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Magazine Chapter 8: Problem 54
Find each partial fraction decomposition. $$\frac{2 x^{2}}{x^{3}+3 x^{2}+3 x+1}$$
Short Answer
Expert verified
\( \frac{2 x^2}{(x+1)^3} = \frac{2}{x+1} - \frac{2}{(x+1)^2} \)
Step by step solution
01
Factor the Denominator
The first step in partial fraction decomposition is to factor the denominator. Here, we need to factor the polynomial \(x^3 + 3 x^2 + 3 x + 1\). Notice that \(x^3 + 3 x^2 + 3 x + 1 = (x + 1)^3\).
02
Set Up Partial Fractions
Since the factor \((x + 1)\) is repeated, we'll write the partial fractions for each power of \((x + 1)\) up to the third power. Therefore, we write: \[ \frac{2 x^{2}}{(x + 1)^3} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3} \]
03
Clear the Denominator
Multiply through by \((x + 1)^3\) to clear the denominator: \[ 2 x^2 = A(x + 1)^2 + B(x + 1) + C \]
04
Expand and Simplify
Expand and simplify the right-hand side: \[ 2 x^2 = A(x^2 + 2x + 1) + B(x + 1) + C \] This simplifies to: \[ 2 x^2 = A x^2 + 2 A x + A + B x + B + C \]
05
Collect Like Terms
Collect the like terms from the expanded polynomial: \[ 2 x^2 = (A)x^2 + (2A + B)x + (A + B + C) \]
06
Solve for Coefficients
Match the coefficients of \(x^2\), \(x\), and the constant term from both sides of the equation: \[ \begin{cases} A = 2 \ 2A + B = 0 \ A + B + C = 0 \ \text{So, } 2 + B = 0 \rightarrow B = -2 \ 2 - 2 + C = 0 \rightarrow C = 0 \ \text{Therefore, } A = 2, B = -2, \text{ and } C = 0 \ \text{Thus, the partial fractions are: } \frac{2}{x + 1} - \frac{2}{(x + 1)^2} \text{.} \ \frac{2 x^{2}}{(x + 1)^3} = \frac{2}{x + 1} - \frac{2}{(x + 1)^{2}} \text{.} \ \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Polynomials
The process of partial fraction decomposition begins with factoring the polynomial in the denominator. Factoring polynomials simplifies and breaks down complex expressions into smaller, more manageable pieces. In our example, the polynomial given is \(x^3 + 3x^2 + 3x + 1\). This particular polynomial is a perfect cube and factors neatly into \((x + 1)^3\). Recognizing patterns in polynomials, like perfect cubes or squares, can help tremendously in factoring them correctly.
A quick method for recognizing perfect cubes involves checking the constant term and the coefficient of each degree term. Practice with different polynomial types can strengthen your understanding and ability to factor quickly and accurately.
A quick method for recognizing perfect cubes involves checking the constant term and the coefficient of each degree term. Practice with different polynomial types can strengthen your understanding and ability to factor quickly and accurately.
Repeated Factors
When factoring polynomials, sometimes you'll encounter repeated factors. In our example, \((x + 1)^3\) is a repeated factor because it appears three times. This means we need to set our partial fractions for each power of \((x + 1)\) up to the third power.
For \(\frac{2x^2}{(x + 1)^3}\), we write the partial fractions as:
\(\frac{2x^2}{(x + 1)^3} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3}\).
By doing so, we account for each instance of the repeated factor, ensuring a complete partial fraction decomposition. Repeated factors require us to write separate fractions for each power to reflect the contribution of each term to the original polynomial.
For \(\frac{2x^2}{(x + 1)^3}\), we write the partial fractions as:
\(\frac{2x^2}{(x + 1)^3} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3}\).
By doing so, we account for each instance of the repeated factor, ensuring a complete partial fraction decomposition. Repeated factors require us to write separate fractions for each power to reflect the contribution of each term to the original polynomial.
Solving Coefficients
The next step after setting up partial fractions is to solve for the coefficients \(A\), \(B\), and \(C\). We start by multiplying through by the common denominator \((x + 1)^3\) to eliminate the fractions:
\(2x^2 = A(x + 1)^2 + B(x + 1) + C\).
Next, we expand and simplify:
\(2x^2 = Ax^2 + 2Ax + A + Bx + B + C\).
Then, collect like terms:
\(2x^2 = Ax^2 + (2A + B)x + (A + B + C)\).
We match the coefficients on both sides to solve for \(A, B, \) and \(C\). Comparing the respective coefficients of \(x^2\), \(x\), and the constant term, we get:
\(A = 2\)
\(2A + B = 0\)
\(A + B + C = 0\).
From these, we solve systematically:
This gives us the final partial fractions:
\(\frac{2}{x + 1} - \frac{2}{(x + 1)^2}\).
\(2x^2 = A(x + 1)^2 + B(x + 1) + C\).
Next, we expand and simplify:
\(2x^2 = Ax^2 + 2Ax + A + Bx + B + C\).
Then, collect like terms:
\(2x^2 = Ax^2 + (2A + B)x + (A + B + C)\).
We match the coefficients on both sides to solve for \(A, B, \) and \(C\). Comparing the respective coefficients of \(x^2\), \(x\), and the constant term, we get:
\(A = 2\)
\(2A + B = 0\)
\(A + B + C = 0\).
From these, we solve systematically:
- \(2 + B = 0 \Rightarrow B = -2\)
- \(2 - 2 + C = 0 \Rightarrow C = 0\)
This gives us the final partial fractions:
\(\frac{2}{x + 1} - \frac{2}{(x + 1)^2}\).
