Problem 39 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{x^{3}-4 x^{2}+9 x-5}{\left(x^{2}-2 x+3\right)^{2}}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is: \( \frac{-5/3}{x^{2}-2x+3} - \frac{2x/3 + 5/6}{(x^{2}-2 x+3)^2} \).

Step 1: Identify the form of the rational expression

The given expression can be represented as \( \frac{P(x)}{Q(x)^n} \) where P(x) is a polynomial of degree 3 and Q(x) is a polynomial of degree 2. Given that n=2, we can realize that Q(x) is a quadratic and is not factorizable. Therefore, we deduce that the expression can be represented as a sum of two terms of the form \( \frac{A}{Q(x)} + \frac{Bx+C}{Q(x)^2} \).

Step 2: Setup the equation and simplify

We know that \( \frac{x^{3}-4 x^{2}+9 x-5}{(x^{2}-2 x+3)^{2}} = \frac{A}{x^{2}-2 x+3} + \frac{Bx+C}{(x^{2}-2 x+3)^2} \). Multiply through by \( (x^{2}-2 x+3)^2 \), we get: \( x^{3}-4 x^{2}+9 x-5 = A*(x^{2}-2 x+3) + B*x +C \). This equation has to hold for all values of x. Therefore you can now choose convenient values of x and find A, B, and C by solving the resulting equations.

Step 3: Solve for A, B and C

Choosing x=0 simplifies the equation and gives A= -5/3. Then, differentiating both sides of the equation twice and solving for x=0 yields B= -2/3 and C= 5/6

Step 4: Substitute A, B and C back into partial fraction decomposition

Substituting A=-5/3, B=-2/3 and C=5/6 into \( \frac{A}{x^{2}-2 x+3} + \frac{Bx+C}{(x^{2}-2 x+3)^2} \) yields the partial fraction decomposition: \( \frac{-5/3}{x^{2}-2x+3} - \frac{2x/3 + 5/6}{(x^{2}-2 x+3)^2} \)