Problem 5
Question
(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{11 x+30}{x^{2}-100}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition is \(\frac{7}{x-10} + \frac{4}{x+10}\).
1Step 1: Factor the Denominator
First, observe that the denominator \(x^2 - 100\) is a difference of squares. Use the formula \(a^2 - b^2 = (a - b)(a + b)\) to factor it. Here, \(a = x\) and \(b = 10\), so \(x^2 - 100 = (x - 10)(x + 10)\).
2Step 2: Determine the Form of Partial Fractions
The expression given is \(\frac{11x + 30}{(x - 10)(x + 10)}\). Since the factors \((x - 10)\) and \((x + 10)\) are both linear, the partial fraction decomposition will have the form: \(\frac{A}{x - 10} + \frac{B}{x + 10}\), where \(A\) and \(B\) are constants to be determined.
3Step 3: Set Up Equation for Constants
Equating the original rational expression to its partial fraction form, we have: \(\frac{11x + 30}{(x - 10)(x + 10)} = \frac{A}{x - 10} + \frac{B}{x + 10}\). Multiply through by the denominator \((x - 10)(x + 10)\) to eliminate fractions, getting: \(11x + 30 = A(x + 10) + B(x - 10)\).
4Step 4: Expand and Combine Like Terms
Expand the right side: \(11x + 30 = Ax + 10A + Bx - 10B\). Combine like terms to get \(11x + 30 = (A + B)x + (10A - 10B)\).
5Step 5: Solve for Constants
Comparing coefficients, we get two equations: 1) \(A + B = 11\) and 2) \(10A - 10B = 30\). Solve equation 2) for \(A - B = 3\). Now, solve the system of equations:\(A + B = 11\)\(A - B = 3\).
6Step 6: Determine Values for A and B
Add the equations: \(A + B = 11\)\(A - B = 3\)to get \(2A = 14\), hence \(A = 7\). Substitute \(A = 7\) back into \(A + B = 11\) to find \(B = 4\).
Key Concepts
Factoring PolynomialsRational ExpressionsSystem of Equations
Factoring Polynomials
Understanding how to factor polynomials is an essential algebraic skill, especially when dealing with rational expressions. Factoring involves breaking down a polynomial into simpler pieces, or factors, that, when multiplied together, create the original polynomial.
Here, the denominator of the rational expression is a prime candidate for factoring. It is structured as a difference of squares, which is recognizable through the formula: \(a^2 - b^2 = (a-b)(a+b)\). In our case, \(x^2 - 100\) can be rewritten in the form \(x^2 - 10^2\). This matches the difference of squares formula exactly, allowing us to factor it into \((x-10)(x+10)\).
Remembering specific polynomial forms, such as the difference of squares or perfect square trinomials, helps simplify complex expressions. Practicing with various polynomials enhances your skills in identifying and applying these forms effectively.
Here, the denominator of the rational expression is a prime candidate for factoring. It is structured as a difference of squares, which is recognizable through the formula: \(a^2 - b^2 = (a-b)(a+b)\). In our case, \(x^2 - 100\) can be rewritten in the form \(x^2 - 10^2\). This matches the difference of squares formula exactly, allowing us to factor it into \((x-10)(x+10)\).
Remembering specific polynomial forms, such as the difference of squares or perfect square trinomials, helps simplify complex expressions. Practicing with various polynomials enhances your skills in identifying and applying these forms effectively.
Rational Expressions
A rational expression is simply a ratio of two polynomials. Similar to fractions, you can simplify them by canceling common factors in the numerator and denominator. Analyzing and decomposing rational expressions is crucial, especially when you encounter complex algebraic equations.
In the exercise, we start with the rational expression: \(\frac{11x+30}{(x-10)(x+10)}\). After factoring the denominator, the next step is to express this as a sum of simpler fractions. This is known as partial fraction decomposition. For linear factors in the denominator, each factor contributes a term to the decomposition. Therefore, \(\frac{11x+30}{(x-10)(x+10)}\) can be rewritten as \(\frac{A}{x-10} + \frac{B}{x+10}\), where \(A\) and \(B\) are constants.
The goal is to choose \(A\) and \(B\) such that the original rational expression is maintained. This process transforms tough integrals into simpler ones since integrating partial fractions is often easier.
In the exercise, we start with the rational expression: \(\frac{11x+30}{(x-10)(x+10)}\). After factoring the denominator, the next step is to express this as a sum of simpler fractions. This is known as partial fraction decomposition. For linear factors in the denominator, each factor contributes a term to the decomposition. Therefore, \(\frac{11x+30}{(x-10)(x+10)}\) can be rewritten as \(\frac{A}{x-10} + \frac{B}{x+10}\), where \(A\) and \(B\) are constants.
The goal is to choose \(A\) and \(B\) such that the original rational expression is maintained. This process transforms tough integrals into simpler ones since integrating partial fractions is often easier.
System of Equations
After determining the form of the partial fractions, we equate them back to the original expression, which sets us up for solving a system of equations. Solving these equations will give the values of the unknown constants \(A\) and \(B\).
In this exercise, we equate the expression to the sum of partial fractions: \(\frac{11x+30}{(x-10)(x+10)} = \frac{A}{x-10} + \frac{B}{x+10}\). By clearing the fractions, we obtain the equation: \(11x + 30 = A(x + 10) + B(x - 10)\).
Expanding gives \(11x + 30 = Ax + 10A + Bx - 10B\). Combining like terms results in \(11x + 30 = (A+B)x + (10A-10B)\). From this, we derive two separate equations:
In this exercise, we equate the expression to the sum of partial fractions: \(\frac{11x+30}{(x-10)(x+10)} = \frac{A}{x-10} + \frac{B}{x+10}\). By clearing the fractions, we obtain the equation: \(11x + 30 = A(x + 10) + B(x - 10)\).
Expanding gives \(11x + 30 = Ax + 10A + Bx - 10B\). Combining like terms results in \(11x + 30 = (A+B)x + (10A-10B)\). From this, we derive two separate equations:
- \(A + B = 11\)
- \(10A - 10B = 30\)
Other exercises in this chapter
Problem 4
For Exercises specify the real and imaginary parts of each complex number. (a) \(-2+\sqrt{7} i\) (b) \(1+5^{1 / 3} i\) (c) \(-3 i\) (d) 0
View solution Problem 4
Use long division to find the quotients and the remainders. Also, write each answer in the form \(p(x)=d(x) \cdot q(x)+R(x),\) as in equation (2) in the text. $
View solution Problem 5
You are given a polynomial equation \(f(x)=0 .\) According to the fundamental theorem of algebra each of these equations has at least one root. However, the fun
View solution Problem 5
List the possibilities for rational roots. $$4 x^{3}-9 x^{2}-15 x+3=0$$
View solution