Math Problem Statement

Reslove into partial fractions

Solution

The expression you've provided is:

x+2(x1)(x2+2)\frac{x+2}{(x-1)(x^2+2)}

To resolve this into partial fractions, we need to express the fraction as a sum of simpler fractions. The general form for this decomposition is:

x+2(x1)(x2+2)=Ax1+Bx+Cx2+2\frac{x+2}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+2}

Here’s how to proceed:

  1. Multiply both sides by the denominator (x1)(x2+2)(x-1)(x^2+2) to eliminate the denominators:

    x+2=A(x2+2)+(Bx+C)(x1)x + 2 = A(x^2 + 2) + (Bx + C)(x-1)

  2. Expand the right-hand side:

    x+2=A(x2+2)+(Bx+C)(x1)x + 2 = A(x^2 + 2) + (Bx + C)(x - 1) =Ax2+2A+Bx2Bx+CxC= Ax^2 + 2A + Bx^2 - Bx + Cx - C =(A+B)x2+(B+C)x+(2AC)= (A + B)x^2 + (-B + C)x + (2A - C)

  3. Now, equate the coefficients of corresponding powers of xx from both sides of the equation:

    For x2x^2: A+B=0A + B = 0

    For x1x^1: B+C=1-B + C = 1

    For the constant term: 2AC=22A - C = 2

  4. Solve the system of equations:

    • From A+B=0A + B = 0, we have B=AB = -A.

    • Substitute B=AB = -A into B+C=1-B + C = 1, giving A+C=1A + C = 1, so C=1AC = 1 - A.

    • Substitute C=1AC = 1 - A into 2AC=22A - C = 2, giving: 2A(1A)=23A1=23A=3A=12A - (1 - A) = 2 \Rightarrow 3A - 1 = 2 \Rightarrow 3A = 3 \Rightarrow A = 1

    • Since A=1A = 1, B=1B = -1, and C=1A=0C = 1 - A = 0.

  5. Substitute back into the partial fraction form:

    x+2(x1)(x2+2)=1x1+xx2+2\frac{x+2}{(x-1)(x^2+2)} = \frac{1}{x-1} + \frac{-x}{x^2+2}

    This can be written as:

    x+2(x1)(x2+2)=1x1xx2+2\frac{x+2}{(x-1)(x^2+2)} = \frac{1}{x-1} - \frac{x}{x^2+2}

This is the partial fraction decomposition of the given expression.

Would you like any further details or have any questions?

Here are some related questions you might find interesting:

  1. How would you perform partial fraction decomposition if the denominator had repeated linear factors?
  2. What happens if the numerator's degree is higher than the denominator's?
  3. How can partial fractions be used in the integration of rational functions?
  4. What is the role of complex roots in partial fraction decomposition?
  5. Can you decompose a fraction if the denominator is already a sum of irreducible quadratic terms?

Tip: When decomposing fractions, always check the degree of the numerator and denominator. If the degree of the numerator is equal to or greater than the denominator, you'll need to perform polynomial division first.

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Math Problem Analysis

Mathematical Concepts

Partial Fractions
Algebra
Equations

Formulas

Partial fraction decomposition

Theorems

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Suitable Grade Level

Grades 11-12