Skip to main content
Ch. 5 - Systems and Matrices
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. 5 - Systems and MatricesProblem 9
Chapter 6, Problem 9

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (4x + 2)/((x + 2)(2x - 1))

Verified step by step guidance
1
Identify the form of the partial fraction decomposition. Since the denominator is factored into linear factors \((x + 2)\) and \((2x - 1)\), set up the decomposition as \(\frac{4x + 2}{(x + 2)(2x - 1)} = \frac{A}{x + 2} + \frac{B}{2x - 1}\), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the common denominator \((x + 2)(2x - 1)\) to clear the fractions. This gives \(4x + 2 = A(2x - 1) + B(x + 2)\).
Expand the right-hand side: \(A(2x - 1) = 2Ax - A\) and \(B(x + 2) = Bx + 2B\). So the equation becomes \(4x + 2 = 2Ax - A + Bx + 2B\).
Group like terms on the right-hand side: \(4x + 2 = (2A + B)x + (-A + 2B)\).
Set up a system of equations by equating the coefficients of corresponding terms from both sides: For the \(x\) terms, \(4 = 2A + B\); for the constant terms, \(2 = -A + 2B\). Solve this system to find the values of \(A\) and \(B\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
8m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions with linear or quadratic denominators. This technique simplifies integration and other algebraic operations by breaking down complex fractions into manageable parts.
Recommended video:
4:07
Decomposition of Functions

Factoring Denominators

Factoring the denominator into linear or irreducible quadratic factors is essential before performing partial fraction decomposition. It allows the rational expression to be split into simpler fractions, each corresponding to a factor in the denominator.
Recommended video:
Guided course
02:58
Rationalizing Denominators

Setting Up and Solving Equations for Coefficients

After expressing the rational function as a sum of unknown coefficients over the factored denominators, you multiply through by the common denominator and equate coefficients of like terms. Solving the resulting system of equations finds the values of these unknowns.
Recommended video:
5:02
Solving Logarithmic Equations
Related Practice
Textbook Question

Solve each system by substitution.

x - 5y = 8

x = 6y

1
views
Textbook Question

Find the dimension of each matrix. Identify any square, column, or row matrices.

[−680041923−571]\(\left\)[ \(\begin{matrix}\) -6 & 8 & 0 & 0 \\ 4 & 1 & 9 & 2 \\ 3 & -5 & 7 & 1 \(\end{matrix}\) \(\right\)]

3
views
Textbook Question

Are the given matrices inverses of each other? (Hint: Check to see whether their products are the identity matrix In.)

[5723]and[3−7−25]\(\left\)[ \(\begin{matrix}\) 5 & 7 \\2 & 3 \(\end{matrix}\) \(\right\)]\(\quad\) \(\text{and}\) \(\quad\]\left\)[ \(\begin{matrix}\) 3 & -7 \\-2 & 5 \(\end{matrix}\) \(\right\)]

1
views
Textbook Question

Use the given row transformation to change each matrix as indicated.

[156−4−12371],4×row 1 added to row 2\(\left\)[ \(\begin{matrix}\) 1 & 5 & 6 \\ -4 & -1 & 2 \\ 3 & 7 & 1 \(\end{matrix}\) \(\right\)], \(\quad\) 4 \(\times\) \(\text{row 1 added to row 2}\)

1
views
Textbook Question

Evaluate each determinant.

∣−1−253∣\(\left\)| \(\begin{matrix}\) -1 & -2 \\ 5 & 3 \(\end{matrix}\) \(\right\)|

Textbook Question

Use the given row transformation to change each matrix as indicated.

[156−23−1470],2×row 1 added to row 2\(\left\)[ \(\begin{matrix}\) 1 & 5 & 6 \\ -2 & 3 & -1 \\ 4 & 7 & 0 \(\end{matrix}\) \(\right\)], \(\quad\) 2 \(\times\) \(\text{row 1 added to row 2}\)

5
views