Problem 18 - College Algebra Essentials

Question

write the partial fraction decomposition of each rational expression. $$ \frac{4 x^{2}-5 x-15}{x(x+1)(x-5)} $$

Step-by-Step Solution

Answer

The partial fraction decomposition of the given rational expression is \[ \frac{-3}{x} - \frac{4}{x+1} + \frac{1}{x-5} \].

Step 1: Setup Partial Fractions

A key point to note is that our objective is to express the given expression in this form: \[ \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-5} \] where A, B, and C are constants that need to be found.

Step 2: Multiplication of Denominators

Next, multiply each of the fractions by the common denominator \(x(x+1)(x-5)\), this gives: \[ 4 x^{2}-5 x-15 = A(x+1)(x-5) + Bx(x-5) + Cx(x+1) \]. This allows for the solution of A, B and C by matching corresponding coefficients

Step 3: Solve for Constants

Setting \(x = 0\), yields \(A = -3\). With \(x = -1\), we get \(B = -4\), and by letting \(x = 5\), we find that \(C=1\).

Step 4: Substituting Constants

Finally, we substitute the constants A, B, and C back into the factual fractions to get the partial fraction decomposition. Hence our final answer is: \[ \frac{-3}{x} + \frac{-4}{x+1} + \frac{1}{x-5} \].

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomial expressions. When dealing with rational expressions, it is important to determine if the expression can be simplified or decomposed further. In partial fraction decomposition, this step is crucial to express a complicated fraction into simpler, addable terms.
Consider the given rational expression: \[ \frac{4x^2 - 5x - 15}{x(x+1)(x-5)} \].
The task is to break down this expression into partial fractions. This process involves rewriting the expression so that the denominator's polynomial is expressed as a product of its linear factors.

• Linear factors are polynomials of the form \(x - r\), where \(r\) is a constant.
• Each linear factor in the denominator will correspond to a separate term in the partial fraction decomposition.

Partial fraction decomposition is especially useful in calculus for integration purposes, as integrating partial fractions is often simpler than dealing with complex fractions.

Algebra

Algebra plays a significant role in partial fraction decomposition. We need to manage polynomial equations and solve for unknown variables, often called constants.
In the step-by-step solution, the rational expression is rewritten by assigning unknown constants \(A\), \(B\), and \(C\) to each term's numerator in the decomposition, corresponding to the roots of the polynomial in the denominator: \[ \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-5} \].
To solve for \(A\), \(B\), and \(C\):

• Multiply through by the denominator \(x(x+1)(x-5)\) to eliminate the fractions.
• This gives the equation: \(4x^2 - 5x - 15 = A(x+1)(x-5) + Bx(x-5) + Cx(x+1)\).
• The strategy here is to choose strategic values of \(x\) to simplify the terms and solve for the constants.

Algebraic manipulation is vital in isolating and determining the constants, forming the foundation of this technique.

Constants in Partial Fractions

Finding constants in partial fraction decomposition is a crucial step. With the equation set up, we strategically choose different values for \(x\) that simplify our equation. This approach isolates constants and simplifies calculations.
Let's recap how the constants \(A\), \(B\), and \(C\) are determined:

• For \(x = 0\), the terms simplify such that only the term with \(A\) remains, leading to \(A = -3\).
• For \(x = -1\), the equation simplifies to focus on \(B\), giving \(B = -4\).
• Finally, choosing \(x = 5\) isolates \(C\) with \(C = 1\).

This problem-solving method is simple yet effective for calculating the unknowns in partial fraction equations.
Once calculated, the constants are substituted back into the original partial fraction format: \[ \frac{-3}{x} + \frac{-4}{x+1} + \frac{1}{x-5} \].
By breaking the steps into manageable actions and applying basic algebraic principles, finding constants becomes straightforward, making the whole process less daunting.