Study S3 A Math Partial Fraction for Proper and Improper Fractions - Geniebook

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.

\(\begin{gather} \xrightarrow {\text {Simplification of Algebraic Fractions}} \end{gather} \)

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

\(\begin{gather} \xleftarrow {\text {Partial Fractions}} \end{gather} \)

 

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, \(x = −1\)

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

 

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

\(\begin{align} \implies1.5 − 14 &= A(2\times 1.5 − 3) + B(1.5 + 1) \\ \implies\quad−12.5 &= 2.5B \\ \implies\qquad\;\;\; B &= −5 \end{align}\)

 

  • 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 \(x = −4\),

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

 

  • Substitute \(x = 1\),

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

 

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

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

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

 

  • 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

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

 

  1. 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!

Keep Learning! Keep Improving!

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