Problem 7
Question
Write out the appropriate form of the partial fraction decomposition of the given rational expression. Do not evaluate the coefficients. $$ \frac{2 x^{3}-x}{\left(x^{2}+1\right)^{2}} $$
Step-by-Step Solution
Verified Answer
\( \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} \)
1Step 1: Identify the Denominator
To set up a partial fraction decomposition, we begin by examining the denominator of the rational expression. The given denominator is \((x^2 + 1)^2\). This indicates the presence of a repeated irreducible quadratic factor.
2Step 2: Set Up the Partial Fractions
Since the denominator contains \((x^2 + 1)\) repeated, we use partial fractions with the form: \[\frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2}\]. Here we assign linear terms over the irreducible quadratic factors.
Key Concepts
Rational ExpressionsDenominator IdentificationIrreducible Quadratic Factor
Rational Expressions
Rational expressions are like fractions but with polynomials on the top and bottom. They represent the ratio of two polynomial expressions just like a fraction represents a ratio of two numbers. Understanding these expressions is essential because they often occur in calculus and algebra.
A rational expression looks like this: \( \frac{p(x)}{q(x)} \), where \(p(x)\) and \(q(x)\) are polynomials and \(q(x) eq 0\). The denominator cannot be zero because dividing by zero is undefined in mathematics.
Rational expressions can be complex, with various terms in numerators and denominators. These can include constant terms, linear terms, or higher degree polynomials. Simplifying, adding, subtracting, multiplying, and dividing these expressions forms the basis for much of algebra and calculus.
When working with rational expressions, always consider the restrictions placed by the denominator. Excluding these values from the domain keeps our math correct and avoids undefined situations.
A rational expression looks like this: \( \frac{p(x)}{q(x)} \), where \(p(x)\) and \(q(x)\) are polynomials and \(q(x) eq 0\). The denominator cannot be zero because dividing by zero is undefined in mathematics.
Rational expressions can be complex, with various terms in numerators and denominators. These can include constant terms, linear terms, or higher degree polynomials. Simplifying, adding, subtracting, multiplying, and dividing these expressions forms the basis for much of algebra and calculus.
When working with rational expressions, always consider the restrictions placed by the denominator. Excluding these values from the domain keeps our math correct and avoids undefined situations.
Denominator Identification
Denominator identification is a crucial step when working with partial fraction decomposition. This involves carefully examining the denominator of a rational expression to determine its factors' structure. Accurate identification sets the foundation for setting up proper partial fractions.
The goal is to break down the denominator into a product of simpler, mutually exclusive factors. These factors can either be linear (e.g., \(x + a\)) or quadratic (e.g., \(x^2 + bx + c\)). Sometimes, these factors repeat in the denominator, which impacts how the corresponding partial fractions should be expressed.
In the provided example, the denominator \((x^2 + 1)^2\) indicates a repeated irreducible quadratic factor. Recognizing this allows us to structure the partial fraction decomposition correctly. For each quadratic factor such as \(x^2 + 1\), there is an associated partial fraction with a linear numerator \(Ax + B\). And for the repeated factor, we use \(\frac{Cx + D}{(x^2 + 1)^2}\), accommodating the repetition by adjusting the power in the denominator.
The goal is to break down the denominator into a product of simpler, mutually exclusive factors. These factors can either be linear (e.g., \(x + a\)) or quadratic (e.g., \(x^2 + bx + c\)). Sometimes, these factors repeat in the denominator, which impacts how the corresponding partial fractions should be expressed.
In the provided example, the denominator \((x^2 + 1)^2\) indicates a repeated irreducible quadratic factor. Recognizing this allows us to structure the partial fraction decomposition correctly. For each quadratic factor such as \(x^2 + 1\), there is an associated partial fraction with a linear numerator \(Ax + B\). And for the repeated factor, we use \(\frac{Cx + D}{(x^2 + 1)^2}\), accommodating the repetition by adjusting the power in the denominator.
Irreducible Quadratic Factor
An irreducible quadratic factor refers to a quadratic expression that cannot be factored over the real numbers into linear factors. Understanding these factors is essential because they dictate the form of partial fractions in decomposition.
For example, a quadratic like \(x^2 + 1\) is irreducible over the reals because it has no real roots (since the discriminant, \(b^2 - 4ac\), is negative). This impacts how we express partial fractions; the numerator must have a degree less than that of the quadratic.
In the provided decomposition: \( \frac{Ax + B}{x^2 + 1} \), the numerator \(Ax + B\) is a linear expression due to the irreducible quadratic factor's presence in the denominator. When such a factor repeats, we adjust by including such expressions with increased powers in the denominator, like \((x^2 + 1)^2\) affecting the partial fraction \( \frac{Cx + D}{(x^2 + 1)^2} \).
For example, a quadratic like \(x^2 + 1\) is irreducible over the reals because it has no real roots (since the discriminant, \(b^2 - 4ac\), is negative). This impacts how we express partial fractions; the numerator must have a degree less than that of the quadratic.
In the provided decomposition: \( \frac{Ax + B}{x^2 + 1} \), the numerator \(Ax + B\) is a linear expression due to the irreducible quadratic factor's presence in the denominator. When such a factor repeats, we adjust by including such expressions with increased powers in the denominator, like \((x^2 + 1)^2\) affecting the partial fraction \( \frac{Cx + D}{(x^2 + 1)^2} \).
- Irreducible quadratic factors keep certain polynomials unfactorable.
- They influence the structure of partial fractions.
- Their presence requires linear terms in the numerator for decomposition.
Other exercises in this chapter
Problem 6
Evaluate the given determinant. In Problem 10 , assume that \(a \neq 0, b \neq 0\). $$ \left|\begin{array}{rr} 0 & -1 \\ 8 & 0 \end{array}\right| $$
View solution Problem 7
Solve the given linear system. State whether the system is consistent, with independent or dependent equations, or whether it is inconsistent. $$ \left\\{\begin
View solution Problem 7
Determine graphically whether the given nonlinear system has any real solutions. $$ \left\\{\begin{array}{l} y-x^{2}=0 \\ x^{2}-y^{2}=4 \end{array}\right. $$
View solution Problem 7
Evaluate the given determinant. In Problem 10 , assume that \(a \neq 0, b \neq 0\). $$ \left|\begin{array}{ll} 4 & 2 \\ 0 & 3 \end{array}\right| $$
View solution