Chapter 26: Problem 35
Resolve \(\frac{1}{x^{2}-7 x+12}\) into partial fractions. (1) \(\frac{1}{x-4}=\frac{1}{x-3}\) (2) \(\frac{1}{x-3}+\frac{1}{x-4}\) (3) \(\frac{2}{x-4}-\frac{1}{x-3}\) (4) \(\frac{1}{x+3}-\frac{1}{x+4}\)
Short Answer
Expert verified
Question: Decompose the fraction \(\frac{1}{x^{2}-7x+12}\) into partial fractions.
Answer: The decomposition of the given fraction into partial fractions is \(\frac{-1}{x-3} + \frac{1}{x-4}\).
Step by step solution
01
Factor the denominator
Factor the quadratic expression in the denominator. In this case, we have \(x^2 - 7x + 12\). We are looking for two terms that multiply to 12 and add up to -7. These terms are -3 and -4. Therefore, we can factor the denominator as \((x-3)(x-4)\).
02
Decompose the fraction
Now, we need to decompose the original fraction into smaller fractions with linear factors in their denominators. We can write \(\frac{1}{(x-3)(x-4)}\) as \(\frac{A}{x-3} + \frac{B}{x-4}\), where A and B are constants to be determined.
03
Remove the denominators
To find A and B, we will try to get rid of the denominators by multiplying both sides of the equation by the original denominator, \((x-3)(x-4)\). This results in: \(1 = A(x-4) + B(x-3)\).
04
Solve for A and B
Next, we will solve for A and B separately. To find A, we will set all the x terms to zero except the terms that involve A. In this case, we can let x = 3:
1 = A(3-4) + B(3-3)
1 = -A
A = -1
Similarly, to find B, we will set all the x terms to zero except the terms that involve B. In this case, we can let x = 4:
1 = A(4-4) + B(4-3)
1 = B
B = 1
Our values for A and B are -1 and 1, respectively.
05
Write the final answer as partial fractions
Now that we have the values for A and B, plug them back into the decomposed fractions: \(\frac{-1}{x-3} + \frac{1}{x-4}\). Thus, the partial fractions decomposition of \(\frac{1}{x^2 - 7x + 12}\) is \(\frac{-1}{x-3} + \frac{1}{x-4}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factor Quadratic Expressions
Factoring quadratic expressions is a foundational skill when decomposing algebraic fractions. A quadratic expression takes the basic form of \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The key is to find two numbers that both add up to \( b \) and multiply to \( ac \). These numbers represent the roots of the expression and can be used to break it down into linear factors.
For example, the quadratic expression \( x^2 - 7x + 12 \) can be factored by identifying numbers that multiply to 12 (the constant term) and add up to -7 (the coefficient of the \( x \) term). These numbers are -3 and -4. This gives us the factors \( x - 3 \) and \( x - 4 \), turning the quadratic expression into a product of linear factors. Factoring is often the first step in solving equations and decomposing fractions.
For example, the quadratic expression \( x^2 - 7x + 12 \) can be factored by identifying numbers that multiply to 12 (the constant term) and add up to -7 (the coefficient of the \( x \) term). These numbers are -3 and -4. This gives us the factors \( x - 3 \) and \( x - 4 \), turning the quadratic expression into a product of linear factors. Factoring is often the first step in solving equations and decomposing fractions.
Linear Factors in Denominators
In the world of algebraic fractions, the presence of linear factors in denominators is particularly important. A linear factor is a term of the first degree, which means it's in the form of \( mx + n \), with \( m \) and \( n \) being constants and \( m \) not equal to zero. When decomposing complex fractions, our goal is to express them as the sum or difference of simpler fractions with these linear factors in their denominators.
Take the exercise where we have the denominator \( x^2 - 7x + 12 \), which was factored into \( (x - 3)(x - 4) \). The presence of these linear factors, \( x - 3 \) and \( x - 4 \), paves the way for partial fraction decomposition, allowing us to rewrite the complex fraction as a combination of simpler fractions, each having one of these linear factors in their respective denominators. This step is pivotal as it simplifies the integration or the calculation of complex algebraic expressions.
Take the exercise where we have the denominator \( x^2 - 7x + 12 \), which was factored into \( (x - 3)(x - 4) \). The presence of these linear factors, \( x - 3 \) and \( x - 4 \), paves the way for partial fraction decomposition, allowing us to rewrite the complex fraction as a combination of simpler fractions, each having one of these linear factors in their respective denominators. This step is pivotal as it simplifies the integration or the calculation of complex algebraic expressions.
Solving for Constants in Partial Fractions
When we break down an algebraic fraction into partial fractions, we introduce constants, usually denoted as \( A \), \( B \), \( C \), etc., which we need to solve for. These constants are numerators in the decomposed form. To find their values, we employ a method where we remove the denominators by multiplying through by the common denominator of all the fractions. This leaves us with a polynomial equation where the constants are the unknowns.
After setting up the equation, we solve for each constant by strategically choosing values for \( x \) that simplify the equation. For instance, if \( x - 3 \) is a denominator, setting \( x = 3 \) will eliminate the terms involving other constants, allowing us to solve for the constant in question. This method of solving for constants in partial fractions is systematic and ensures that we find the correct values that make our decomposed equation equivalent to the original fraction.
After setting up the equation, we solve for each constant by strategically choosing values for \( x \) that simplify the equation. For instance, if \( x - 3 \) is a denominator, setting \( x = 3 \) will eliminate the terms involving other constants, allowing us to solve for the constant in question. This method of solving for constants in partial fractions is systematic and ensures that we find the correct values that make our decomposed equation equivalent to the original fraction.
Algebraic Fraction Decomposition
Algebraic fraction decomposition, also known as partial fraction decomposition, is a technique used to simplify complex rational expressions into a sum or difference of simpler fractions. This is particularly useful when dealing with integration or finding antiderivatives in calculus. The goal is to break down a fraction with a polynomial in the denominator into a set of fractions whose denominators are the factors of the original polynomial, and whose numerators are constants to be determined.
The process involves several steps, starting with factoring the denominator, setting up a decomposed form with undetermined coefficients, and then solving for those coefficients, as illustrated in our example. The result is an expression that is easier to work with and can often be used to perform calculations that would be more complex with the original fraction.
The process involves several steps, starting with factoring the denominator, setting up a decomposed form with undetermined coefficients, and then solving for those coefficients, as illustrated in our example. The result is an expression that is easier to work with and can often be used to perform calculations that would be more complex with the original fraction.