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Expand the quotients in Exercises \(1-8\) by partial fractions. $$\frac{x+4}{(x+1)^{2}}$$

Short Answer

Expert verified
\( \frac{x+4}{(x+1)^{2}} = \frac{1}{x+1} + \frac{3}{(x+1)^{2}} \)

Step by step solution

01

Recognize the Form

The function \( \frac{x+4}{(x+1)^2} \) is a rational expression with a repeated linear factor in the denominator, \( (x+1)^2 \). Therefore, its partial fraction decomposition will include two terms having \( (x+1) \) raised to different powers.
02

Set Up the Partial Fractions

The partial fraction decomposition for \( \frac{x+4}{(x+1)^2} \) is set up as follows: \[ \frac{x+4}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} \]where \( A \) and \( B \) are constants that need to be determined.
03

Clear the Denominator

Multiply both sides of the equation by \( (x+1)^2 \) to eliminate the fractions:\[ x+4 = A(x+1) + B \]
04

Expand and Combine Like Terms

Expand the equation:\[ x+4 = Ax + A + B \]This results in matching the coefficients of the polynomial on the left side with those on the right side.
05

Solve for Constants

Match the coefficients on both sides of the equation:- For the coefficient of \( x \), we have \( A = 1 \).- For the constant term, equate \( A + B = 4 \).Using \( A = 1 \), solve for \( B \):\[ 1 + B = 4 \Rightarrow B = 3 \].
06

Write the Partial Fraction Decomposition

After finding the values of \( A \) and \( B \), write down the decomposition:\[ \frac{x+4}{(x+1)^2} = \frac{1}{x+1} + \frac{3}{(x+1)^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Expressions
Rational expressions are fractions where both the numerator and the denominator are polynomials. They play a significant role in mathematics and can represent a wide range of functions. The key aspect of working with rational expressions is understanding how to manipulate them just like regular fractions. You can add, subtract, multiply, and divide them.

For example, consider the rational expression \( \frac{x+4}{(x+1)^2} \). The numerator, \( x+4 \), is a simple linear polynomial, while the denominator, \( (x+1)^2 \), is a polynomial raised to a power. This makes the expression a rational one.

When dealing with rational expressions, checking for domain restrictions is important. Here, the term \( (x+1)^2 \) cannot be zero. This means \( x eq -1 \) to keep the denominator non-zero and the expression valid.
Partial Fraction Decomposition
Partial fraction decomposition is a method used to express a complex rational expression as a sum of simpler fractions. This is particularly useful when integrating or simplifying expressions. It involves rewriting a rational expression like \( \frac{x+4}{(x+1)^2} \) into a sum of fractions that are easier to work with.

The process begins with identifying the form of the denominator. If it includes repeated or distinct factors, you will set up a corresponding expression with unknown coefficients. In our example, since the factor \( (x+1) \) is repeated, the setup becomes \( \frac{A}{x+1} + \frac{B}{(x+1)^2} \). This allows us to break down the complex expression into simpler parts.
  • Each term corresponds to a component of the denominator raised to different powers.
  • Solving this involves clearing the denominator by multiplying through, allowing you to match and solve for constants like \( A \) and \( B \) through equating coefficients.
This decomposition is especially helpful in calculus and higher-level algebra for simplifying integration and solving equations.
Repeated Linear Factors
When approaching partial fraction decomposition, encountering repeated linear factors in the denominator is common. A repeated linear factor, like \( (x+1)^2 \) in our example, indicates that the factor \( (x+1) \) is multiplied by itself.

To handle these, the partial fraction decomposition includes separate terms for each power of the factor. For \( (x+1)^2 \), the terms \( \frac{A}{x+1} \) and \( \frac{B}{(x+1)^2} \) are used. This is because the effects of the factor occur at multiple levels, requiring distinct expressions.
  • Each term's coefficient is determined by expanding and matching coefficients.
  • Repeated factors often lead to a series of simpler equations that can be solved linearly.
Understanding how to break down repeated linear factors into simpler expressions aids in problem-solving and makes computations more manageable, especially in integration where polynomial division might be otherwise required.

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