Problem 9
Question
Use long division to write \(f(x)\) as a sum of a polynomial and a proper rational function. $$ f(x)=\frac{x^{3}+3 x^{2}+3 x+1}{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
The function \(f(x)\) can be written as \(x + 2 + \frac{3x - 1}{x^{2} + 1}\).
1Step 1: Set Up the Long Division
Write the division in the long division format. Place the dividend, \(x^{3} + 3x^{2} + 3x + 1\), inside the division bracket, and the divisor, \(x^{2} + 1\), outside the bracket.
2Step 2: Perform Initial Division
Divide the leading term of the dividend, \(x^{3}\), by the leading term of the divisor, \(x^{2}\), to get the first term of the quotient, which is \(x\). Multiply \(x\) by the entire divisor, \(x^{2} + 1\), resulting in \(x^{3} + x\).
3Step 3: Subtract and Bring Down Next Term
Subtract \(x^{3} + x\) from the original dividend \(x^{3} + 3x^{2} + 3x + 1\). This gives \(2x^{2} + 3x + 1\) as the new dividend. Bring down the remainder, which is \(+1\), to form \(2x^{2} + 3x + 1\).
4Step 4: Repeat Division Process
Divide the new leading term \(2x^{2}\) by \(x^{2}\) to get \(2\). Multiply \(2\) by the divisor \(x^{2} + 1\), obtaining \(2x^{2} + 2\).
5Step 5: Subtract Again
Subtract \(2x^{2} + 2\) from \(2x^{2} + 3x + 1\). This results in a remainder of \(3x - 1\).
6Step 6: Express the Result
The solution to the long division is \(x + 2\) with a remainder of \(3x - 1\). Therefore, the expression is written as \(x + 2 + \frac{3x - 1}{x^{2} + 1}\).
Key Concepts
Polynomial DivisionRational FunctionsCalculus Techniques
Polynomial Division
Polynomial division is similar to regular long division that you might know from arithmetic with numbers, but it involves polynomials instead of digits. The goal is to divide one polynomial (the dividend) by another polynomial (the divisor) to get a quotient and a remainder. In this context, it helps to simplify complex polynomial expressions by breaking them into more manageable parts.
When you perform polynomial division, you start by dividing the leading term (the term with the highest power) of the dividend by the leading term of the divisor. Here, you are essentially seeing how many times the divisor can "fit" into the dividend. The result gives you the first term of the quotient.
When you perform polynomial division, you start by dividing the leading term (the term with the highest power) of the dividend by the leading term of the divisor. Here, you are essentially seeing how many times the divisor can "fit" into the dividend. The result gives you the first term of the quotient.
- Set up the division problem by writing down your polynomials: the dividend inside the division bracket and the divisor outside.
- Divide the leading terms to find the first term of the quotient.
- Multiply the entire divisor by this quotient term, and subtract the result from the original dividend.
- Bring down the next term of the dividend and repeat the process until you've addressed all the terms.
Rational Functions
Rational functions are expressions that involve the division of two polynomials. The function takes the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). These functions appear often in calculus and algebra, and they can behave quite differently depending on their composition.
In the context of this exercise, after performing polynomial long division, the function \( f(x) \) was rewritten as a sum of a polynomial and a proper rational function (a rational function where the degree of the numerator is less than the degree of the denominator). This proper rational function is crucial because it often gives insights into the behavior of the function, such as vertical asymptotes or points where the function is undefined.
Thus, by dividing polynomials and rewriting rational functions in this manner, you simplify the expressions, making it easier to understand and analyze the characteristics of these complex functions in mathematical analysis.
In the context of this exercise, after performing polynomial long division, the function \( f(x) \) was rewritten as a sum of a polynomial and a proper rational function (a rational function where the degree of the numerator is less than the degree of the denominator). This proper rational function is crucial because it often gives insights into the behavior of the function, such as vertical asymptotes or points where the function is undefined.
Thus, by dividing polynomials and rewriting rational functions in this manner, you simplify the expressions, making it easier to understand and analyze the characteristics of these complex functions in mathematical analysis.
Calculus Techniques
Calculus techniques often involve working with polynomials and rational functions extensively, especially when dealing with limits, derivatives, and integrals. Long division helps precursors to these techniques, such as simplifying expressions and identifying limits.
For example, knowing how to use polynomial division allows you to tackle integrals involving rational functions more efficiently. You simplify the function first and then apply integration techniques without dealing with a complex polynomial in the numerator directly.
For example, knowing how to use polynomial division allows you to tackle integrals involving rational functions more efficiently. You simplify the function first and then apply integration techniques without dealing with a complex polynomial in the numerator directly.
- Understand how simplified expressions can make differentiation and integration more straightforward.
- Identify asymptotic behavior more easily, as simplifying rational functions can make it more transparent.
- Examine limit values of functions at infinity, as breaking down rational functions simplifies determining the limit behavior as well.
Other exercises in this chapter
Problem 9
All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{-\infty}^{\infty
View solution Problem 9
In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x^{2} e^{x} d x $$
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In Problems 1-16, evaluate each indefinite integral by making the given substitution. $$ \int e^{2 x+3} d x, \text { with } u=2 x+3 $$
View solution Problem 10
Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int_{1}^{2} x \ln (x+3) d x $$
View solution