Problem 84
Question
Find the partial fraction decomposition: \(\frac{7 x^{2}-5 x+30}{x^{3}-8}\)
Step-by-Step Solution
Verified Answer
\( \frac{4}{x - 2} + \frac{3x - 7}{x^2 + 2x + 4} \)
1Step 1: Factor the denominator
Recognize that the denominator can be factored as a difference of cubes: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4)\]
2Step 2: Set up the partial fraction decomposition
Express the fraction as a sum of partial fractions. Given the factors of the denominator, the decomposition is: \[ \frac{7x^2 - 5x + 30}{x^3 - 8} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + 2x + 4} \]
3Step 3: Multiply through by the denominator
Remove the denominators by multiplying both sides by \( x^3 - 8 \): \[ 7x^2 - 5x + 30 = A(x^2 + 2x + 4) + (Bx + C)(x - 2) \]
4Step 4: Expand and collect like terms
Distribute and combine like terms on the right side: \[ 7x^2 - 5x + 30 = A(x^2 + 2x + 4) + Bx(x - 2) + C(x - 2) \] \[ = Ax^2 + 2Ax + 4A + Bx^2 - 2Bx + Cx - 2C \] \[ 7x^2 - 5x + 30 = (A + B)x^2 + (2A - 2B + C)x + (4A - 2C) \]
5Step 5: Set up and solve the system of equations
Match the coefficients from both sides to set up a system of equations: 1. \( A + B = 7 \) 2. \( 2A - 2B + C = -5 \) 3. \( 4A - 2C = 30 \) Solve this system to find \( A, B, \) and \( C \).
6Step 6: Solve for constants \( A, B, \) and \( C \)
First solve equation 3 for A: \[ 4A - 2C = 30 \implies 2A - C = 15 \] So: \[ C = 2A - 15 \] Substituting \( C \) into equation 2: \[ 2A - 2B + (2A - 15) = -5 \implies 4A - 2B - 15 = -5 \implies 4A - 2B = 10 \] Therefore: \[ 2A - B = 5 \] Solve for B from the system: \[ A + B = 7 \implies B = 7 - A \] Substituting \(B\) in 2A - B equation: \[ 2A - (7 - A) = 5 \implies 3A - 7= 5 \implies 3A = 12 \implies A = 4 \] Using \( A \) value to find \( B \) and \( C \): \[ B = 7 - 4 = 3 \] \[ C = 2(4) - 15 = 8 - 15 = -7 \] Therefore, \( A = 4, B = 3, C = -7 \).
7Step 7: Write the final decomposition
Insert the constants back into the partial fractions: \[ \frac{7x^2 - 5x + 30}{x^3 - 8} = \frac{4}{x - 2} + \frac{3x - 7}{x^2 + 2x + 4} \]
Key Concepts
Partial FractionsDifference of CubesSystem of EquationsPolynomial Factorization
Partial Fractions
Partial fraction decomposition is a technique used to break down a complex rational expression into simpler fractions that are easier to handle. This helps in integrating the expression, solving equations, or simplifying complex algebraic expressions.
To decompose a fraction, we express the original fraction as a sum of simpler fractions, determined by the factors in the denominator.
For instance, in the exercise, the fraction \(\frac{7x^2 - 5x + 30}{x^3 - 8}\) is decomposed into \(\frac{4}{x - 2} + \frac{3x - 7}{x^2 + 2x + 4}\). By breaking it down, we simplify the process of working with the fraction.
To decompose a fraction, we express the original fraction as a sum of simpler fractions, determined by the factors in the denominator.
For instance, in the exercise, the fraction \(\frac{7x^2 - 5x + 30}{x^3 - 8}\) is decomposed into \(\frac{4}{x - 2} + \frac{3x - 7}{x^2 + 2x + 4}\). By breaking it down, we simplify the process of working with the fraction.
Difference of Cubes
Recognizing and factoring special polynomial forms is crucial. The 'difference of cubes' formula is one such form:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
In our exercise, the denominator is \(x^3 - 8\), which is a difference of cubes where \(x = a\) and \(2 = b\).
Thus, \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4)onumber \]
This factorization helps in setting up the partial fraction decomposition accurately.
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
In our exercise, the denominator is \(x^3 - 8\), which is a difference of cubes where \(x = a\) and \(2 = b\).
Thus, \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4)onumber \]
This factorization helps in setting up the partial fraction decomposition accurately.
System of Equations
Once the partial fractions are set up, you'll need to solve for unknown constants. This involves setting up a system of equations based on the equalities formed by matching coefficients.
For the given expression, after expanding and matching coefficients, we get:
For the given expression, after expanding and matching coefficients, we get:
- \( A + B = 7 \)
- \( 2A - 2B + C = -5 \)
- \( 4A - 2C = 30 \)
Solving these equations simultaneously gives the values for \( A \), \( B \), and \( C \). Correctly solving these systems is essential to finding the correct partial fractions.
Polynomial Factorization
Factorization of polynomials is a key step in partial fraction decomposition. It simplifies the problem by breaking down complex expressions.
In our problem, we factor the cubic polynomial in the denominator: \( x^3 - 8 \), using the difference of cubes formula.
Factoring results in two simpler factors: \( x - 2 \) and \( x^2 + 2x + 4 \).
This allows us to decompose the fraction into simpler, more manageable parts: \( \frac{4}{x - 2} \) and \( \frac{3x - 7}{x^2 + 2x + 4} \).
Understanding and recognizing factorization patterns help solve complex rational expressions efficiently.
In our problem, we factor the cubic polynomial in the denominator: \( x^3 - 8 \), using the difference of cubes formula.
Factoring results in two simpler factors: \( x - 2 \) and \( x^2 + 2x + 4 \).
This allows us to decompose the fraction into simpler, more manageable parts: \( \frac{4}{x - 2} \) and \( \frac{3x - 7}{x^2 + 2x + 4} \).
Understanding and recognizing factorization patterns help solve complex rational expressions efficiently.
Other exercises in this chapter
Problem 82
Find the 85 th term of the sequence \(5,12,19,26, \ldots\)
View solution Problem 83
Find the area bounded by the graphs of \(y=\frac{3}{5} x+\frac{12}{5}, y=-x+4,\) and \(y=-\sqrt{16-x^{2}}\)
View solution Problem 80
Simplify: \(\sqrt{108}-\sqrt{147}+\sqrt{363}\)
View solution