Problem 32 - Algebra and Trigonometry

Question

Find the partial fraction decomposition of the rational function. \(\frac{-2 x^{2}+5 x-1}{x^{4}-2 x^{3}+2 x-1}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{2x + 1}{x^2 + 1} - \frac{1}{x - 1}\).

Step 1: Check the Degree

Firstly, compare the degree of the numerator and the denominator. The degree of the numerator, \(-2x^2 + 5x - 1\), is 2. The degree of the denominator, \(x^4 - 2x^3 + 2x - 1\), is 4. Since the degree of the numerator is less than the degree of the denominator, we proceed with the partial fraction decomposition without any division.

Step 2: Factor the Denominator

Attempt to factor the polynomial in the denominator \(x^4 - 2x^3 + 2x - 1\). Using synthetic division or polynomial division, we find that \(x^4 - 2x^3 + 2x - 1 = (x^2 + 1)(x - 1)^2\). These are the factors to use in the decomposition.

Step 3: Set Up Partial Fraction Decomposition Form

Since the denominator factors into \((x^2 + 1)(x - 1)^2\), we set up the partial fraction format as: \(\frac{-2x^2 + 5x - 1}{(x^2 + 1)(x - 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}\).

Step 4: Clear the Denominator

Multiply both sides by the factored denominator \((x^2 + 1)(x - 1)^2\) to eliminate the denominators: \(-2x^2 + 5x - 1 = (Ax + B)(x - 1)^2 + C(x^2 + 1)(x - 1) + D(x^2 + 1)\).

Step 5: Expand and Collect Like Terms

Expand the right-hand side and collect like terms:- \((Ax + B)(x - 1)^2 = (Ax + B)(x^2 - 2x + 1)\) expands to \(Ax^3 - 2Ax^2 + Ax + Bx^2 - 2Bx + B)\)- \(C(x^2 + 1)(x - 1) = C(x^3 - x^2 + x - 1)\)- \(D(x^2 + 1) = Dx^2 + D\)Collect all these to form a single polynomial.

Step 6: Solve for Coefficients

Compare coefficients on both sides of the equation to solve for \(A\), \(B\), \(C\), and \(D\): 1. For \(x^3\) terms set coefficient equal to zero as \(-2x^3 + 0x^2 + 0x + 0\) implies no \(x^3\) component.2. Continue solving: \(A - 2C = 0\) \(-2A + B + C + D = -2\) \(A - 2B + C = 5\) \(B - C + D = -1\).Solve the linear system to find: \(A = 2\), \(B = 1\), \(C = -1\), \(D = 0\).

Step 7: Write the Decomposition

Substitute the values of \(A\), \(B\), \(C\), and \(D\) back into the partial fraction format: \(\frac{-2x^2 + 5x - 1}{(x^2 + 1)(x - 1)^2} = \frac{2x + 1}{x^2 + 1} - \frac{1}{x - 1}\).

Key Concepts

Rational Functions

A rational function is expressed as the ratio of two polynomials. In mathematics, it is expressed in the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials. The degree of a polynomial is the highest power of the variable within it. Understanding the degree of both the numerator and the denominator is crucial in partial fraction decomposition.
Partial fraction decomposition involves breaking down a complex rational function into simpler fractions, making integration or further algebraic manipulation easier. If the degree of the numerator is less than the degree of the denominator, directly proceed to partial fraction decomposition without any need for division.
This technique helps in simplifying the rational function into more manageable expressions, especially when further mathematical analyses or integrations are required.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials. It is especially useful when dividing by a linear factor of the form \(x-c\). This method streamlines the process and reduces the time spent compared to long division.
By using synthetic division in our exercise, we transformed \(x^4 - 2x^3 + 2x - 1\) into \((x^2 + 1)(x - 1)^2\). This step is essential for factoring polynomials, thus aiding in the decomposition of a rational function into partial fractions. Unlike polynomial division, synthetic division does not require writing out the variables, which simplifies the arithmetic involved.

Polynomial Division

Polynomial division, akin to long division of numbers, involves dividing one polynomial by another. It helps in simplifying the expression and is vital when the degree of the numerator is equal to or greater than the denominator. This divides the polynomial into a quotient and a remainder.
In our solution, before decomposing into partial fractions, it's key to determine whether the polynomial division is needed. Since the numerator degree is lower than the denominator, we skipped this process. However, in cases where it's needed, polynomial division prepares the rational expression adequately for further decomposition.

Linear System of Equations

A linear system of equations is a collection of one or more linear equations involving the same set of variables. Solving these systems is crucial for finding unknown constants in partial fraction decomposition.
In the partial fraction setup, the equation \(-2x^2 + 5x - 1 = (Ax + B)(x - 1)^2 + C(x^2 + 1)(x - 1) + D(x^2 + 1)\) leads to several equations based on coefficients of \(x\). Solving the system:

• \(A - 2C = 0\)
• \(-2A + B + C + D = -2\)
• \(A - 2B + C = 5\)
• \(B - C + D = -1\)

Single out specific values for \(A, B, C,\) and \(D\). These values are then substituted back, completing the decomposition. This step is vital for deriving the simplest form of each component in the rational function.