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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 41

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

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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 \(\left(x - 2\right)\left(x^{2} + 2x + 4\right)\).
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: \(\frac{4x^{2} + 3x + 14}{(x - 2)(x^{2} + 2x + 4)} = \frac{A}{x - 2} + \frac{Bx + C}{x^{2} + 2x + 4}\).
Multiply both sides of the equation by the denominator \((x - 2)(x^{2} + 2x + 4)\) to clear the fractions. This gives: \(4x^{2} + 3x + 14 = A(x^{2} + 2x + 4) + (Bx + C)(x - 2)\).
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: \(Ax^{2} + 2Ax + 4A + Bx^{2} - 2Bx + Cx - 2C\).
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 \(C\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
4:07
Decomposition of Functions

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.
추천 영상:
07:30
Introduction to Factoring Polynomials

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.
추천 영상:
07:11
Example 1