# Partial Fractions

## Summary

Remember these formulas of partial fractions for different types of fractions:

Types of fractionsForm of the partial fractions $\frac { px\quad +\quad q }{ (x\quad +\quad a)(x\quad +\quad b) }$ $\frac { A }{ x\quad +\quad a } \quad +\quad \frac { B }{ x\quad +\quad b }$ $\frac { px\quad +\quad q }{ { (x\quad +\quad a) }^{ 2 } }$ $\frac { A }{ x\quad +\quad a } \quad +\quad \frac { B }{ { (x\quad +\quad a) }^{ 2 } }$ $\frac { p{ x }^{ 2 }\quad +\quad qx\quad +\quad r }{ (x\quad +\quad a)(x\quad +\quad b)(x\quad +\quad c) }$ $\frac { A }{ x\quad +\quad a } \quad +\quad \frac { B }{ x\quad +\quad b } \quad +\quad \frac { C }{ x\quad +\quad c }$ $\frac { p{ x }^{ 2 }\quad +\quad qx\quad +\quad r }{ { (x\quad +\quad a) }^{ 2 }(x\quad +\quad b) }$ $\frac { A }{ x\quad +\quad a } \quad +\quad \frac { B }{ { (x\quad +\quad a) }^{ 2 } } \quad +\quad \frac { C }{ x\quad +\quad c }$ $\frac { p{ x }^{ 2 }\quad +\quad qx\quad +\quad r }{ (x\quad +\quad a)({ x }^{ 2 }\quad +\quad bx\quad +\quad c) }$ $\frac { A }{ x\quad +\quad a } \quad +\quad \frac { Bx\quad +\quad C }{ { x }^{ 2 }\quad +\quad bx\quad +\quad c }$

We can recall from GCSE’s that to transform a function consisting of many fractions into a single fraction, we take LCM (lowest common factor) of the entire function i.e:

if we have a function $f\left( x \right) \quad =\quad \frac { 2 }{ x\quad +\quad 1 } \quad +\quad \frac { 3 }{ 2x\quad +\quad 3 }$, we take its LCM and make it into a single fraction: $f\left( x \right) \quad =\quad \frac { 2 }{ x\quad +\quad 1 } \quad +\quad \frac { 3 }{ 2x\quad +\quad 3 }$ $=\quad \frac { 2(2x\quad +\quad 3)\quad +\quad 3(x\quad +\quad 1) }{ (x\quad +\quad 1)(2x\quad +\quad 3) }$ $=\quad \frac { 4x\quad +\quad 6\quad +\quad 3x\quad +\quad 3 }{ (x\quad +\quad 1)(2x\quad +\quad 3) }$ $(x\quad +\quad 1)(2x\quad +\quad 3)$

Now the question is what do we do when we want to reverse this process and split a single fraction into two or more fractions. Well, the process of breaking a single fraction into multiple fractions is known as splitting into ”partial fractions”. It could be both sum or difference of two or more fractions.

There are three different types of fractions:

1.     Where a fraction consists of only linear factors in the denominator.

2.     Where there are repeated factors in the denominator of the fraction.

3.     Where there are quadratic factors in the denominator of the fraction.

We will go through each one of the types with the methods used to solve them along with examples below.

#### 1.     Linear factors in the denominator

This included both proper fractions and improper fractions. Let’s have a look at the proper fractions first.

#### Example #1

Q. Find the partial fractions of $\frac { 7x\quad +\quad 9 }{ (x\quad +\quad 1)(2x\quad +\quad 3) }$

Note: This is the same function that resulted by taking LCM of fractions in the beginning of this article.

Solution:

Since the denominator has linear factors, there required partial fractions will be: $\frac { A }{ x\quad +\quad 1 } \quad +\quad \frac { B }{ 2x\quad +\quad 3 }$

Hence: $\frac { 7x\quad +\quad 9 }{ (x\quad +\quad 1)(2x\quad +\quad 3) } \quad =\quad \frac { A }{ x\quad +\quad 1 } \quad +\quad \frac { B }{ 2x\quad +\quad 3 } \quad \quad \Rightarrow \quad \quad (i)$ $7x\quad +\quad 9\quad =\quad A(2x\quad +\quad 3)\quad +\quad B(x\quad +\quad 1)\quad \quad \quad \Rightarrow \quad \quad equation1$

First find the 2 values of x: $2x\quad +\quad 3\quad =\quad 0$ $2x\quad =\quad -3$ $x\quad =\quad \frac { -3 }{ 2 }$

and $x\quad +\quad 1\quad =\quad 0$ $x\quad =\quad -1$

Substitute each value of x in equation 1, one at a time.

So to find the value of A put x = -1 in equation 1, $7(-1)\quad +\quad 9\quad =\quad A(2(-1)\quad +\quad 3)\quad +\quad B(-1\quad +\quad 1)$ $-7\quad +\quad 9\quad =\quad A(-2\quad +\quad 3)\quad +\quad B(0)$ $2\quad =\quad A(1)$ $A\quad =\quad 2$

So to find the value of B put $x\quad =\quad \frac { -3 }{ 2 }$  in equation 1: $7(\frac { -3 }{ 2 } )\quad +\quad 9\quad =\quad A(2(\frac { -3 }{ 2 } )\quad +\quad 3)\quad +\quad B(\frac { -3 }{ 2 } \quad +\quad 1)$ $\frac { -21 }{ 2 } \quad +\quad 9\quad =\quad A(0)\quad +\quad B(\frac { -1 }{ 2 } )$ $\frac { -3 }{ 2 } \quad =\quad B(\frac { -1 }{ 2 } )$ $B\quad =\quad 3$

Substituting the values of A and B in equation (i) above gives us our partial fractions:

Ans: $\frac { 2 }{ x\quad +\quad 1 } \quad +\quad \frac { 3 }{ 2x\quad +\quad 3 }$

For such proper fractions whose denominators are linear factors we can also use a cover up method.

Cover up method

This is basically a shortcut of finding the partial fractions, where we don’t have to do long calculations like we did in the above example i.e let’s do the above example now with the cover up method. You will see how quickly we can find the results.

#### Example #2

Q. Find the partial fractions of $\frac { 7x\quad +\quad 9 }{ (x\quad +\quad 1)(2x\quad +\quad 3) }$  using the cover up methods.

Solution:

As we know the partial fraction expression would be: $\frac { 7x\quad +\quad 9 }{ (x\quad +\quad 1)(2x\quad +\quad 3) } \quad =\quad \frac { A }{ x\quad +\quad 1 } \quad +\quad \frac { B }{ 2x\quad +\quad 3 }$

To find A we consider the left hand side of the equation: $\frac { 7x\quad +\quad 9 }{ (x\quad +\quad 1)(2x\quad +\quad 3) }$

We cover up one factor in the denominator first i.e cover up (x + 1)

Since we have covered up (x + 1), the value of x in this case is -1.

We get $A\quad =\quad \frac { 7x\quad +\quad 9 }{ (2x\quad +\quad 3) }$ , substitute the value of x: $A\quad =\quad \frac { 7(-1)\quad +\quad 9 }{ (2(-1)\quad +\quad 3) } \quad \quad \Rightarrow \quad \quad A\quad =\quad 2$

Similarly to find B we cover up (2x + 3) and find that $x\quad =\quad \frac { -3 }{ 2 }$  in this case.

Substituting this value of x we get: $B\quad =\quad \frac { 7(\frac { -3 }{ 2 } )\quad +\quad 9 }{ (2(\frac { -3 }{ 2 } )\quad +\quad 3) } \quad \quad \Rightarrow \quad \quad B\quad =\quad 3$

Therefore our partial fractions are $\frac { 2 }{ x\quad +\quad 1 } \quad +\quad \frac { 3 }{ 2x\quad +\quad 3 }$

Now that we have understood how we find partial fractions for proper fractions, we move on to improper fractions.

#### Partial fractions of Improper fractions

Improper fractions are fractions whose degree of denominator is equal to or less than the degree of its numerator i.e: $\frac { { x }^{ 3 }\quad +\quad 3x }{ (x\quad +\quad 1)(x\quad +\quad 2) }$  or $\frac { { x }^{ 5 }\quad +\quad x\quad +\quad 2 }{ (x\quad +\quad 3)(x\quad +\quad 1) }$

these are both considered as improper fractions.

To find work out the partial fractions, we must have the function as a proper fraction. Therefore, we convert all improper fractions into proper ones before we decompose them into partial fractions. We do this by dividing the numerator by its denominator till it becomes a proper fractions. This is done through algebraic long division. Algebraic long division has been explained in detail in the article ”Algebraic long division”. Let’s work out an example now.

#### Example #3

Q. Find the partial fractions of $\frac { { x }^{ 2 }\quad +\quad 3x\quad +\quad 1 }{ (x\quad -\quad 1)(x\quad +\quad 2) }$

Solution:

We can see that the above function is an improper fractions as the degree of numerator is equal to degree of the denominator. Hence, we must carry out long division to convert it into a proper fraction.

After the long division the fraction becomes:  $\frac { { x }^{ 2 }\quad +\quad 3x\quad +\quad 1 }{ (x\quad -\quad 1)(x\quad +\quad 2) } \quad =\quad 1\quad +\quad \frac { 2x\quad +\quad 3 }{ (x\quad -\quad 1)(x\quad +\quad 2) }$

Now $\frac { 2x\quad +\quad 3 }{ (x\quad -\quad 1)(x\quad +\quad 2) }$  is a proper fraction, we can therefore split it into partial fractions. $\frac { 2x\quad +\quad 3 }{ (x\quad -\quad 1)(x\quad +\quad 2) } \quad =\quad \frac { A }{ x\quad -\quad 1 } \quad +\quad \frac { B }{ x\quad +\quad 2 }$

Now using the cover up method we find the values of A and B.

Cover up x – 1; $\frac { 2x\quad +\quad 3 }{ x\quad +\quad 2 }$

Substitute x = 1; $\frac { 2(1)\quad +\quad 3 }{ 1\quad +\quad 2 } \quad \quad \quad \quad \Rightarrow \quad \quad A\quad =\quad \frac { 5 }{ 3 }$

Cover up x + 2; $\frac { 2x\quad +\quad 3 }{ x\quad -\quad 1 } \quad$

Substitute x = -2; $\frac { 2(-2)\quad +\quad 3 }{ -2\quad -\quad 1 } \quad \quad \quad \quad \Rightarrow \quad \quad B\quad =\quad \frac { 1 }{ 3 }$

Therefore, the required partial fractions are: $\frac { 5 }{ 3(x\quad -\quad 1) } \quad +\quad \frac { 1 }{ 3(x\quad +\quad 2) }$      Ans

#### 2.    Repeated factors in the denominator

When a square term occurs in a denominator i.e ${ (x\quad +\quad m) }^{ 2 }$,  we consider two separate constants for such expressions.

#### Example #4

Q. Express $\frac { x }{ (x\quad -\quad 1){ (x\quad -\quad 2) }^{ 2 } }$ in partial fractions

Solution: $\frac { x }{ (x\quad -\quad 1){ (x\quad -\quad 2) }^{ 2 } } \quad =$ $=\quad \frac { A }{ x\quad -\quad 1 } \quad +\quad \frac { B }{ x\quad -\quad 2 } \quad +\quad \frac { C }{ { (x\quad -\quad 2) }^{ 2 } }$ $x\quad =\quad A{ (x\quad -\quad 2) }^{ 2 }\quad +\quad B(x\quad -\quad 2)(x\quad -\quad 1)\quad +\quad C(x\quad -\quad 1)\quad \quad \quad \Rightarrow \quad \quad eq1$

When x = 1: $1\quad =\quad A{ (1\quad -\quad 2) }^{ 2 }\quad +\quad B(1\quad -\quad 2)(1\quad -\quad 1)\quad +\quad C(1\quad -\quad 1)$ $1\quad =\quad A{ (-1) }^{ 2 }\quad +\quad B(0)\quad +\quad C(0)$ $A\quad =\quad 1$

When x = 2: $2\quad =\quad A{ (2\quad -\quad 2) }^{ 2 }\quad +\quad B(2\quad -\quad 2)(2\quad -\quad 1)\quad +\quad C(2\quad -\quad 1)$ $2\quad =\quad A{ (0) }^{ 2 }\quad +\quad B(0)\quad +\quad C(1)$ $C\quad =\quad 2$

To find B simplify eq 1: $x\quad =\quad A{ (x\quad -\quad 2) }^{ 2 }\quad +\quad B(x\quad -\quad 2)(x\quad -\quad 1)\quad +\quad C(x\quad -\quad 1)$ $x\quad =\quad A({ x }^{ 4 }\quad -\quad 4x\quad +\quad 2)\quad +\quad B({ x }^{ 2 }\quad -\quad 3x\quad +\quad 2)\quad +\quad C(x\quad -\quad 1)$ $x\quad =\quad A{ x }^{ 2 }\quad -\quad 4Ax\quad +\quad 2A\quad +\quad { Bx }^{ 2 }\quad -\quad 3Bx\quad +\quad 2B\quad +\quad Cx\quad -\quad C$

Comparing all the coefficients of ${ x }^{ 2 }$:

0 = A + B  we know  A = 1

0 = 1 + B

B =  -1

Hence, the required partial fractions are: $\frac { 1 }{ x\quad -\quad 1 } \quad +\quad \frac { 1 }{ x\quad -\quad 2 } \quad +\quad \frac { 2 }{ { (x\quad -\quad 2) }^{ 2 } }$       Ans

#### 3.    Quadratic factors in the denominator

In the case, where a fraction has a quadratic factor in the denominator which cannot be simplified further, then that denominator will have a linear numerato in its partial fraction i.e:

If we have a function $\frac { { x }^{ 2 }\quad +\quad 4x\quad -\quad 2 }{ (x\quad -\quad 1)({ x }^{ 2 }\quad +\quad 1) }$  we will write it as $\frac { A }{ x\quad -\quad 1 } \quad +\quad \frac { Bx\quad +\quad C }{ { x }^{ 2 }\quad +\quad 1 }$.

We will then follow the same process as above to find the values of A and after that we compare the coefficients of x to find the value of B and C.