Perform the partial fraction decomposition of the following rational function: \(\frac{3x^2-2x+1}{x^3-x^2}\).
Answer
Step 7: Substituting the values of A, B, and C into the partial fractions, we get: \(\frac{3x^2-2x+1}{x^3-x^2} = \frac{1}{x} + \frac{1}{x^2} + \frac{2}{x-1}\).
Step 1 ：Step 1: To perform the partial fraction decomposition, we first factorize the denominator. Here, \(x^3-x^2\) can be factorized to \(x^2(x-1)\).
Step 2 ：Step 2: Once we have the factors, we can write the rational function as a sum of simpler fractions. We get: \(\frac{3x^2-2x+1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\) where A, B, and C are constants to be determined.
Step 3 ：Step 3: Clearing the fractions, we get: \(3x^2-2x+1 = A(x^2)(x-1) + B(x)(x-1) + C(x^2)\).
Step 4 ：Step 4: Expanding and collecting like terms, we get: \(3x^2-2x+1 = (A+C)x^2 + (-A+B)x + A\).
Step 5 ：Step 5: Comparing coefficients of the powers of x, we can set up three equations: \(A+C=3\), \(-A+B=-2\), and \(A=1\).
Step 6 ：Step 6: Solving these equations, we find: \(A=1\), \(B=1\), and \(C=2\).
Step 7 ：Step 7: Substituting the values of A, B, and C into the partial fractions, we get: \(\frac{3x^2-2x+1}{x^3-x^2} = \frac{1}{x} + \frac{1}{x^2} + \frac{2}{x-1}\).
