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Write the partial fraction decomposition of each rational expression. $$\frac{2 x^{2}-x+2}{\left(x^{2}+1\right)^{2}}$$

Short Answer

Expert verified
The partial fraction decomposition of the given rational expression is \( \frac{1}{x^{2}+1} \).

Step by step solution

01

Identify the form of partial fraction decomposition

In order to decompose a fraction, one must first determine the form of the decomposition. It depends on the factors in the denominator. Our denominator here is \( (x^{2}+1)^{2} \), which is a quadratic term squared. So the form of partial fractions will be \( A/x^{2}+1 + B/(x^{2}+1)^{2} \).
02

Write the equation using the identified form

Next, setup the equation according to the form determined above. The given function equals to \( \frac{2 x^{2}-x+2}{\left(x^{2}+1\right)^{2}} = \frac{A}{x^{2} + 1} + \frac{B}{(x^{2}+1)^{2}} \). Therefore, we can write it as \( 2x^{2} - x + 2 = A(x^{2} + 1) + B \).
03

Find the values of the coefficients A and B

In order to determine the coefficients A and B, we need to solve the equation from step 2. From the given equation, by comparing coefficients, we get two equations: \(2A = 2\) (which leads to \(A = 1\)) and \(-A + B = -1\) (which leads to \(B = 0\)).
04

Write the partial fraction decomposition

Once values of coefficients are found, substitute these values into the identified form in our first step. By substitifying A by 1 and B by 0, we obtain the final result : \( \frac{1}{x^{2}+1} + \frac{0}{(x^{2}+1)^{2}} \), which simplifies to \( \frac{1}{x^{2}+1} \) as our partial fraction decomposition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Expressions
Rational expressions are fractions that have a polynomial in both the numerator and the denominator. In simpler terms, they're like fractions you work with when dealing with numbers, but instead of numbers, you have polynomials. Understanding rational expressions is key when diving into topics like partial fraction decomposition. They form the backbone of more complex calculations involving fractions in algebra. For any given rational expression, we can potentially simplify it by dividing or factoring. Here's a simple process to understand rational expressions:
  • Examine both the numerator and the denominator.
  • Factor both parts if possible to potentially simplify.
  • If factors cancel each other out, simplify to the reduced form.
Understanding this process will aid in tasks like partial fraction decomposition, where a complex rational expression is broken into a sum of simpler fractions.
Quadratic Terms
Quadratic terms are expressions of the form \( ax^2 + bx + c \), where \(a eq 0\). These terms are commonly found in mathematics, especially in algebra when dealing with polynomials.In the context of partial fraction decomposition, quadratic terms present unique challenges and requirements. When decomposing fractions, identifying these quadratic terms is essential because they determine the type of partial fractions you'll work with.When encountering a squared quadratic term, like \((x^2 + 1)^2\), you should keep these points in mind:
  • Recognize the term and its power, as it affects the decomposition process.
  • The power of the quadratic term often dictates the number of terms in the partial fraction decomposition.
  • For each power of a quadratic term, you'll need to assign separate coefficients in the decomposition formula.
By understanding the nature of quadratic terms, you can better handle partial fraction decomposition involving complex denominators.
Coefficient Comparison
Coefficient comparison is a method used to determine unknown values in mathematical equations, especially in problems involving polynomials and fractions. During the partial fraction decomposition process, this technique becomes vital.Here's how coefficient comparison works:
  • Expand both sides of an equation after setting the decomposed form equal to the original expression.
  • Compare coefficients from both sides for similar powers of \(x\).
  • Develop equations based on these comparisons to solve for unknown coefficients, like \(A\) and \(B\) in our decomposition task.
In our example problem, we compared coefficients from \(2x^2 - x + 2 = A(x^2 + 1) + B\). This comparison method allowed us to solve for \(A = 1\) and \(B = 0\). This simple yet powerful approach ensures we correctly decompose rational expressions into their constituent partial fractions.

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Most popular questions from this chapter

A farmer has 90 acres available for planting corn and soybeans. The cost of seed per acre is \(\$ 4\) for corn and \(\$ 6\) for soybeans. To harvest the crops, the farmer will need to hire some temporary help. It will cost the farmer \(\$ 20\) per acre to harvest the corn and \(\$ 10\) per acre to harvest the soybeans. The farmer has \(\$ 480\) available for seed and \(\$ 1400\) available for labor. His profit is \(\$ 120\) per acre of corn and \(\$ 150\) per acre of soybeans. How many acres of each crop should the farmer plant to maximize the profit?

A supply function for widgets is modeled by \(P(q)=a q+b,\) where \(q\) is the number of widgets supplied and \(P(q)\) is the total price of \(q\) widgets, in dollars. It is known that 200 widgets can be supplied for \(\$ 40\) and 100 widgets can be supplied for \(\$ 25 .\) Use a system of linear equations to find the constants \(a\) and \(b\) in the expression for the supply function.

For the given matrices \(A, B,\) and \(C,\) evaluate the indicated expression. $$\begin{aligned}&A=\left[\begin{array}{rr}3 & -8 \\\2 & 4\end{array}\right] ; \quad B=\left[\begin{array}{rr}-6 & 0 \\\0 & -6\end{array}\right] ; \quad C=\left[\begin{array}{rr}3 & 5 \\\\-2 & 6\end{array}\right]\\\&(A+2 B) C\end{aligned}$$

Consider the following system of equations. $$\left\\{\begin{array}{l}6 u+6 v-3 w=-3 \\\2 u+2 v-w=-1\end{array}\right.$$ (a) Show that each of the equations in this system is a multiple of the other equation. (b) Explain why this system of equations has infinitely many solutions. (c) Express \(w\) as an equation in \(u\) and \(v\) (d) Give two solutions of this system of equations.

If \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]\) and \(B=\left[\begin{array}{cc}2 & 2 a+b \\ b-a & 6\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=B A ?\)

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