Partial fractions

In a nutshell

An algebraic fraction can be written in the form $\dfrac{P(x)}{Q(x)}$ , where $P(x)$ and $Q(x)$ are polynomials. The order of a polynomial is given by the highest power of $x$ . If the order of $P(x)$ is less than the order of $Q(x)$ , then it is a proper fraction, and can be split up into two or more partial fractions. The denominators of the partial fractions are based on the linear factors of $Q(x)$ .

Partial fractions

When splitting a algebraic fraction into partial fractions, there are two methods, which are:

Substitution
Equating coefficients

Substitution

procedure

1.	Factorise the denominator into linear factors.
2.	Set $A$ and $B$ as numerators of your partial fractions.
3.	Multiply up the fractions.
4.	In order to find the values of $A$ and $B$ , substitute $x=\rm root \ of \ the \ factor$ .

Example 1

Split $\cfrac{7x+4}{x^2-49}$ into partial fractions using the substitution method.

First, factorise the denominator into linear factors.

$\cfrac{7x+4}{x^2-49}=\cfrac{7x+4}{(x+7)(x-7)}$ 

Now, set $\it A$ and $\it B$ as numerators of the partial fractions.

$\cfrac{7x+4}{(x+7)(x-7)} = \cfrac{A}{x+7}+\cfrac{B}{x-7}$

Multiply up the fractions to eliminate the denominators.

$\begin{aligned}\dfrac{7x+4}{(x+7)(x-7)} &= \dfrac{A}{x+7}+\dfrac{B}{x-7}\\\dfrac{7x+4}{(x+7)(x-7)} &=\cfrac{A(x-7)}{(x+7)(x-7)} + \cfrac{B(x+7)}{(x+7)(x-7)} \\\\7x+4 & = A(x-7) + B(x+7)\end {aligned}$

In order to find $A$ , substitute $x=-7$ , so $B$ is eliminated from the equation.

$\begin{aligned}{7(-7)+4} &={A(-7-7) + B(-7+7)} \\-49 +4 &= -14A \\ A&= \underline{\dfrac {45}{14}}\end{aligned}$

Repeat the process using $x=7$ in order to find $B$ .

$\begin{aligned}7 \times 7 + 4 &= A ( 7-7) + B (7+7) \\ 53 &= 14B \\ B&= \underline{ \dfrac{53}{14}}\end{aligned}$

$53= 14B$

$\underline{B = \cfrac{53}{14}}$

Therefore, $\underline{\dfrac{7x+4}{x^2-49}=\dfrac{45}{14(x+7)} + \dfrac{53}{14(x-7)}}$

Equating coefficients

procedure

1.	Factorise the denominator into linear factors.
2.	Set $A$ and $B$ as numerators of your partial fractions.
3.	Multiply up the fractions.
4.	Expand all brackets and equate the coefficients of $x$ and the constant terms, to find the values of $A$ and $B$ .

Example 2

Split $\cfrac{7x+4}{x^2-49}$ into partial fractions using the equating coefficients method.

First, factorise the denominator into linear factors.

$\cfrac{7x+4}{x^2-49} = \cfrac{7x+4}{(x+7)(x-7)}$ 

Now, set $\it A$ and $\it B$ as numerators of the partial fractons.

$\cfrac{7x+4}{(x+7)(x-7)}=\cfrac{A}{x+7}+\cfrac{B}{x-7}$

Multiply up the fractions to eliminate the denominators.

$\begin{aligned}\dfrac{7x+4}{(x+7)(x-7)} &= \dfrac{A}{x+7}+\dfrac{B}{x-7}\\\dfrac{7x+4}{(x+7)(x-7)} &=\cfrac{A(x-7)}{(x+7)(x-7)} + \cfrac{B(x+7)}{(x+7)(x-7)} \\\\7x+4 & = A(x-7) + B(x+7)\end {aligned}$

Next, expand all the brackets, to find the coefficients.

$7x+4 = Ax - 7A + Bx + 7B$ 

$7x+4 = x(A+B)+7(B-A)$ 

Equate coefficients of x.

$7x = x(A+B)$ 

And constant terms.

$4 = 7 (B-A)$ 

And solve both equations simultaneously.

$\begin{cases}7x = x(A+B) \\4= 7 (B-A) \end{cases}$

Giving:

$\underline{A=\cfrac{45}{14}} \hspace{10mm}\underline{B=\cfrac{53}{14}}$

Note: You will have to split your fractions into as many terms ( $A, \ B, \ C$ ) as factors in the denominator you have.