Maths > Partial Fractions > 2.0 Types of Partial fraction
Partial Fractions
1.0 Introduction
2.0 Types of Partial fraction
2.1 When the denominator has non-repeated linear factors
2.2 When the denominator has repeated linear factors
2.3 When the denominator has non-repeated quadratic factors
2.4 When the denominator has repeated quadratic factors
2.1 When the denominator has non-repeated linear factors
2.2 When the denominator has repeated linear factors
2.3 When the denominator has non-repeated quadratic factors
2.4 When the denominator has repeated quadratic factors
Let $\frac{{p(x)}}{{q(x)}}$ be any rational function, in which polynomial $q(x)$ has non-repeated linear factor.
Mathematically, $$q(x) = (x - {a_1})(x - {a_2})...(x - {a_n})$$
So, $\frac{{p(x)}}{{q(x)}}$ can be expressed as partial fraction as shown below. $$\frac{{p(x)}}{{q(x)}} = \frac{{{A_1}}}{{(x - {a_1})}} + \frac{{{A_2}}}{{(x - {a_2})}} + ... + \frac{{{A_n}}}{{(x - {a_n})}}$$ Where $A_1,\ A_2,\ ...\ A_n$ are the constants which can be determined by equating LHS and RHS.
Question 1. Resolve $\frac{{2{x^2}}}{{(x - 1)(x - 2)(x - 3)}}$ into partial fractions.
Solution: The above equation is a proper fraction because the denominator is a 3 degree polynomial where as the numerator is a 2 degree polynomial.
So, the rational fraction can be expressed as the sum of partial fraction as shown below, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{2{x^2}}}{{(x - 1)(x - 2)(x - 3)}} = \frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{x - 3}} \\ \Rightarrow \frac{{2{x^2}}}{{(x - 1)(x - 2)(x - 3)}} = \frac{{A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)}}{{(x - 1)(x - 2)(x - 3)}} \\\end{aligned} \end{equation} $$
Equating the numerator of LHS and RHS we get, $$ \Rightarrow A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2) = 2{x^2}$$
For $(x-1)=0$ or $x=1$, $$\begin{equation} \begin{aligned} \Rightarrow A(1 - 2)(1 - 3) + B(1 - 1)(1 - 3) + C(1 - 1)(1 - 2) = 2{(1)^2} \\ \Rightarrow 2A = 2 \\ \Rightarrow A = 1 \\\end{aligned} \end{equation} $$
For $(x-2)=0$ or $x=2$, $$\begin{equation} \begin{aligned} \Rightarrow A(2 - 2)(2 - 3) + B(2 - 1)(2 - 3) + C(2 - 1)(2 - 2) = 2{(2)^2} \\ \Rightarrow - B = 8 \\ \Rightarrow B = - 8 \\\end{aligned} \end{equation} $$
For $(x-3)=0$ or $x=3$, $$\begin{equation} \begin{aligned} \Rightarrow A(3 - 2)(3 - 3) + B(3 - 1)(3 - 3) + C(3 - 1)(3 - 2) = 2{(3)^2} \\ \Rightarrow 2C = 18 \\ \Rightarrow C = 9 \\\end{aligned} \end{equation} $$
Therefore, $$ \Rightarrow \frac{{2{x^2}}}{{(x - 1)(x - 2)(x - 3)}} = \frac{1}{{x - 1}} - \frac{8}{{x - 2}} + \frac{9}{{x - 3}}$$
Question 2. Resolve $\frac{{3{x^3} - 4{x^2} - 6x - 15}}{{{x^2} - 2x - 3}}$ into partial fractions.
Solution: The above equation is an improper fraction because the denominator is a 2 degree polynomial where as the numerator is a 3 degree polynomial.
So, the improper fraction can be expressed as the sum of partial fraction by actual division,
When ${\left( {3{x^3} - 4{x^2} - 6x - 15} \right)}$ is divided by ${\left( {{x^2} - 2x - 3} \right)}$ we get,
As, $\left( {Divedend = Divisor \times Quotient + \operatorname{Re} mainder} \right)$. So, $$ \Rightarrow 3{x^3} - 4{x^2} - 6x - 15 = (3x + 2)\left( {{x^2} - 2x - 3} \right) + (7x - 9)$$ or $$\begin{equation} \begin{aligned} \Rightarrow \frac{{\left( {3{x^3} - 4{x^2} - 6x - 15} \right)}}{{\left( {{x^2} - 2x - 3} \right)}} = \frac{{(3x + 2)\left( {{x^2} - 2x - 3} \right) + (7x - 9)}}{{{x^2} - 2x - 3}} \\ \Rightarrow 3x + 2 + \frac{{7x - 9}}{{{x^2} - 2x - 3}}\quad ...(i) \\\end{aligned} \end{equation} $$ Now, we will express $\left( {\frac{{7x - 9}}{{{x^2} - 2x - 3}}} \right)$ as the sum of partial functions.
Factorising the denominator as, $${x^2} - 2x - 3 = (x + 1)(x - 3)$$
As we know, $\left( {\frac{{7x - 9}}{{{x^2} - 2x - 3}}} \right)$ is a proper function. So, it can be expressed the sum of partial function as,$$\begin{equation} \begin{aligned} \Rightarrow \frac{{7x - 9}}{{(x + 1)(x - 3)}} = \frac{A}{{x + 1}} + \frac{B}{{x - 3}} \\ \Rightarrow \frac{{A(x - 3) + B(x + 1)}}{{(x + 1)(x - 3)}} = \frac{{7x - 9}}{{(x + 1)(x - 3)}} \\\end{aligned} \end{equation} $$ Equating the numerator of LHS and RHS we get, $$ \Rightarrow A(x - 3) + B(x + 1) = 7x - 9$$
For $(x-3)=0$ or $x=3$, $$\begin{equation} \begin{aligned} \Rightarrow A(3 - 3) + B(3 + 1) = 7(3) - 9 \\ \Rightarrow 4B = 12 \\ \Rightarrow B = 3 \\\end{aligned} \end{equation} $$
For $(x+1)=0$ or $x=-1$, $$\begin{equation} \begin{aligned} \Rightarrow A(-1 - 3) + B(-1 + 1) = 7(-1) - 9 \\ \Rightarrow -4A = -16 \\ \Rightarrow A= 4 \\\end{aligned} \end{equation} $$
Therefore, $$ \Rightarrow \frac{{7x - 9}}{{(x + 1)(x - 3)}} = \frac{4}{{x + 1}} + \frac{3}{{x - 3}}\quad ...(ii)$$
From equation $(i)$ and $(ii)$ we get, $$\frac{{3{x^3} - 4{x^2} - 6x - 15}}{{{x^2} - 2x - 3}} = 3x + 2 + \frac{4}{{x + 1}} + \frac{3}{{x - 3}}$$
