid is utilized in order to produce "Elmer's glue" material. Inks, paints and coatings are also created via a reaction involving acetic acid. It is frequently utilized as a polar solvent in lab research settings and thereby has been involved in certain medical practices. As mentioned above acetic acid is also present in vinegar which has a variety of household uses; however, the acetic acid is diluted in water to a greater degree than in a research lab.