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Magazine Chapter 6: Problem 22
Find the partial fraction decomposition for each rational expression. $$\frac{3}{x(x+1)\left(x^{2}+1\right)}$$
Short Answer
Expert verified
\(\frac{-1}{x} + \frac{-1}{x+1} + \frac{x+3}{x^2+1}\)
Step by step solution
01
Identify the Factorization
The first step is to identify the denominator's factorization: \[ x(x+1)(x^2+1) \]It is already factored into linear factors \(x\) and \((x+1)\), and an irreducible quadratic factor \((x^2+1)\).
02
Set Up Partial Fraction Decomposition
Write the expression as a sum of partial fractions. For linear factors, use a constant numerator. For quadratic factors, use a linear numerator. Thus:\[\frac{3}{x(x+1)(x^{2}+1)} = \frac{A}{x} + \frac{B}{x+1} + \frac{Cx+D}{x^{2}+1}\]
03
Clear the Denominator
Multiply both sides by the common denominator \(x(x+1)(x^2+1)\) to eliminate the fractions:\[ 3 = A(x+1)(x^2+1) + Bx(x^2+1) + (Cx+D)x(x+1) \]
04
Expand and Collect Terms
Expand the right side and collect like terms:\[3 = A(x^3 + x + x^2 + 1) + B(x^3 + x) + (Cx^2 + Cx + Dx + D)(x+1)\]Further expand and arrange terms by powers of \(x\).
05
Compare Coefficients
Set the coefficients of like terms equal on both sides of the equation. This gives a system of equations:\[Ax^3 + Ax^2 + Bx^3 + Ax + Bx + Cx^3 + Cx^2 + Dx + D = 3\]Equate coefficients for each power of \(x\):- Coefficient of \(x^3\): \(A + B + C = 0\)- Coefficient of \(x^2\): \(A + C = 0\)- Coefficient of \(x\): \(A + B + D = 0\)- Constant term: \(D = 3\)
06
Solve the System of Equations
Solve the system of equations derived in the previous step:1. From \(D = 3\), we have: \(D = 3\).2. Substitute \(D\) in the third equation: \(A + B + 3 = 0 \Rightarrow A + B = -3\).3. From the second equation: \(A + C = 0 \Rightarrow C = -A\).4. The system of equations becomes: - \(A + B = -3\) - \(A + B - A = 0 \Rightarrow B = A\)5. Solving, if \(A = -1\), then \(B = -1\) and \(C = 1\).
07
Write the Final Decomposition
Substitute \(A, B, C, D\) back into the partial fractions:\[\frac{3}{x(x+1)(x^{2}+1)} = \frac{-1}{x} + \frac{-1}{x+1} + \frac{x+3}{x^{2}+1}\]Thus, this is the partial fraction decomposition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorization
Factorization is a key step in partial fraction decomposition. It involves breaking down a complex expression into simpler elements. In the denominator of a rational expression, factorization helps identify how the expression can be decomposed. For example, with the expression \(x(x+1)(x^2+1)\):
- We identify it as a product of two linear factors \(x\) and \((x+1)\),
- and one irreducible quadratic factor \((x^2+1)\).
Linear Factors
Linear factors are expressions of the form \(ax + b\). In partial fraction decomposition, each linear factor in the denominator corresponds to a fraction with a constant numerator. This is because dividing a polynomial by a linear term can be expressed simply.
- For example, if the denominator includes \(x\) and \(x+1\), each of these factors will lead to separate terms with constants like \(A\) and \(B\), respectively.
- This results in terms such as \(\frac{A}{x}\) and \(\frac{B}{x+1}\).
Irreducible Quadratic Factors
Irreducible quadratic factors, like \(x^2 + 1\), cannot be factored further using real numbers. In partial fraction decomposition, each irreducible quadratic factor in the denominator demands a numerator of a linear form \(Cx + D\).
- Such terms are more complex than those for linear factors, since the numerator itself needs to account for higher degree polynomials.
- For instance, the factor \(x^2 + 1\) in our expression leads to a term \(\frac{Cx + D}{x^2+1}\).
System of Equations
A crucial part of the partial fraction decomposition process is solving the system of equations, which arises from equating the original and expanded expressions. The expression is cleared of fractions by multiplying through by the common denominator, allowing us to match coefficients of powers of \(x\) across both sides:
- Each coefficient comparison results in an equation.
- For example, in this situation, equating coefficients created equations such as \(A + B + C = 0\) and \(D = 3\).
- Solving these equations, typically a system if two or more equations, helps in determining the values of unknowns like \(A, B, C,\) and \(D\).
