In this mini-lesson, we will explore the process of decomposition of partial fractions with partial fractions examples and interactive questions.

Jonathan was working on the addition of rational expressions.

He found the sum of two rational expressions \(\dfrac{1}{x+1}+ \dfrac{3}{x-2}\) to be \(\dfrac{4x+1}{(x+1)(x-2)}\).

Suddenly he got a thought, is there any way to write \(\dfrac{4x+1}{(x+1)(x-2)}\) as the sum of two fractions \(\dfrac{1}{x+1}+ \dfrac{3}{x-2}\)?

He asked his teacher about this and the answer from his teacher was "Partial Fractions".

Jonathan was curious to know how to do partial fraction decomposition (or) partial fractions expansion.

So he read this page and understood this process completely.

**Lesson Plan**

**What is Meant By Partial Fractions?**

When a rational expression is split into the sum of two or more rational expressions, the rational expressions that are a part of the sum are called the partial fractions.

Every factor of the denominator of a rational expression corresponds to a partial fraction.

For example, in the above figure, \(\dfrac{4 x+1}{(x+1)(x-2)}\) has two factors in the denominator and hence there are two partial fractions, one with the denominator \(x+1\) and the other with the denominator \(x-2\).

**What Are General Formulas of Partial Fractions?**

In the above example, the numerators of partial fractions are 1 and 3.

Is the numerator of a partial fraction always a constant?

No, it depends upon the denominator.

- If the denominator is linear, the numerator is constant.
- If the denominator is quadratic, the numerator is linear.

It means, the numerator's degree of a partial fraction is always just 1 less than the denominator's degree.

Here are partial fractions formulas (here, all variables apart from \(x\) are constants).

Factor in denominator |
Partial fraction |
---|---|

\(ax+b\) (non-repeated linear factor) |
\(\dfrac{A}{ax+b}\) |

\((ax+b)^n\) (repeated linear factors) |
\( \begin{align}&\dfrac{A_1}{ax+b}\\[0.2cm]&+\dfrac{A_2}{(ax+b)^2}\\[0.2cm]&+...\\[0.2cm]&+ \dfrac{A_n}{(ax+b)^n}\end{align}\) |

\(ax^2+bx+c\) |
\(\dfrac{Ax+B}{ax^2+bx+c}\) |

\((ax^2+bx+c)^n\) (repeated irreducible quadratic factors) |
\( \begin{align}&\dfrac{A_1x+B_1}{ax^2+bx+c}\\[0.2cm]&+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}\\[0.2cm]&+...\\[0.2cm]&+ \dfrac{A_nx+B_n}{(ax^2+bx+c)^n}\end{align}\) |

**Examples**

\( \begin{align} \frac{4}{(x\!-1)(x\!+5)} &= \frac{A}{x\!-1}+ \frac{B}{x\!+5}\\[0.3cm] \frac{3x\!+1}{(2x\!-1)(x\!+2)^2}&= \frac{A}{2x\!-1}+ \frac{B}{x\!+2}+\\ \frac{C}{(x\!+2)^2}\\[0.3cm] \frac{2x\!-3}{(x\!-2)(x^2\!+1)} &= \frac{A}{x\!-2}+ \frac{Bx\!+C}{x^2\!+1} \end{align}\)

In all these examples, \(A, B,\) and \(C\) are constants to be determined. Let's learn how to find these constants.

- The numerator's degree of a partial fraction is always just 1 less than the denominator's degree.
- When a partial fraction has repeated factors of the form \((ax+b)^n\) or \((ax^2+bx+c)^n\), they correspond to \(n\) different partial fractions where the denominators of the partial fractions have exponents \(1, 2, 3, ..., n\).
- The above partial fractions formulas do not depend upon the numerator of the given rational expression.
- Before applying the above formulas, factorize the denominator as much as possible. Otherwise, the answer won't be accurate.

**What Is the Decomposition of Partial Fractions? **

As we have seen earlier, the decomposition of partial fractions is writing a rational expression as the sum of two or more partial fractions.

Let us learn this process of partial fractions decomposition (or expansion) by an example.

**Example**

Find the partial fraction expansion of the expression\[\dfrac{4x+12}{x^2+4x}\]

**Solution**

**Always remember to factor the denominator as much as possible before doing the partial fraction decomposition.**

\[\dfrac{4x+12}{x^2+4x} = \dfrac{4x+12}{x(x+4)}\]

The denominator has non-repeated linear factors.

So, every factor corresponds to a constant in the numerator while writing the partial fractions.

Let us assume that:

\[ \dfrac{4x+12}{x(x+4)} = \dfrac{A}{x}+ \dfrac{B}{x+4} \,\,\, \rightarrow (1) \]

The LCD (Least Common Denominator) of the sum (on the right side) is \(x(x+4)\).

Multiplying both sides by \(x(x+4)\), \[ 4x+12 = A(x+4)+Bx \,\,\, \rightarrow (2)\]

Now we have to solve it for \(A\) and \(B\).

**For that, we set each linear factor to zero.**

Substitute \(x+4=0 \text{ (or) }x=-4\) in (2): \[ \begin{align} 4(-4)+12 &= A(0) +B (-4)\\[0.2cm] -4&=-4B\\[0.2cm] B&=1 \end{align}\]

Substitute \(x=0\) in (2): \[ \begin{align} 4(0)+12 &= A(0+4)+B(0) \\[0.2cm] 12&=4A\\[0.2cm] A&=3 \end{align}\]

Substitute the values of \(A\) and \(B\) in (1), we get the partial fractions decomposition of the given expression: \[ \dfrac{4x+12}{x(x+4)} = \dfrac{3}{x}+ \dfrac{1}{x+4}\]

**Partial Fractions Calculator**

Here is the "Partial Fractions Calculator".

You can enter any rational expression. This calculator decomposes it into partial fractions.

**What Are Partial Fractions of Improper Fraction?**

When we have to decompose an improper fraction into partial fractions, we first should do the long division.

**Example**

Find the partial fraction decomposition of the expression \[\frac{x^{3}+4 x^{2}-2 x-5}{x^{2}-4 x+4}\]

**Solution**

Here, the degree of the numerator (3) is greater than the degree of the denominator (2).

So the given fraction is improper.

So we have to do the long division first.

Then write the given fraction as \(\text{Quotient}+ \dfrac{\text{Remainder}}{\text{Divisor}}\).

Then we get: \[\begin{align}&\frac{x^{3}+4 x^{2}-2 x-5}{x^{2}-4 x+4}\\[0.2cm]&=x+8+\frac{26 x-37}{x^{2}-4 x+4}\,\,\, \rightarrow (1)\end{align}\]

Here, the fraction on the right side is a proper fraction and hence it can be split into partial fractions using:

\[\begin{align}&\frac{26 x-37}{x^{2}-4 x+4} = \frac{26 x-37}{(x-2)^2}\\[0.2cm]&= \dfrac{A}{x-2}+ \dfrac{B}{(x-2)^2}\,\,\, \rightarrow (2)\end{align}\]

Can you try to solve for \(A\) and \(B\) here? (Hint: Set each of \(x-2\) and \(x\) one by one to zero to get \(A\) and \(B\)).

You should get \(A=26\) and \(B=15\).

Substituting these values in (2):

\[\frac{26 x-37}{x^{2}-4 x+4} = \dfrac{26}{x-2}+ \dfrac{15}{(x-2)^2}\]

Substituting this in (1):

\[\begin{align}&\frac{x^{3}+4 x^{2}-2 x-5}{x^{2}-4 x+4}\\[0.2cm]&=x+8+ \dfrac{26}{x-2}+ \dfrac{15}{(x-2)^2}\end{align}\]

**Solved Examples**

Here are more partial fractions examples.

Example 1 |

Can we help Rachel decompose the following expression into partial fractions?

\[\dfrac{x^{4}+x^{3}+x^{2}+1}{x^{2}+x-2}\]

**Solution**

When we factorize the denominator, we get: \[x^2+x-2=(x+2)(x-1)\]

The degree of the numerator (4) is greater than that of the denominator (2).

So it is an improper fraction.

We do the long division first.

So the given fraction can be written as:

\[\begin{align}&\dfrac{x^{4}+x^{3}+x^{2}+1}{x^{2}+x-2} \\[0.2cm]&= x^2+3 + \dfrac{-3x+7}{(x+2)(x-1)} \,\,\, \rightarrow (1)\end{align}\]

Now we will decompose \(\dfrac{-3x+7}{(x+2)(x-1)}\) into partial fractions using:

\[\dfrac{-3x+7}{(x+2)(x-1)} = \dfrac{A}{x+2}+\dfrac{B}{x-1}\,\, \rightarrow (2)\]

Multiplying both sides by the LCD \((x+2)(x-1)\),

\[-3x+7 = A(x-1)+B(x+2)\,\,\,\, \rightarrow (3)\]

Substitute \(x-1=0\) (or \(x=1\)) in (3): \[ -3+7 = 3B \Rightarrow B= \dfrac{4}{3}\]

Substitute \(x+2=0\) (or \(x=-2\)) in (3): \[ 6+7 = -3A \Rightarrow A= \dfrac{-13}{3}\]

Substitute these values in (2):

\[\dfrac{-3x+7}{(x+2)(x-1)} = -\dfrac{13}{3(x+2)}+\dfrac{4}{3(x-1)}\]

Substitute this in (1), the partial fractions decomposition of the given expression is:

\(\therefore\) \(x^2+3-\dfrac{13}{3(x+2)}+\dfrac{4}{3(x-1)}\) |

Example 2 |

Can we help Mia decompose the following rational expression into partial fractions?

\[\frac{4 x^{3}+x+2}{x^{2}\left(x^{2}+1\right)}\]

**Solution**

Look at the denominator.

We have \(x^2\). It means the linear factor \(x\) is repeating.

\(x^2+1\) is an irreducible (can't be factorized) quadratic factor.

So the given fraction can be decomposed as follows:

\[\begin{align}&\frac{4 x^{3}+x+2}{x^{2}\left(x^{2}+1\right)}\\[0.2cm]& = \dfrac{A}{x}+ \dfrac{B}{x^2}+ \dfrac{Cx+D}{x^2+1}\\[0.2cm]&\,\,\rightarrow (1)\end{align}\]

Multiplying both sides by the LCD \(x^2(x^2+1)\):

\[\begin{align}&4x^3+x+2\\[0.2cm]& = Ax(x^2+1)+ B(x^2+1) + (Cx+D)x^2\\[0.2cm] &\,\,\rightarrow (2) \end{align}\]

Setting the linear factor \(x\) to 0, i.e., \(x=0\), we get: \[2=B\]

Now we do not have any other linear factors to set to zero.

So we will expand the right-hand side expression. Then we will compare the coefficients of \(x^3,x^2,x\), and constant.

By comparing the coefficients of \(x^3\), we get: \(4 = A+C\).

By comparing the coefficients of \(x^2\), we get: \(0 = B+D\).

By comparing the coefficients of \(x\), we get: \(1=A\).

By comparing the constants, we get: \(2 = B\).

By solving these equations, we get: \[A=1\\B=2\\C=3\\D=-2\]

Substitute all these values in (1), the given expression becomes:

\(\therefore\) \(\dfrac{1}{x}+\dfrac{2}{x^{2}}+\dfrac{3 x-2}{x^{2}+1}\) |

If the denominator has non-repeated linear factors:

- The constants can be obtained by setting each linear factor to zero.

If the denominator has either repeated linear factors and/or irreducible quadratic factors:

- Set the linear factors to zero to find the value of some constants.
- Set \(x=0\) to get at least one another constant.
- Compare the coefficients of \(x^3, x^2, ...\), etc to find the other constants.

**Interactive Questions**

**Here are a few activities for you to practice. **

**Select/type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

The mini-lesson targeted the fascinating concept of Partial Fractions. The math journey around Partial Fractions starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

**About Cuemath**

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Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

**Frequently Asked Questions (FAQs) **

## 1. How do you know how to add partial fractions?

While writing an expression as the sum of partial fractions, keep the following points in mind:

- The numerator's degree of a partial fraction is always just 1 less than the denominator's degree.
- When a partial fraction has repeated factors of the form \((ax+b)^n\) (or) \((ax^2+bx+c)^n\), they correspond to \(n\) different partial fractions where the denominators of the partial fractions have exponents \(1, 2, 3, ..., n\).

For more information, go to "What are General Formulas of Partial Fractions?" section of this page.

To add to partial fractions, we just make their denominators same and add.

For example: \[\begin{align}&\dfrac{3}{x}+ \dfrac{1}{x+4} \\[0.2cm]

&=\dfrac{3}{x}\cdot \dfrac{x+4}{x+4}+ \dfrac{1}{x+4} \cdot \dfrac{x}{x}\\[0.2cm]

&= \dfrac{3x+12+x}{x(x+4)}\\[0.2cm]

&=\dfrac{4x+12}{x^2+4x} \end{align}\]

## 2. How do you differentiate partial fractions?

We differentiate the partial fractions using the quotient rule as they are fractions.

## 3. How do you solve a repeated root partial fraction?

When a partial fraction has repeated factors of the form \((ax+b)^n\) or \((ax^2+bx+c)^n\), they correspond to \(n\) different partial fractions where the denominators of the partial fractions have exponents \(1, 2, 3, ..., n\).

For example, if the denominator is of the form \((ax+b)^n\), then the corresponding partial fractions should be of the form \(\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+...\).