Partial Fractions

In this article, we will be learning about partial fractions for proper fractions. More specifically, we will be concentrating on the first two cases listed below.

• Case 1:

Partial fractions with distinct linear factors in the denominator.

• Case 2:

Proper fractions with a repeated linear factor in the denominator.

• Case 3:

Proper fractions with a quadratic factor that cannot be factorised in the denominator.
 

What are Partial Fractions?

A proper algebraic fraction may be expressed in two or more partial fractions.

The process from left to right shows the result of subtracting two algebraic fractions.

The reverse process from right to left shows the ‘splitting’ of an algebraic fraction into two or more partial fractions.

Case 1: Proper Fractions with Distinct Linear Factors in Denominator

An algebraic fraction where the denominator is a product of two distinct linear factors may be expressed in partial fractions as shown below.

\(\begin{align*} \frac{px+q}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d} \end{align*}\), where \(A\) and \(B\) are constants.

The denominator is a product of \((ax+b)\) and \((cx+d)\). \((ax+b)\) is one linear factor and \((cx+d)\) is another linear factor. 

Question 1:

Express \(\begin{align*} \frac{(x-14)}{(x+1)(2x-3)} \end{align*}\) in partial fractions.

Solution:

• Let \(\begin{align*} \frac{(x-14)}{(x+1)(2x-3)}=\frac{A}{(x+1)}+\frac{B}{(2x-3)} \end{align*}\).

We need to find the value of \(A\) and of \(B\). 

• Multiply the equation throughout by the denominator \((x + 1)(2x – 3)\). 

\(\begin{align*} \frac{(x-14)}{(x+1)(2x+3)} \times {(x+1)(2x-3)} \;=\; \frac{A}{(x+1)} \times {(x+1)(2x-3)} \;+\; \frac {B}{(2x-3)} \times {(x+1)(2x-3)} \end{align*}\)

• We will be left with the following equation,

\(\begin{align*} (x-14)=A(2x-3)+B(x+1) \end{align*}\)

• Then, we need to perform smart substitution. Substitute \(x=-1\), so that the\( B(x + 1)\) term is eliminated.

So, \(x = −1\)

\(\begin{align*} & −1−14 = A(2(−1)−3) \\ & −15 = A(−5) \\ & A = \frac{-15}{-5 } \\ &= 3 \end{align*}\)

• Then, substitute \(x = 1.5\) to eliminate the \(A(2x – 3)\) term.

\(\begin{align} &1.5 − 14 = A(2\times 1.5 − 3) + B(1.5 + 1) \\ &−12.5 = 2.5B \\ & B = −5 \end{align}\)

• Lastly, substitute the value of \(A\) and of \(B\) into the first equation.

Therefore, \(\begin{align*} \frac{(x-14)}{(x+1)(2x+3)}=\frac {3}{(x+1)}-\frac{5}{(2x-3)} \end{align*}\).

Case 2: Proper Fractions with Repeated Linear Factors in the Denominator

An algebraic fraction where the denominator contains a linear factor that is squared may be expressed in partial fractions as shown below.

\(\begin{align*} \frac {px+q}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2} \end{align*} \), where \(A\) and \(B\) are constants. 

Question 2:

Express \(\begin{align*} \frac {x}{(x-1)(x+4)^2} \end{align*} \) in partial fractions.

Solution:

• Let \(\begin{align*} \frac {x}{(x-1)(x+4)^2}=\frac{A}{(x-1)}+\frac{B}{(x+4)}+\frac{C}{(x+4)^2} \end{align*}\)

• Multiply the equation throughout by the denominator \((x – 1)(x + 4)^2\). 

\(x = A(x+4) + B(x−1)(x+4) + C(x−1)\)

• Substitute \(x = −4\),

\(\begin{align*} &−4 = C(−4−1) \\ &−4 = −5C \\ & C = \frac{4}{5} \end{align*} \)

• Substitute \(x = 1\),

\(\begin{align*} & 1 = A(1+4)^2+ 0 + 0 \\ &1 = 25A \\ & A = \frac{1}{25} \end{align*}\)

• For the third substitution, choose any number (or a number that makes the equation easy to manipulate).  

Substitute\(\begin{align*} x = 0, \;C = \frac{4}{5}, \;A = \frac {1}{25}, \end{align*}\)

\(\begin{align*} & 0 = \frac{1}{25}\times(0+4)^2 + B(0-1)(0+4) + \frac{4}{5} \times (0-1) \\ & 0 =\frac{16}{25} + B(−4) − \frac{4}{5} \\ & \frac{4}{5} − \frac{16}{25} = − 4B \\ & \frac{4}{25} = − 4B \\ & B = − \frac{1}{25} \end{align*} \)

• Lastly, substitute the values of \(A\), \(B\) and \(C\) into the first equation. 

\(\begin{align*} \frac{x}{(x-1)(x+4)^2}=\frac{1}{25(x-1)}-\frac{1}{25(x+4)}+\frac{4}{5(x+4)^2} \end{align*}\)

Conclusion

In this article, we have covered two cases for splitting an algebraic fraction into partial fractions:

1. A proper algebraic fraction with distinct linear factors

\(\begin{align*} \frac{px+q}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d} \end{align*}\)

2. A proper fraction with repeated linear factors

\(\begin{align*} \frac {px+q}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2} \end{align*} \)

The only difference between these two forms is a square in the denominator. However, please take note of it as the subsequent steps will differ quite a bit!

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