Question

Resolve \(\frac{2 x^{2}+8 x+13}{(x+1)^{4}}\) into partial fractions. (1) \(\frac{2}{(x+1)^{2}}+\frac{5}{(x+1)^{3}}+\frac{7}{(x+1)^{4}}\) (2) \(\frac{1}{(x+1)^{2}}+\frac{5}{(x+1)^{3}}+\frac{7}{(x+1)^{4}}\) (3) \(\frac{2}{(x+1)^{2}}+\frac{4}{(x+1)^{3}}+\frac{7}{(x+1)^{4}}\) (4) \(\frac{4}{(x+1)^{2}}+\frac{5}{(x+1)^{3}}+\frac{7}{(x+1)^{4}}\)

Short answer

Answer: \(\frac{2x^{2}+8x+13}{(x+1)^{4}} = \frac{2}{(x+1)^{2}} + \frac{4}{(x+1)^{3}} + \frac{7}{(x+1)^{4}}\)

Step-by-step solution

Write the general form of the partial fraction decomposition

To decompose the given function into partial fractions, we will write it as a sum of terms with denominators of the form \((x+1)^{n}\), where n is an integer. The general form will be: $$\frac{2x^{2}+8x+13}{(x+1)^{4}} = \frac{A}{(x+1)} + \frac{B}{(x+1)^{2}} + \frac{C}{(x+1)^{3}} + \frac{D}{(x+1)^{4}}$$

Clear the denominators

Now let's clear the denominators by multiplying both sides by \((x+1)^{4}\): \(2x^{2}+8x+13 = A(x+1)^{3} + B(x+1)^{2} + C(x+1) + D\)

Set up a system of equations

Now we will find the values of A, B, C, and D. To do this, we will substitute some convenient values of x into the equation. Since the factors are \((x+1)\), we will use x = -1 to find D, as the other terms will be zero: \(2(-1)^{2} + 8(-1) + 13 = D \Rightarrow D = 7\) Next, we can differentiate both sides of the equation with respect to x and substitute convenient values to find A, B, and C.

Differentiate and substitute values

First derivative: \(4x + 8 = 3A(x+1)^{2} + 2B(x+1) + C\) At x = -1: \(4(-1) + 8 = 3A(0) + 2B(0) + C \Rightarrow C = 4\) Second derivative: \(4 = 6A(x+1) + 2B\) At x = -1: \(4 = 6A(0) + 2B \Rightarrow B=2\) Third derivative: \(0 = 6A\) This implies that A = 0.

Write the final answer

Now, we can write down the partial fractions decomposition using the values of A, B, C, and D found in the previous steps: $$\frac{2 x^{2}+8 x+13}{(x+1)^{4}} = \frac{0}{(x+1)} + \frac{2}{(x+1)^{2}} + \frac{4}{(x+1)^{3}} + \frac{7}{(x+1)^{4}}$$ Simplifying, we get: $$\frac{2 x^{2}+8 x+13}{(x+1)^{4}} = \frac{2}{(x+1)^{2}} + \frac{4}{(x+1)^{3}} + \frac{7}{(x+1)^{4}}$$ So, the correct answer is (3).

Key concepts

The Essence of Algebra

Algebra is like a bridge that connects numbers and operations with symbols. It's a way to describe patterns, solve puzzles about numbers, and predict the unknown. In algebra, we deal with equations and variables. Variables are symbols, often letters like x and y, which stand for numbers that can vary. We manipulate these variables and equations to find values that make certain statements true.

When you look at a problem like the decomposition of rational expressions, you're seeing algebra in action. It's all about breaking down complex, intimidating expressions into simpler, more digestible pieces, exactly the way a large puzzle is solved piece by piece. Understanding algebra is fundamental to mastering concepts such as partial fractions decomposition.

Rational Expressions and Their Decomposition

Rational expressions are fractions involving polynomials in their numerator and denominator, like the expression \(\frac{2x^{2}+8x+13}{(x+1)^{4}}\) in our exercise. Decomposing such expressions into partial fractions means breaking them down into simpler quotient forms. This is important because simpler forms are easier to integrate or differentiate, which is a common task in calculus.

Partial fraction decomposition involves finding a set of coefficients that, when plugged into a prescribed format, reconstruct the original expression. We usually deal with polynomials where the degree of the numerator is less than that of the denominator. If not, polynomial division is employed first. The goal is to convert the complex fraction into a sum of simpler fractions, where each one represents a part of the original whole.

Polynomial Division as a Prerequisite

Polynomial division is a technique used when the numerator's degree is not less than that of the denominator. In such cases, before attempting partial fractions decomposition, you must divide the polynomials to get a quotient and a remainder. This remainder then becomes the subject of partial fractions.

Think of it as dividing a pizza into equal slices so everyone gets the same size piece. Polynomial division ensures that the 'slices' of our mathematical 'pizza' are properly sized before we proceed with partial fractions. Students should also note that polynomial division helps simplify expressions, making subsequent calculus operations much smoother.Understanding how to perform polynomial division, like long division with numbers, is a crucial step towards mastering rational expressions and their decomposition.