Problem 9 - Precalculus

Question

In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{11 x^{2}-5 x-10}{5 x^{3}-5 x^{2}} $$

Step-by-Step Solution

Answer

\(\frac{15}{x} + \frac{10}{x^2} - \frac{4}{5(x-1)}\)

Step 1: Express the Denominator in Factored Form

First, factor the denominator of the given rational expression. The denominator is \(5x^3 - 5x^2\). We can factor out \(5x^2\), which gives us: \(5x^2(x-1)\).

Step 2: Set Up the Partial Fraction Decomposition

Since the denominator is \(5x^2(x-1)\), the partial fraction decomposition will take the form: \[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} \]

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \(5x^2(x-1)\) to clear the fractions: \[ 11x^2 - 5x - 10 = A x(x-1) + B(x-1) + C 5x^2 \]

Step 4: Expand and Equate Coefficients

Expand the right-hand side and equate coefficients with the corresponding terms on the left-hand side. The equation becomes: \[ 11x^2 - 5x - 10 = Ax^2 - Ax + Bx - B + C5x^2 \] Simplifying, group and compare coefficients for each power of \(x\).

Step 5: Set Up System of Equations

You will have three equations: 1. Coefficient of \(x^2\): \(11 = A + 5C\)2. Coefficient of \(x\): \(-5 = B - A\)3. Constant: \(-10 = -B\)

Step 6: Solve the System of Equations

Solve for \(B\) in equation 3: \(B = 10\). Substitute \(B\) into equation 2: \(-5 = 10 - A\) implies \(A = 15\). Substitute \(B = 10\) and \(A = 15\) into equation 1 to find \(C\): \(11 = 15 + 5C\), leading to \(C = -\frac{4}{5}\).

Step 7: Write the Partial Fraction Decomposition

The partial fraction decomposition based on the values of \(A\), \(B\), and \(C\) is: \[ \frac{15}{x} + \frac{10}{x^2} - \frac{4}{5(x-1)} \]

Key Concepts

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Just like regular fractions, rational expressions can be simplified, factored, and decomposed into simpler fragments. This is especially useful in calculus and algebra to make complex problems easier to solve. Consider a rational expression as a fraction

• The numerator could be any polynomial, represented generally as a sum of terms such as \( ax^2 + bx + c \).
• The denominator is also a polynomial and must not equal zero to avoid division by zero.

Partial fraction decomposition is a method used to break down a complex rational expression into simpler fractions with linear or quadratic factors in the denominators. By doing this, solving equations or integrating becomes more straightforward.

Factoring Polynomials

Factoring is the process of breaking down a complex expression into a product of simpler expressions. For instance, take the polynomial in our denominator, \(5x^3 - 5x^2\). Here, the first step to simplifying is to factor out the greatest common factor (GCF):

• The GCF in this case is \(5x^2\) since it is the highest degree term that appears in each component of the polynomial.
• When you factor \(5x^2\) out, the simplified expression becomes \(5x^2(x-1)\).

This process of factoring enables us to clearly see the individual linear factors involved in the expression. These factors form the basis for setting up our partial fraction decomposition.

System of Equations

In partial fraction decomposition, a system of equations helps us find the coefficients of the residues in simpler fractions. Once we factor the denominator and express the partial fraction form as \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}\), we derive a set of equations by equating coefficients from both sides. Here's how it works:

• Once the denominator is cleared, expand the right-hand side and compare it term-by-term with the left-hand side.
• This allows you to set up equations for the coefficients of each power of \(x\).

In solving the exercise, the system of equations becomes:

• \(11 = A + 5C\)
• \(-5 = B - A\)
• \(-10 = -B\)

This method systematically determines the unknown coefficients \(A\), \(B\), and \(C\).

Comparing Coefficients

Comparing coefficients is a technique used to determine the values of unknown variables in polynomials or rational expressions. It involves aligning expressions and equating corresponding coefficients of like terms from both sides of an equation. Here’s how it’s applied:

• Begin by expanding both the original polynomial and the polynomial formed from the partial fractions.
• Group like terms—those involving the same power of \(x\).

In our example exercise, the left-hand side expression is \(11x^2 - 5x - 10\). Equate it to the expanded right-hand side, which simplifies to:

• \(Ax^2 - Ax + Bx - B + C5x^2\).

By matching each coefficient of powers of \(x\) on both sides, you get equations like \(11 = A + 5C\), etc. Solving these equations provides the values for \(A\), \(B\), and \(C\), completing the decomposition process.