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
2.2 When the denominator has 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, and polynomial $q(x)$ has repeated linear factors.
Mathematically, $q(x) = {(x - a)^n}$, In this case fraction $\frac{{p(x)}}{{q(x)}}$ is expressed as the sum of partial as shown below. $$\frac{{p(x)}}{{q(x)}} = \frac{{p(x)}}{{{{(x - a)}^n}}} = \frac{{{A_1}}}{{(x - a)}} + \frac{{{A_2}}}{{{{(x - a)}^2}}} + ... + \frac{{{A_{n - 1}}}}{{{{(x - a)}^{n - 1}}}} + \frac{{{A_n}}}{{{{(x - a)}^n}}}$$ where, $A_1$, $A_2$ ... $A_n$ are the constants which can be determined by equating the LHS & RHS.
Question 3. Resolve $\left( {\frac{{5{x^2} + 12x + 4}}{{{{(x + 1)}^3}}}} \right)$ into partial fractions.
Solution: As we know that the above fraction has repeated linear factors. So, it can be expressed as the sum of partial fraction as below. $$\begin{equation} \begin{aligned} \Rightarrow \left( {\frac{{5{x^2} + 12x + 4}}{{{{(x + 1)}^3}}}} \right) = \frac{A}{{(x + 1)}} + \frac{B}{{{{(x + 1)}^2}}} + \frac{C}{{{{(x + 1)}^3}}} \\ \Rightarrow \frac{{A{{(x + 1)}^2} + B(x + 1) + C}}{{{{(x + 1)}^3}}} = \left( {\frac{{5{x^2} + 12x + 4}}{{{{(x + 1)}^3}}}} \right) \\\end{aligned} \end{equation} $$ Equating the numerators we get, $$ \Rightarrow A{(x + 1)^2} + B(x + 1) + C = 5{x^2} + 12x + 4$$ For $(x+1)=0$ or $x=-1$, $$ \Rightarrow C = - 3$$ For $x=0$, $$\begin{equation} \begin{aligned} \Rightarrow A + B + C = 4 \\ \Rightarrow A + B = 7\quad ...(i)\quad (As,\;C = - 3) \\\end{aligned} \end{equation} $$ For $x=1$, $$\begin{equation} \begin{aligned} \Rightarrow 4A + 2B + C = 21 \\ \Rightarrow 4A + 2B = 24\quad \quad (As,\;C = - 3) \\ \Rightarrow 2A + B = 12\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$ \Rightarrow A = 5\quad \& \quad B = 2$$ Therefore, $$ \Rightarrow \left( {\frac{{5{x^2} + 12x + 4}}{{{{(x + 1)}^3}}}} \right) = \frac{5}{{(x + 1)}} + \frac{3}{{{{(x + 1)}^2}}} - \frac{3}{{{{(x + 1)}^3}}}$$
