Chapter 11.5, Problem 39AYU

To determine

To find: The partial fraction decomposition of the given rational function.

Answer to Problem 39AYU

Solution:

$- \frac{3}{x - 1} - \frac{1}{{\left(x - 1\right)}^2} + \frac{4}{\left(x - 2\right)}$

Explanation of Solution

Given:

$\frac{x^2}{x^3 - 4x^2 + 5x - 2}$

Formula used:

a. If $Q(x) = (x - a_1)\dots \dots (x - a_n)$ then partial fraction decomposition of $\text{P}\left(\text{x}\right)/\text{Q}\left(\text{x}\right)$ takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a_1\right)} + \frac{A_2}{\left(x - a_2\right)} + \mathrm{\cdots }\frac{A_n}{\left(x - a_n\right)}$

where $A_1, A_2\dots A_n$ are to be determined.

b. If $Q$ has repeated linear factor of the form ${\left(x - a\right)}^n, \text{n} \ge 2$ an integer, then, in the partial fraction decomposition of $P\left(x\right)/Q\left(x\right)$, takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a\right)} + \frac{A_2}{{\left(x - a\right)}^2} + \cdots + \frac{A_n}{{\left(x - a\right)}^n}$

where the numbers $A_1, A_2\mathrm{...}{A}_n$ are to be determined.

Calculation:

The denominator of the rational function $x^3 - 4x^2 + 5x - 2\text{can}$ be factored as $\left(x - 1\right)\left(x - 1\right)\left(x - 2\right)$

The denominator has repeated linear factor x and non repeated linear factor $\left(\text{x} - 1\right)$. Therefore the partial fraction decomposition takes the form

$\begin{array}{l}\frac{x^2}{x^3 - 4x^2 + 5x - 2} = \frac{A}{x - 1} + \frac{B}{{\left(x - 1\right)}^2} + \frac{C}{\left(x - 2\right)}\hfill \\ \begin{array}{ll}\Rightarrow x^2\hfill & = A\left(x - 1\right)\left(x - 2\right) + B\left(x - 2\right) + C{\left(x - 1\right)}^2\hfill \\ \hfill & = A\left(x^2 + 2 + 3x\right) + B\left(x - 2\right) + C\left(x^2 + 1 - 2x\right)\hfill \end{array}\hfill \end{array}$   -----Eq (1)

Equating the co-efficients on both sides in Eq(1),

$x^2, \Rightarrow A + C = 1$   -----Eq(2)

$x, \Rightarrow 3A + B - 2C = 0$   -----Eq(3)

$constant \Rightarrow A - 2B + C = 0$   -----Eq(4)

Solving for $A,B, C$ from Eq(2) – Eq(4) we get

$A = - 3,B = - 1,C = 4,$

The partial fraction decomposition of $\frac{x^2}{x^3 - 4x^2 + 5x - 2}$ is $- \frac{3}{x - 1} - \frac{1}{{\left(x - 1\right)}^2} + \frac{4}{\left(x - 2\right)}$.