Problem 35 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the rational expression is \(\frac{1}{x} + \frac{5x - 2}{x^{2}+x+1}\)

Step 1: Factorize the Denominator of the Rational Expression

Begin by factorizing the denominator of the given expression. Here, \(x^{3}+x^{2}+x+1 = x(x^{2}+x+1+1)\). However, \(x^{2}+x+1\) isn't factorisable using real numbers. So the factored form of the denominator is \(x(x^{2}+x+1)\).

Step 2: Decompose into Partial Fraction

The next step is to decompose the fraction. We rewrite the rational function as the sum of partial fractions. Since the denominator was previously factored as \(x(x^{2}+x+1)\), the decomposition of the given rational expression can be expressed as: \(\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1} = \frac{A}{x} + \frac{Bx+C}{x^{2}+x+1}\).

Step 3: Equate Coefficients and Solve

Clear the fraction by multiplying both sides by \(x^{3}+x^{2}+x+1\), resulting in: \(6x^{2}-x+1 = A(x^{2}+x+1) + (Bx+C)x\). Compare the coefficients on both sides and solve for A, B, and C. Coefficients of \(x^{2}\) on left side: 6; on right side: A+B. So, A + B = 6. Coefficients of \(x\) on left side: -1; on right side: A + C. So, A + C = -1. Constants on left side: 1; on right side: A. So, A = 1. Substitute A = 1 into the other equations to get B = 5 and C = -2. The partial fraction decomposition of the given rational expression is therefore given by: \(\frac{1}{x} + \frac{5x - 2}{x^{2}+x+1}\)