Problem 40 - College Algebra

Question

Find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. \(\frac{x^{2}+2 x+40}{x^{3}-125}\)

Step-by-Step Solution

Answer

\(\frac{2}{x-5} - \frac{x + 13}{x^2+5x+25}\)

Step 1: Identify the Denominator Factorization

First, recognize the factorization required for the denominator. The denominator is of the form \(x^3-125\), which can be factored using the difference of cubes formula \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Hence, it becomes \((x-5)(x^2+5x+25)\).

Step 2: Set Up the Partial Fractions

Express the original fraction as a sum of partial fractions. Since \(x-5\) is a linear factor and \(x^2+5x+25\) is an irreducible quadratic factor, the decomposition will be: \(\frac{A}{x-5} + \frac{Bx+C}{x^2+5x+25}\).

Step 3: Multiply Through by the Common Denominator

Eliminate the denominators by multiplying through by the common denominator, \((x-5)(x^2+5x+25)\), to obtain: \(x^2 + 2x + 40 = A(x^2 + 5x + 25) + (Bx + C)(x - 5)\).

Step 4: Expand and Simplify

Expand the right-hand side of the equation: \(A(x^2 + 5x + 25) = Ax^2 + 5Ax + 25A\) and \((Bx + C)(x - 5) = Bx^2 - 5Bx + Cx - 5C\). Combine these to get \(Ax^2 + 5Ax + 25A + Bx^2 - 5Bx + Cx - 5C\).

Step 5: Collect Like Terms

Collect terms in the expanded expression: \((A + B)x^2 + (5A - 5B + C)x + (25A - 5C)\).

Step 6: Match Coefficients

Compare coefficients with the left-hand side \(x^2 + 2x + 40\) to get the equations: \(A + B = 1\), \(5A - 5B + C = 2\), and \(25A - 5C = 40\).

Step 7: Solve the System of Equations

Solve the system of equations obtained: From \(A + B = 1\), solve for \(B = 1 - A\). Substitute into the other equations, \(5A - 5(1 - A) + C = 2\) simplifies to \(10A + C = 7\), and \(25A - 5C = 40\). Solving these gives \(A = 2\), \(B = -1\), and \(C = -13\).

Step 8: Write the Partial Fraction Decomposition

Substitute the values of \(A, B,\) and \(C\) back into the partial fraction decomposition: \(\frac{2}{x-5} - \frac{x + 13}{x^2+5x+25}\).

Key Concepts

Irreducible Quadratic Factors

In mathematics, an irreducible quadratic factor is a polynomial of the second degree that cannot be factored further into real or simpler linear factors. This is crucial for partial fraction decomposition. Partial fractions represent a rational expression as a sum of simpler fractions, making integration easier. When a denominator includes an irreducible quadratic expression, it contributes a specific term to the decomposition:

• A term in the form \(\frac{Bx + C}{ax^2 + bx + c}\), indicating the presence of an irreducible quadratic factor.

Being irreducible means the quadratic expression has no real roots (indicating a positive discriminant or simply being complex).
This factorization assists in simplifying computations involving integration or algebraic manipulation, transforming complex rational expressions into manageable pieces.

Difference of Cubes

The difference of cubes is a handy algebraic identity for expressing one cubic term minus another. Specifically, the difference of cubes formula simplifies \(a^3 - b^3\) into the product of linear and quadratic terms, as follows:

• \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

This formula is essential when working with cubic polynomials.
In the problem context, it simplifies \(x^3 - 125\) into \((x-5)(x^2 + 5x + 25)\), making it possible to use partial fraction decomposition.
This decomposition allows us to break down complex rational expressions into simpler components for further analysis or integration.

System of Equations

A system of equations is a collection of equations with multiple variables that are solved simultaneously. In partial fraction decomposition, once you have set up the equation by equating the original polynomial numerator to the expanded form of your partial fractions' sum, a system of equations arises from aligning coefficients:

• In this scenario, after expanding \(A(x^2 + 5x + 25) + (Bx + C)(x-5)\), you collect like terms to create individual equations by matching the coefficients of \(x^2, x,\) and constant terms.
• For instance, the equations extracted from: \((A + B)x^2 + (5A - 5B + C)x + (25A - 5C)\) matched with \(x^2 + 2x + 40\).

The solution involves using algebraic methods to find the values of unknowns like \(A, B,\) and \(C\) that satisfy all the equations simultaneously.
This analytical step ensures the correct coefficients for each term of the partial fraction decomposition.

Algebraic Expressions

Algebraic expressions form the basis of algebra and include constants, variables, and operations such as addition, subtraction, multiplication, and division. Understanding how these expressions work is key to mastering calculus problems like partial fraction decomposition. Here are a few attributes:

• An algebraic expression may represent part or all of a particular polynomial or rational expression needed in partial fraction decomposition.
• In the given exercise, algebraic expression manipulations involve factoring, expanding, and matching coefficients.

Recognizing that expressing the numerator \(x^2 + 2x + 40\) in terms of decomposed parts involves turning simpler algebraic expressions into a whole.
This highlights how different algebraic components come together to solve more complex mathematical problems efficiently.