Math Problem Statement

x3 /(x2 + 7x + 6) partial fraction decomposition

Solution

To decompose the rational expression x3x2+7x+6\frac{x^3}{x^2 + 7x + 6} into partial fractions, we need to follow these steps:

Step 1: Factor the denominator

We begin by factoring the quadratic denominator:

x2+7x+6=(x+1)(x+6)x^2 + 7x + 6 = (x + 1)(x + 6)

Step 2: Set up the partial fraction decomposition form

Since the denominator factors as (x+1)(x+6)(x + 1)(x + 6), we can express the fraction as:

x3(x+1)(x+6)=Ax+1+Bx+6\frac{x^3}{(x + 1)(x + 6)} = \frac{A}{x + 1} + \frac{B}{x + 6}

where AA and BB are constants we need to determine.

Step 3: Multiply through by the denominator

To eliminate the denominators, multiply both sides by (x+1)(x+6)(x + 1)(x + 6):

x3=A(x+6)+B(x+1)x^3 = A(x + 6) + B(x + 1)

Step 4: Expand and collect like terms

Expand both sides:

x3=A(x+6)+B(x+1)=Ax+6A+Bx+Bx^3 = A(x + 6) + B(x + 1) = A \cdot x + 6A + B \cdot x + B x3=(A+B)x+(6A+B)x^3 = (A + B) \cdot x + (6A + B)

Step 5: Set up equations by comparing coefficients

Now, compare the coefficients of xx on both sides:

  • From the x3x^3 term: There is no x3x^3 term on the right-hand side, which means A+BA + B must equal zero:

    A+B=0A + B = 0

  • From the constant term: The constant term on the right-hand side is 6A+B6A + B, and on the left-hand side, the constant is 0. So:

    6A+B=06A + B = 0

Step 6: Solve the system of equations

Now, we solve the system of two equations:

  1. A+B=0A + B = 0
  2. 6A+B=06A + B = 0

From the first equation, B=AB = -A. Substitute this into the second equation:

6AA=0    5A=0    A=06A - A = 0 \implies 5A = 0 \implies A = 0

Since A=0A = 0, from the first equation, B=0B = 0.

Step 7: Conclusion

In this case, the partial fraction decomposition simplifies to:

x3(x+1)(x+6)=0\frac{x^3}{(x + 1)(x + 6)} = 0

This indicates there is no non-trivial partial fraction decomposition for this expression, and the original fraction remains as-is.


Do you have any questions or would you like further clarification?

Here are some related questions to explore:

  1. How does partial fraction decomposition work for higher-degree polynomials?
  2. Can partial fraction decomposition be applied to improper fractions?
  3. What happens when the degree of the numerator is greater than the denominator?
  4. How do you apply partial fraction decomposition to repeated factors?
  5. How does partial fraction decomposition apply to rational functions with complex roots?

Tip: Always check if the degree of the numerator is higher than the denominator. If so, perform polynomial division first!

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Math Problem Analysis

Mathematical Concepts

Partial Fractions
Algebra
Polynomial Factorization

Formulas

Partial Fraction Decomposition
Polynomial Expansion

Theorems

Factor Theorem

Suitable Grade Level

Grades 10-12