Problem 12 - College Algebra

Question

Find the decomposition of the partial fraction for the nonrepeating linear factors. \(\frac{x+1}{x^{2}+7 x+10}\)

Step-by-Step Solution

Answer

\( \frac{x+1}{(x+2)(x+5)} = \frac{-1/3}{x+2} + \frac{4/3}{x+5} \).

Step 1: Factor the Denominator

First, we need to factor the quadratic in the denominator. Given the quadratic expression \(x^2 + 7x + 10\), we look for two numbers that multiply to 10 and add up to 7. The numbers 2 and 5 fit this requirement, so we can factor the expression as: \( (x + 2)(x + 5) \).

Step 2: Set Up the Partial Fraction

Since our denominator factors to \( (x + 2)(x + 5) \), we can express the partial fraction decomposition as the sum of two fractions: \( \frac{x+1}{(x+2)(x+5)} = \frac{A}{x+2} + \frac{B}{x+5} \). Here, \(A\) and \(B\) are constants that we need to find.

Step 3: Clear the Denominator

Multiply through by the common denominator \((x+2)(x+5)\) to eliminate the fractions: \( x + 1 = A(x+5) + B(x+2) \).

Step 4: Expand and Simplify

Expand the right side: \( A(x+5) + B(x+2) = Ax + 5A + Bx + 2B \). Combine like terms: \( Ax + Bx + 5A + 2B = (A + B)x + (5A + 2B) \).

Step 5: Set Up a System of Equations

To find \(A\) and \(B\), match the coefficients from both sides of the equation \( x + 1 = (A + B)x + (5A + 2B) \). This gives us the system of equations: \( A + B = 1 \) and \( 5A + 2B = 1 \).

Step 6: Solve the System of Equations

Solve \( A + B = 1 \) and \( 5A + 2B = 1 \). From \( A + B = 1 \), we have \( B = 1 - A \). Substitute \( B = 1 - A \) in the second equation: \( 5A + 2(1-A) = 1 \). Simplify: \( 5A + 2 - 2A = 1 \), resulting in \( 3A = -1 \). Thus, \( A = -\frac{1}{3} \). Substituting back, \( B = 1 - (-\frac{1}{3}) = \frac{4}{3} \).

Step 7: Write the Final Decomposition

Now that we have \( A = -\frac{1}{3} \) and \( B = \frac{4}{3} \), we can complete the partial fraction decomposition: \( \frac{x+1}{(x+2)(x+5)} = \frac{-1/3}{x+2} + \frac{4/3}{x+5} \).

Key Concepts

Nonrepeating Linear Factors

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions. When dealing with nonrepeating linear factors, the focus is on denominators that contain distinct, non-repeating linear terms.
For instance, in the expression \((x+2)(x+5)\), both \(x+2\) and \(x+5\) are nonrepeating linear factors. Each linear factor represents a unique linear equation, ensuring they do not repeat in the factorization.

The process involves setting up the decomposition as separate fractions for each linear factor. That means you will write a partial fraction in the form:

• \( \frac{A}{x+2} \)
• \( \frac{B}{x+5} \)

where \(A\) and \(B\) are constants determined by solving a system of equations that arises from equating the original expression to the sum of these fractions.
Nonrepeating linear factors simplify the process because each factor independently contributes a term in the decomposition, making it straightforward to identify and solve for the constants.

Factoring Quadratics

Factoring quadratics is an essential skill for partial fraction decomposition. This technique allows you to simplify a quadratic expression by expressing it as a product of linear factors. The goal is to break down the quadratic polynomial into two binomials.
An example is the quadratic equation \(x^2 + 7x + 10\). To factor this expression, look for two numbers whose product equals the constant term (10) and whose sum equals the linear coefficient (7).

For \(x^2 + 7x + 10\), these numbers are 2 and 5:

• \(10 = 2 \times 5\)
• \(7 = 2 + 5\)

The quadratic can therefore be factored as \((x + 2)(x + 5)\). Factoring is easier when the quadratic fits perfectly into this product form, allowing for straightforward simplification into smaller components. When this factorization is complete, the expression is ready for partial fraction decomposition.

System of Equations

A system of equations is used to determine the constants in a partial fraction decomposition. This involves comparing coefficients from both sides of an equation to establish relationships between unknowns.

To find the values of \(A\) and \(B\) in the decomposition process, you begin by clearing any denominators and expanding the expression. Align coefficients on both sides to set up simultaneous equations. For example, from the expression:
\[ x + 1 = A(x + 5) + B(x + 2) \]
you expand to obtain:

• \( Ax + Bx + 5A + 2B = (A + B)x + (5A + 2B) \)

This results in the equations:

• \( A + B = 1 \)
• \( 5A + 2B = 1 \)

By solving these linear equations, you determine the values of \(A\) and \(B\).

Simplifying this system of equations is crucial for identifying the correct coefficients for the partial fractions. The solution involves basic algebraic manipulation and substitution to solve for the unknowns, ultimately leading to the completed decomposition.