Problem 27 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{x^{2}+1}{x^{3}+x^{2}} $$

Step-by-Step Solution

Answer

\(\frac{-1}{x} + \frac{1}{x^2} + \frac{2}{x+1}\).

Step 1: Factor the Denominator

Start by factoring the denominator, \(x^3 + x^2\). We can factor out the greatest common factor, which is \(x^2\). So, \(x^3 + x^2 = x^2(x + 1)\).

Step 2: Set Up the Partial Fractions

Write the rational function as a sum of partial fractions based on the factors in the denominator. After factoring, we have \(x^2(x + 1)\). So, the partial fractions will be \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\).

Step 3: Multiply Through by the Denominator

To eliminate the fractions, multiply both sides by the common denominator \(x^2(x + 1)\). This yields: \[ x^2 + 1 = A(x)(x+1) + B(x + 1) + C(x^2)\].

Step 4: Expand and Rearrange the Right-hand Side

Expand the right-hand side:\[ A(x^2 + x) + Bx + B + Cx^2 = (A + C)x^2 + Ax + Bx + B\]. Rearranging gives us: \[ (A + C)x^2 + (A + B)x + B\].

Step 5: Equate Coefficients

Equate the coefficients from both sides of the equation:- For \(x^2\): \(A + C = 1\)- For \(x\): \(A + B = 0\)- Constant term: \(B = 1\)

Step 6: Solve the System of Equations

From \(B = 1\), substitute \(B\) into \(A + B = 0\) to get \(A + 1 = 0\) which implies \(A = -1\). Substitute \(A = -1\) into \(A + C = 1\) to get \(-1 + C = 1\), thus \(C = 2\).

Step 7: Write the Partial Fraction Decomposition

The partial fraction decomposition is:\[\frac{-1}{x} + \frac{1}{x^2} + \frac{2}{x+1}\].

Key Concepts

Rational Functions

Rational functions are a fascinating category of mathematical expressions. Simply put, they are fractions where the numerator and the denominator are both polynomials. In our task, we have a rational function given by \(\frac{x^{2}+1}{x^{3}+x^{2}}\). Here, \(x^{2} + 1\) is the numerator, and \(x^{3} + x^{2}\) represents the denominator.One important skill when working with rational functions is the ability to simplify them. This can often involve factoring the polynomials in the numerator and the denominator, and then potentially canceling out common factors. This leads to a clearer, simpler expression which we can analyze further.Understanding rational functions is crucial because they appear in various branches of mathematics, including calculus where they are often used to describe curves and surfaces. Their properties, such as asymptotes and intercepts, provide valuable information when graphing or applying these functions practically.

Factoring Polynomials

Factoring polynomials is a fundamental skill that is essential when dealing with rational functions. It involves breaking down a polynomial into simpler "factor" polynomials that, when multiplied together, give the original polynomial. Let's take a closer look using our problem's denominator, \(x^3 + x^2\).### Factoring Step-by-Step

• Identify Common Factors: In \(x^3 + x^2\), we can see that \(x^2\) is a common factor. By factoring it out, we obtain \(x^2(x + 1)\).
• Result: The expression is now factored into \(x^2\) and \((x + 1)\).

This factorization is crucial as it allows us to decompose the original rational function into partial fractions. This makes it easier to handle, especially in integration tasks later on. Practicing factoring polynomials enhances your problem-solving skills and helps you see the interconnectedness of algebraic expressions.

Systems of Equations

Solving systems of equations is a key aspect when performing partial fraction decomposition because you need to find specific coefficients for the partial fractions. In the given solution, we encounter a system of equations derived from equating the coefficients of polynomials.### The Process

• Set Up Equations: By matching the coefficients from the expanded expression \( (A + C)x^2 + (A + B)x + B \) to the original equation \(x^2 + 1\), we establish relationships:
• For \(x^2\): \(A + C = 1\)
• For \(x\): \(A + B = 0\)
• For the constant term: \(B = 1\)

### Solving the System

• Start by solving the simplest equation, \(B = 1\).
• Substituting \(B = 1\) into \(A + B = 0\) gives \(A = -1\).
• Use \(A = -1\) in \(A + C = 1\) to find \(C = 2\).

Solving these equations gives us the required coefficients \(A, B,\) and \(C\) for the partial fraction decomposition. Understanding how to solve systems of equations effectively is a vital skill, enabling you to decompose complex expressions into more manageable parts.