# Partial Fractions - Linear Factors

**Partial fraction decomposition** is a technique used to write a rational function as the sum of simpler rational expressions. In certain cases, a rational function can be expressed as the sum of fractions whose denominators are linear binomials. For example,

\[\frac{2}{x^2-1} \implies \frac{1}{x-1} - \frac{1}{x+1}.\]

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## Linear Factor Partial Decomposition Forms

The most simple case of partial fraction decomposition is when

- there is a rational expression with a polynomial numerator and denominator,
- the degree of the numerator is less than the degree of the denominator, and
- the denominator can be factored into linear binomial factors.

Find the partial fraction decomposition form of the rational expression

\[\frac{2}{x^2+5x+6}.\]

Note that the numerator is a constant and that the denominator can be factored:

\[\frac{2}{x^2+5x+6}=\frac{2}{(x+2)(x+3)}.\]

This means that partial decomposition form will contain a sum of rational expressions, each of which contains a constant numerator and a denominator with one of these linear binomial factors:

\[\frac{2}{x^2+5x+6}=\frac{A}{x+2}+\frac{B}{x+3},\]

where \(A\) and \(B\) are constants. \(_\square\)

If the degree of the numerator is greater than or equal to the degree of the denominator, then polynomial division can be applied to write an equivalent expression that is the sum of a polynomial and a rational expression.

Find the partial fraction decomposition form of the rational expression

\[\frac{x^2+1}{x^2-1}.\]

Applying polynomial division gives the equivalent expression

\[\frac{x^2+1}{x^2-1}=1+\frac{2}{x^2-1}.\]

The denominator polynomial can be factored:

\[x^2-1=(x+1)(x-1).\]

Then the partial decomposition form of the rational expression is

\[\frac{x^2+1}{x^2-1}=1+\frac{A}{x+1}+\frac{B}{x-1},\]

where \(A\) and \(B\) are constants. \(_\square\)

The methods shown here to decompose rational expressions do not work for every problem. If a rational expression contains a denominator factor with multiplicity greater than 1, then one must use a different method: partial fractions--repeated factors.

Find the partial decomposition form of the rational expression

\[\frac{4x}{x^3+4x^2+5x+2}.\]

The roots of the denominator can be found with the rational root theorem. The denominator is factored as

\[x^3+4x^2+5x+2=(x+1)^2(x+2).\]

Note that the \((x+1)\) factor has multiplicity 2. This means that the repeated factors method must be used to decompose the rational expression. \(_\square\)

Although it was shown that a quadratic factor can be decomposed even if it cannot be factored, this might not always be the most efficient way to find the partial fraction decomposition. It is often better to use quadratic or higher degree factors.

Some decompositions can be performed with quadratic or higher degree factors.

Find the partial fraction decomposition form of the rational expression

\[\frac{1}{x^3-1}.\]

The denominator can be factored as a difference of cubes:

\[x^3-1=(x-1)(x^2+x+1).\]

The quadratic term cannot be factored any further. The quadratic formula could be used to find the complex roots of the quadratic. However, this would be an inelegant way to decompose the expression. There is a much simpler partial fraction decomposition using a quadratic factor. \(_\square\)

Although the method linked above is usually preferred for polynomials that do not factor, nevertheless, sometimes it is desirable to obtain a partial fraction decomposition with linear factors. To do so, one must find the roots of the polynomial, either through the quadratic formula, rational root theorem, or some other method.

Find the partial fraction decomposition form of the rational expression

\[\frac{5x-1}{x^2+x-1}.\]

The numerator has a lesser degree than the denominator, but the denominator cannot be factored. However, the roots of the polynomial can be found using the quadratic formula. The roots are

\[x=\frac{-1\pm\sqrt{5}}{2}.\]

This gives a factorization of the denominator:

\[\begin{align} x^2+x-1 &= \left(x-\frac{-1+\sqrt{5}}{2}\right)\left(x-\frac{-1-\sqrt{5}}{2}\right) \\ &= \left(\frac{1}{2}\right) \left(\vphantom{\frac{1}{2}} 2x-(1+\sqrt{5})\right) \left(\frac{1}{2}\right) \left(\vphantom{\frac{1}{2}} 2x-(-1-\sqrt{5})\right) \\ &= \frac{1}{4}\left(2x+1-\sqrt{5}\right)\left(2x+1+\sqrt{5}\right). \end{align}\]

Then, the partial fraction decomposition will have the form

\[\frac{5x-1}{x^2+x-1} = 4\left(\frac{A}{2x+1-\sqrt{5}}+\frac{B}{2x+1+\sqrt{5}}\right),\]

where \(A\) and \(B\) are constants. Note that the denominators are still linear, even though they contain irrational numbers. \(_\square\)

This kind of approach can be used even if the denominator polynomial has complex roots.

## Solving for the Coefficients

Finding the partial fraction decomposition form is only part of the goal of partial fraction decomposition. The ultimate goal is to calculate the values of the numerators so that the partial fraction decomposition is equivalent to the original expression.

Returning to the example introduced in the previous section:

Find the partial fraction decomposition of the rational expression

\[\frac{2}{x^2+5x+6}.\]

Recall from the previous section that the partial fraction decomposition form is

\[\frac{2}{x^2+5x+6}=\frac{A}{x+2}+\frac{B}{x+3}.\]

The goal now is to find the values of \(A\) and \(B\) so that these expressions are equivalent. Begin by combining the fractions on the right-hand side of the equation:

\[\begin{align} \frac{A}{x+2}+\frac{B}{x+3} &= \frac{A(x+3)+B(x+2)}{(x+2)(x+3)} \\ \\ &=\frac{(A+B)x+3A+2B}{x^2+5x+6}. \end{align}\]

Note that the denominator of this expression is the same as the denominator of the original expression. This means that the numerators must be equal:

\[2=(A+B)x+3A+2B.\]

There is no \(x\) term in the original numerator. Therefore,

\[A+B=0.\]

The other part of the numerator must be equal to 2:

\[3A+2B=2.\]

Solving this system of equations gives the values of \(A\) and \(B\) that will cause the partial fraction decomposition to be equivalent to the original expression: \(A=2\) and \(B=-2.\) Then the partial fraction decomposition is

\[\frac{2}{x^2+5x+6}=\frac{2}{x+2}-\frac{2}{x+3}.\ _\square\]

Using the above example,

\[\frac{1}{x^3-1} =\frac{1}{(x-1)(x^2+x+1)} \implies \frac{1}{(x-1)(x^2+x+1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}. \]

Multiplying both sides of the equation by \((x-1)(x^2+x+1),\) we obtain

\[ 1 = A(x^2 + x + 1) + (Bx + C)(x - 1) = A(x^2 + x + 1) + Bx^2 - Bx + Cx - C. \]

Equating coefficients, we obtain

\[\begin{align} A + B &= 0 \\ A - B + C &= 0 \\ A - C &= 1. \end{align}\]

Since \(A + B = 0 \implies B = -A,\) replacing \( B = -A \) in the second equation gives \( 2A + C = 0. \)

Solving the system

\[\begin{align} 2A + C &= 0 \\ A - C &= 1, \end{align}\]

we have

\[\begin{align} A = \frac{1}{3}, C = \frac{-2}{3} &\Rightarrow B = \frac{-1}{3} \\ &\Rightarrow \frac{1}{x^3 - 1} = \frac{1}{3} \left(\frac{1}{x - 1} - \frac{x + 2}{x^2 + x + 1}\right).\ _\square \end{align}\]

The previous examples gives an intuitive walk-through for how to solve for the coefficients of a partial fraction decomposition. There are more efficient methods available to solve for these coefficients, however.

Recall that in order for expressions to be equivalent, they must be equal for *any* value of a variable in those expressions. Thus, one can select a *particular* value of a variable in order to more efficiently compute the coefficients of a partial fraction decomposition.

Find the partial fraction decomposition of the rational expression

\[\frac{x^2+1}{x^2-1}.\]

Recall from the previous section that the partial fraction decomposition had the form

\[\frac{x^2+1}{x^2-1}=1+\frac{2}{x^2-1}=1+\frac{A}{x+1}+\frac{B}{x-1}.\]

Combining the fractions gives

\[1+\frac{2}{x^2-1}=1+\frac{A(x-1)+B(x+1)}{x^2-1}.\]

Since the denominators are the same, the numerators must be equal:

\[2=A(x-1)+B(x+1).\]

Regardless of the value of \(x\), the two sides of this equation should be equal. Some clever choices for the value \(x\) will give a quick solution for \(A\) and \(B.\)

Let \(x=1,\) which gives

\[\begin{align} 2 &= A(1-1)+B(1+1) \\ 2 &= 2B \\ B &= 1. \end{align}\]

Let \(x=-1,\) which gives

\[\begin{align} 2 &= A(-1-1)+B(-1+1) \\ 2 &= -2A \\ A &= -1. \end{align}\]

Then the partial fraction decomposition is

\[\frac{x^2+1}{x^2-1}=1-\frac{1}{x+1}+\frac{1}{x-1}.\ _\square\]

This method can be generalized:

Variable selection method for computing coefficients:Given a partial fraction decomposition form

\[\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots+\frac{A_n}{x-a_n},\]

combine these rational expressions and set the combined rational expression equal to the original expression. The denominators should be equal, so the numerators will be equal as well. Write the equation of equal numerators.

Then for each \(a_i,\) substituting \(x=a_i\) into this equation will give the value for \(A_i.\)

Note that this method "breaks the rules" somewhat, because these particular values for \(x\) will cause the denominator of a rational expression to equal \(0.\) Even so, this method is sufficient to compute the values of the coefficients of a partial fraction decomposition. The more mathematically rigorous basis for this method is the limit method.

## Limit Method

Main Article: Partial Fractions - Limit Method

The following method is less efficient than many of the other methods to find the coefficients of a partial fraction decomposition. However, it forms the basis for some of these more efficient methods.

Find the partial fraction decomposition of the rational expression

\[\frac{1}{x^2+9x+14}.\]

The denominator can be factored as \((x+2)(x+7).\) This gives the partial fraction decomposition form

\[\frac{1}{x^2+9x+14}=\frac{A}{x+2}+\frac{B}{x+7}.\]

Observe what happens when we take the limit of the right-hand side as \(x\) approaches \(-2.\) The first rational expression will approach infinity, while the second rational expression will approach a constant. Therefore, the limit of this sum is equal to the limit of just the first rational expression.

\[\begin{align} \lim_{x \rightarrow -2}\left( \frac{A}{x+2}+\frac{B}{x+7} \right) &= \lim_{x \rightarrow -2} \frac{A}{x+2} \\ \\ \lim_{x \rightarrow -2}\frac{1}{x^2+9x+14} &= \lim_{x \rightarrow -2} \frac{A}{x+2}. \end{align}\]

Multiply both sides of this equation by the \((x+2)\) factor, and then evaluate the limit:

\[\begin{align} \lim_{x \rightarrow -2}\frac{1}{x+7} &= \lim_{x \rightarrow -2} A \\ \\ \frac{1}{5} &= A. \end{align}\]

This same process is used to compute \(B.\) This time, the limit is taken as \(x\) approaches \(-7:\)

\[\begin{align} \lim_{x \rightarrow -7}\left( \frac{A}{x+2}+\frac{B}{x+7} \right) &= \lim_{x \rightarrow -7} \frac{B}{x+7} \\ \\ \lim_{x \rightarrow -7}\frac{1}{x^2+9x+14} &= \lim_{x \rightarrow -7} \frac{B}{x+7} \\ \\ \lim_{x \rightarrow -7}\frac{1}{x+2} &= \lim_{x \rightarrow -7} B \\ \\ -\frac{1}{5} &= B. \end{align}\]

Thus, the partial fraction decomposition is

\[\frac{1}{x^2+9x+14}=\frac{1}{5(x+2)}-\frac{1}{5(x+7)}.\ _\square\]

## Heaviside's Cover-up Method

Main Article: Partial Fractions - Cover up Rule

We will first see an example before discussing the method: \[\dfrac{x−7}{(x−1)(x + 2)}=\dfrac A {x−1} + \dfrac B {x+2}.\] To get \(A\), cover up \((x - 1)\) on the left side and substitute \((x - 1) = 0,\) or \(x=1,\) in the remaining \(\frac{x−7}{x + 2}: \) \[\dfrac{1−7}{1+2} = - 2 = A. \] Similarly, for \(B\), substitute \(x = -2 \ \) in \( \frac{x−7}{x−1} \) to get \[\dfrac{-2−7}{-2-1} = 3 = B. \]

So we finally have \( \frac{x−7}{(x−1)(x + 2)} = \frac{-2}{x−1} + \frac{3}{x+2}. \)

In other words, to find \(A\), we are multiplying both sides by the denominator of \(A\) and get \[\left. \dfrac{x−7}{ {\color{red}(x−1)}(x + 2)} \times {\color{red}(x−1)} = \dfrac{A}{{\color{red}(x−1)}}\times {\color{red}(x−1)} + \dfrac{B}{x+2} \times {\color{red}(x−1)}\right|_{x=1}.\] By substituting \( (x - 1) = 0, \) i.e. \( x = 1, \) we now get the value of \(A\).

## Additional Examples and Problem Solving

Express \(\frac{2}{1-x^2}\) as partial fractions.

Solution 1:

The denominator \((1-x^2)\) can be factorized as \((1-x)(1+x)\). Therefore, let \(A\) and \(B\) be constants such that \[ \frac{2}{1-x^2} = \frac{A}{1-x} + \frac{B}{1+x}. \] Multiply both sides by \((1-x^2)\) to get \[\begin{align} 2 &= A (1+x) + B (1-x) \\ &= (A+B) + (A-B) x. \end{align}\] By comparing like terms, we have the following set of equations: \[\begin{cases}A+B=2\\A-B=0.\end{cases}\] We solve it and get \(A=1\) and \(B=1\). Therefore, \[\frac{2}{1-x^2} = \frac{1}{1-x} + \frac{1}{1+x}.\ _\square\]

Solution 2:

We may find \(A\) and \(B\) by inserting two differentvalidvalues to \(x\) and solving the simultaneous equations in \(A\) and \(B\).Substitute \( x=2 \implies 2 = 3A - B, \) and \( x = -2 \implies 2 = - A +3B. \)

So \( 4A=4B \implies A=B \) and \( A = 1,\) as above.

Note the word "valid." You cannot take \( x=1 \) or \( x=-1 \) here.Often \( x=0, x=1, \) and \( x=-1 \) are good choices.

Always try to take values for \( x \) that simplify calculations.

Say we have \(\frac 2 {x-3} \). Then \( x=1 \) is a good choice if it suits the other expression, or \( x=5.\ _\square\)\[\]

Solution 3:

Heaviside's cover-up method: (cannot be used for quadratic factors)We have \[\begin{align} \dfrac{2}{(1-x)(1+x)} &= \dfrac{A}{1-x} + \dfrac{B}{1+x} \\\\ \left. A=\dfrac{2}{1+1}\right|_{x= 1}\ &=\dfrac 2 2 \\&=1 \\\\ \left. B=\dfrac{2}{1-(-1)}\right|_{x=-1}\ &=\dfrac 2 2 \\&=1, \end{align}\] which is the same as above. \(_\square\)

Write \(\dfrac{2x+3}{(x+2)(x+5)}\) in the form of partial fractions.

First let's write the given rational fraction as follows: \[\dfrac{2x+3}{(x+2)(x+5)} = \dfrac{A}{x+2} + \dfrac{B}{x+5}.\] By taking the LCM, the right side can also be written as \[\dfrac{2x+3}{(x+2)(x+5)} = \dfrac{A(x+5)+B(x+2)}{(x+2)(x+5)}.\] Now cancelling the denominators of LHS and RHS, we get \[A(x+5) + B(x+2) = 2x + 3. \] To find the value of the term containing \(B\), we must eliminate the term containing \(A\) value.

Substitute \(x=-5\) for the whole term \(A(x+5)\) to be equal to \(0\), then \[B(-5+2) = 2(-5) + 3 \implies B = \frac{7}{3}. \] Now, to find the value of \(A\), we must eliminate \(B\). This means \(x=-2,\) which gives \[A(-2+5) = 2(-2) + 3 \implies A = -\frac{1}{3}. \]Therefore, the required partial fractions are \[\dfrac{2x+3}{(x+2)(x+5)} = -\dfrac{1}{3(x+2)} + \dfrac{7}{3(x+5)}.\ _\square\]

**Cite as:**Partial Fractions - Linear Factors.

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