Problem 21
Question
\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{x+14}{x^{2}-2 x-8} $$
Step-by-Step Solution
Verified Answer
\(\frac{x+14}{x^2-2x-8} = \frac{3}{x-4} - \frac{2}{x+2}\).
1Step 1: Factor the Denominator
First, we need to factor the quadratic polynomial in the denominator, \(x^2 - 2x - 8\). We do this by finding two numbers that multiply to \(-8\) and add up to \(-2\). These numbers are \(-4\) and \(2\) because \(-4 \cdot 2 = -8\) and \(-4 + 2 = -2\). So, the factored form is \((x-4)(x+2)\).
2Step 2: Set Up the Partial Fraction Decomposition
Next, write the expression using partial fractions. Since the denominator is factored as \((x-4)(x+2)\), our decomposition will have the form:\[\frac{x+14}{x^2 - 2x - 8} = \frac{A}{x-4} + \frac{B}{x+2}.\]
3Step 3: Clear the Denominator
Multiply both sides by \((x-4)(x+2)\) to clear the fractions:\[x + 14 = A(x+2) + B(x-4).\]
4Step 4: Expand and Simplify
Expand the right-hand side:\[x + 14 = Ax + 2A + Bx - 4B.\]Combine like terms, we have:\[x + 14 = (A + B)x + (2A - 4B).\]
5Step 5: Form a System of Equations
Equate the coefficients of the same powers of \(x\) in both sides of the equation:1. For the \(x\)-term: \(A + B = 1\).2. For the constant term: \(2A - 4B = 14\).
6Step 6: Solve the System of Equations
To solve the system:1. From \(A + B = 1\), we can express \(A\) as \(A = 1 - B\).2. Substitute \(A = 1 - B\) into \(2A - 4B = 14\):\[2(1 - B) - 4B = 14\].3. Simplify and solve:\[2 - 2B - 4B = 14\].\[2 - 6B = 14\].\[-6B = 12\].\[B = -2\].4. Substitute \(B = -2\) back into \(A = 1 - B\):\[A = 1 - (-2) = 3\].
7Step 7: Write the Partial Fraction Decomposition
Substitute \(A = 3\) and \(B = -2\) into the partial fraction decomposition:\[\frac{3}{x-4} - \frac{2}{x+2}.\]
Key Concepts
Rational FunctionsFactoring PolynomialsSystem of EquationsAlgebra
Rational Functions
Rational functions are expressions that involve ratios of polynomials. They take the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). In the case of our exercise, the rational function is \( \frac{x+14}{x^2 - 2x - 8} \).
Rational functions are intriguing because they can model many real-world phenomena and often appear in calculus problems involving limits, continuity, and integration. One common technique used with rational functions is partial fraction decomposition, which simplifies the integration process by breaking down a complex rational function into simpler fractions.
To utilize rational functions effectively, it is essential to understand how to manipulate these expressions through factoring, which is a key step in preparing a function for partial fraction decomposition.
Rational functions are intriguing because they can model many real-world phenomena and often appear in calculus problems involving limits, continuity, and integration. One common technique used with rational functions is partial fraction decomposition, which simplifies the integration process by breaking down a complex rational function into simpler fractions.
To utilize rational functions effectively, it is essential to understand how to manipulate these expressions through factoring, which is a key step in preparing a function for partial fraction decomposition.
Factoring Polynomials
Factoring polynomials is a crucial skill needed to work with rational functions, especially when performing partial fraction decomposition. It involves expressing a polynomial as a product of its factors, which can simplify calculations and reveal the roots of the polynomial.
In the exercise, we need to factor the polynomial \( x^2 - 2x - 8 \), which serves as the denominator of the rational function. To factor a quadratic polynomial like this, we look for two numbers that multiply to produce the constant term (\(-8\)) and add to the coefficient of the linear term (\(-2\)). Here, these numbers are \(-4\) and \(2\).
So, the expression \(x^2 - 2x - 8\) is factored as \((x-4)(x+2)\). Recognizing this factored form is vital because it helps us set up the structure for partial fraction decomposition.
In the exercise, we need to factor the polynomial \( x^2 - 2x - 8 \), which serves as the denominator of the rational function. To factor a quadratic polynomial like this, we look for two numbers that multiply to produce the constant term (\(-8\)) and add to the coefficient of the linear term (\(-2\)). Here, these numbers are \(-4\) and \(2\).
So, the expression \(x^2 - 2x - 8\) is factored as \((x-4)(x+2)\). Recognizing this factored form is vital because it helps us set up the structure for partial fraction decomposition.
System of Equations
When decomposing a rational function into partial fractions, a system of equations is often used to determine unknown coefficients within the decomposition.
In our example, once the denominator \(x^2 - 2x - 8\) is factored into \((x-4)(x+2)\), the expression is set up as \(\frac{A}{x-4} + \frac{B}{x+2}\). To find \(A\) and \(B\), we multiply throughout by \((x-4)(x+2)\), which removes the fractions:
\[ x + 14 = A(x+2) + B(x-4) \]
Expanding and rearranging leads to a linear equation that relates coefficients on both sides, resulting in the system:
In our example, once the denominator \(x^2 - 2x - 8\) is factored into \((x-4)(x+2)\), the expression is set up as \(\frac{A}{x-4} + \frac{B}{x+2}\). To find \(A\) and \(B\), we multiply throughout by \((x-4)(x+2)\), which removes the fractions:
\[ x + 14 = A(x+2) + B(x-4) \]
Expanding and rearranging leads to a linear equation that relates coefficients on both sides, resulting in the system:
- \( A + B = 1 \)
- \( 2A - 4B = 14 \)
Algebra
Algebra is fundamental to understanding and solving problems involving rational functions and partial fractions. It is the language of these expressions and consists of rules and techniques for simplifying and solving mathematical expressions.
The process of partial fraction decomposition extensively uses algebraic skills. For instance, solving the system of equations involves isolating variables and combining like terms, both essential algebraic techniques. In the equation \( A + B = 1 \), using substitution, we express \(A\) in terms of \(B\), i.e., \( A = 1 - B \), and substitute into another equation to solve for one variable first.
Once equations are solved, the algebraic manipulation also involves substituting back to verify solutions, ensuring correct values that satisfy the original rational function. Mastering these algebraic concepts allows you to confidently decompose rational functions into simpler fractions.
The process of partial fraction decomposition extensively uses algebraic skills. For instance, solving the system of equations involves isolating variables and combining like terms, both essential algebraic techniques. In the equation \( A + B = 1 \), using substitution, we express \(A\) in terms of \(B\), i.e., \( A = 1 - B \), and substitute into another equation to solve for one variable first.
Once equations are solved, the algebraic manipulation also involves substituting back to verify solutions, ensuring correct values that satisfy the original rational function. Mastering these algebraic concepts allows you to confidently decompose rational functions into simpler fractions.
Other exercises in this chapter
Problem 21
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The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x+y+z
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Find the inverse of the matrix if it exists. $$ \left[\begin{array}{rrr}{0} & {-2} & {2} \\ {3} & {1} & {3} \\ {1} & {-2} & {3}\end{array}\right] $$
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The matrices \(A, B, C, D, E, F, G\) and \(H\) are defined as follows. $$ A=\left[\begin{array}{rr}{2} & {-5} \\ {0} & {7}\end{array}\right] \quad B=\left[\begi
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