Write the partial fraction decomposition of each rational expression. (4x²+3x+14)/(x³ - 8)

Verified step by step guidance

1. Recognize that the denominator is a difference of cubes: \(x^{3} - 8\) can be factored using the formula \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\). Here, \(a = x\) and \(b = 2\), so factor the denominator as \(\left(x - 2\right)\left(x^{2} + 2x + 4\right)\).
2. Set up the partial fraction decomposition form. Since \(x - 2\) is a linear factor, assign a constant numerator \(A\) over it. Since \(x^{2} + 2x + 4\) is an irreducible quadratic factor, assign a linear numerator \(Bx + C\) over it. So, write: \(\frac{4x^{2} + 3x + 14}{(x - 2)(x^{2} + 2x + 4)} = \frac{A}{x - 2} + \frac{Bx + C}{x^{2} + 2x + 4}\).
3. Multiply both sides of the equation by the denominator \((x - 2)(x^{2} + 2x + 4)\) to clear the fractions. This gives: \(4x^{2} + 3x + 14 = A(x^{2} + 2x + 4) + (Bx + C)(x - 2)\).
4. Expand the right-hand side by distributing \(A\) and then expanding \((Bx + C)(x - 2)\). Combine like terms to express the right side as a polynomial in standard form: \(Ax^{2} + 2Ax + 4A + Bx^{2} - 2Bx + Cx - 2C\).
5. Group like terms by powers of \(x\) and equate the coefficients of corresponding powers of \(x\) on both sides. This will give a system of equations to solve for \(A\), \(B\), and \(C\).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions with denominators that are factors of the original denominator. This technique simplifies integration and other algebraic operations by breaking down complex rational expressions.

Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of its factors. For the denominator x³ - 8, recognizing it as a difference of cubes allows factoring into (x - 2)(x² + 2x + 4), which is essential for setting up the partial fractions correctly.

Setting Up Partial Fractions for Irreducible Quadratics

When the denominator includes irreducible quadratic factors like x² + 2x + 4, the corresponding partial fraction has a linear numerator (Ax + B). This ensures the decomposition accounts for all possible numerators and allows solving for unknown coefficients.