This post is based on a section called Integration by Partial Fractions in Morris Kline’s book on Calculus [1]. The first step in understanding partial fractions is learning about polynomials.

Definition of a Polynomial

From Wikipedia, the following statement defines a polynomial [2].

In mathematics, a polynomial is an expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

The word “non-negative” is important because this means a polynomial cannot have a term such as $x^{-1} = 1/x$. However, the term $x^0 \equiv 1$ could be in a polynomial.

From reference [1], “a polynomial is…

$a_0 x^n + a_1 x^{n-1} + … + a_{n-1} x + a_n$

wherein $a_0, a_1, …, a_n$ are constants.” The subscript of the left-most constant is $0$ whereas the superscript of the left-most variable is $n$. And $n$ is the highest integer in this expression. Moving from left to right, the superscripts decrease in integer steps, and the subscripts increase in integer steps. The right-most term does not show the variable $x$ because $x^0 \equiv 1$.

Ratio of Polynomials

Even though a ratio such as $\frac{1}{x}$ is not considered a polynomial, it is possible to write a ratio or quotient of two polynomials. For instance, the expression


is actually a ratio of two polynomials. The numerator is $1 = x^0$, which (as noted above) is a polynomial. And the denominator, which is $x = x^1$, is a polynomial as well.


Recall that [2]:

the degree of a polynomial is the largest degree of any term with nonzero coefficient.

The exponent or superscript on a variable $x$ such as $a_1 x^{n-1}$ is the degree of $x$ [2]. In this case the degree of $a_1 x^{n-1}$ is $n-1$.


Ratios of polynomials can be characterized by how the degree of the numerator’s polynomial compares to the degree of the denominator’s polynomial. The first example provided in reference [1] is the following ratio of polynomials:

$\frac{x^4 + 4x^3 + x^2 + 6}{x^2 + 2x + 1}$.

The degree of the numerator’s polynomial is $4$, and the degree of the denominator’s polynomial is $2$. This means that the degree of numerator’s polynomial is higher or greater than the degree of the denominator’s polynomial. Because the polynomial degree is higher in the numerator, it is useful–for the purpose of partial fraction decomposition–to divide the polynomials with polynomial division:

$(x^4 + 4x^3 + x^2 + 6) ÷ (x^2 + 2x + 1)$.

The polynomial with the higher degree is divided by the polynomial with the lower degree. Interestingly, nothing has really changed since the expression

$\frac{x^4 + 4x^3 + x^2 + 6}{x^2 + 2x + 1}$

is another way to write $(x^4 + 4x^3 + x^2 + 6) ÷ (x^2 + 2x + 1)$.

Polynomial Division

Dividing two polynomials “is entirely analogous to long division of ordinary numbers” [1].

Example of Polynomial Division


$(x^4 + 4x^3 + x^2 + 6) ÷ (x^2 + 2x + 1)$ is $x^2 + 2x – 4$ with a remainder of $6x + 10$.


Let $R$ signify a remainder. In numerical division, for example, $\frac{16}{5} = 3 R 1 = \frac{3*5}{5} + \frac{1}{5} = 3 + \frac{1}{5}$. Comparing the second and fourth expressions in the previous string of equalities, both expressions have a $3$, so the remainder of $1$ written as $ R 1$ can be handled by adding to the previous number (e.g. $3$) a fraction of $\frac{1}{5}$. Notice that $\frac{1}{5}$ has the numerical remainder (e.g. $1$) in the numerator and the divisor (e.g. $5$) in the denominator. To reiterate, $\frac{16}{5} = 3 + \frac{1}{5}$. Applying this process to polynomials,

$\frac{x^4 + 4x^3 + x^2 + 6}{x^2 + 2x + 1} = x^2 + 2x – 4 + \frac{6x + 10}{x^2 + 2x + 1} $.

The Degree of the Numerator’s Polynomial is Less than the Degree of the Denominator’s Polynomial

Consider a ratio of two polynomials. If the degree of the numerator’s polynomial is less than the degree of the denominator’s polynomial, the ratio of these two polynomials can be expressed differently. The following ratio falls into this category [1].

$ \frac{x^2 + 2x + 4}{x^3 – 2 x^2 – 5x + 6}$

So that the denominator can be factored, it is useful to find the roots of the polynomial in the denominator. A root of a polynomial of one variable such as $x$ is a value of $x$ that makes the entire polynomial equal to zero. From algebra, “if $a$ is a root of a polynomial…then $x-a$ is a factor of the polynomial. To each root there is a factor” [1].

From reference [1], the roots of $x^3 – 2 x^2 – 5x + 6$ are $1, -2,$ and $3$. Therefore,

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

Method of Partial Fractions

This method entails using constants $A$, $B$, and $C$–to be determined. In particular:

$\frac{x^2 + 2x + 4}{(x-1) (x+2) (x-3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3}$.

This method can be examined; on the right side, get a common denominator:

$\frac{x^2 + 2x + 4}{(x-1) (x+2) (x-3)} = \frac{A(x+2)(x-3)}{(x-1)(x+2)(x-3)} + \frac{B(x-1)(x-3)}{(x-1)(x+2)(x-3)} + \frac{C(x-1)(x+2)}{(x-1)(x+2)(x-3)}$.


$\frac{x^2 + 2x + 4}{(x-1) (x+2) (x-3)} = \frac{A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)}{(x-1)(x+2)(x-3)}$.

If the constants $A$, $B$, and $C$ can be found such that the above equality is met, then the method of partial fractions works. The constants $A$, $B$, and $C$ can be found by setting the numerator on the left side equal to the numerator on the right side–the equality between the two numerators holds for all values of $x$ including the roots of the denominator’s polynomial, $1, -2,$ and $3$:

$x^2 + 2x + 4 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) $.

After solving for $A$, $B$, and $C$ by substituting each of the three roots (i.e. $1, -2,$ and $3$), one obtains:

$\frac{x^2 + 2x + 4}{(x-1) (x+2) (x-3)} = -\frac{\frac{7}{6}}{x-1} + \frac{\frac{4}{15}}{x+2} + \frac{\frac{19}{10}}{x-3}$.


[1] Morris Kline. Calculus, An Intuitive and Physical Approach (Second Edition). Dover Publications, Inc. 1998.


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