Problem 26 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{9 x^{2}-9 x+6}{2 x^{3}-x^{2}-8 x+4} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{-2}{x-1} + \frac{1}{x+2} + \frac{5}{2x-2} \).

Step 1: Factor the Denominator

First, factor the polynomial in the denominator, which is \(2x^{3} - x^{2} - 8x + 4\). Begin by looking for possible roots of the denominator's polynomial using methods like the Rational Root Theorem or synthetic division.

Step 2: Identify Real Roots and Factor Further

Assume \(x=1\) is a root. By synthetic division or substitution, confirm this. Perform polynomial division to factor the denominator further based on the known root.

Step 3: Factor Completely

After identifying \(x=1\) is a root, divide the original polynomial by \(x-1\) (using synthetic or long division) to find that \(2x^3 - x^2 - 8x + 4 = (x-1)(2x^2 + x - 4)\). Further factor \(2x^2 + x - 4\) to \((x+2)(2x-2)\), leaving us \((x-1)(x+2)(2x-2)\) as the complete factorization.

Step 4: Setup Partial Fractions

Now, express the original rational function as a sum of fractions with the factored denominator. Write it as \(\frac{9x^2-9x+6}{(x-1)(x+2)(2x-2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{2x-2}\).

Step 5: Solve for Coefficients

Multiply through by the denominator to clear the fractions: \(9x^2 - 9x + 6 = A(x+2)(2x-2) + B(x-1)(2x-2) + C(x-1)(x+2)\). Equate coefficients on both sides of the equation, then solve the system of equations for \(A\), \(B\), and \(C\).

Step 6: Write the Decomposition

Once \(A\), \(B\), and \(C\) are found, substitute them back to write the full partial fraction decomposition of the function.

Key Concepts

Rational Functions

Rational functions are expressions formed by dividing two polynomials. A polynomial is simply a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In a rational function, the variable is part of the numerator or the denominator of a fraction.

Consider the function given in the exercise:

• Numerator: \(9x^2 - 9x + 6\)
• Denominator: \(2x^3 - x^2 - 8x + 4\)

The main goal here is to break down the function into simpler fractions, called partial fractions, which are easier to work with, especially in integration and solving differential equations. Partial fraction decomposition represents a rational function as a sum of simpler fractions, making complex algebra operations more manageable.

These fractions often reveal important characteristics about the function, such as its asymptotes and potential points of discontinuity. To perform partial fraction decomposition, the denominator must be factored correctly, as the form of the decomposition is dictated by its factors.

Polynomial Factoring

Polynomial factoring is the process of breaking down a polynomial into its simplest 'factor' components, which are polynomials of lower degrees. This is a fundamental skill in algebra, vital for partial fraction decomposition.

In our exercise, the polynomial in the denominator is \(2x^3 - x^2 - 8x + 4\). Factoring begins with finding the roots of the polynomial using methods such as:

• Rational Root Theorem
• Trial and error substitution
• Synthetic division

Once a root is identified, the polynomial can be divided by the corresponding linear factor, as seen with the factor \((x-1)\) in our denominator. After confirming \(x = 1\) as a root, we use polynomial division to divide the entire polynomial by \(x-1\), simplifying it further. Eventually, the denominator was completely factored to \((x-1)(x+2)(2x-2)\).

These factors facilitate the construction of partial fractions. Each factor from the denominator becomes the denominator of one of the new simpler fractions, crucial for setting up the partial fraction decomposition.

Synthetic Division

Synthetic division is a streamlined version of long division specific to polynomials, especially useful when performing polynomial operations. It's a quicker, more straightforward process, particularly effective when dividing polynomials by linear factors.In our example, after suspecting that \(x=1\) is a root, synthetic division helps to confirm this assumption. The steps involved are:

• Using the suspected root (1 in this case) as the divisor in the synthetic division setup
• Arranging coefficients of the polynomial: 2, -1, -8, and 4
• Performing the synthetic division to confirm that the result gives a remainder of zero

The zero remainder confirms that \(x-1\) is indeed a factor of the polynomial. Once the divisor successfully divides the polynomial, synthetic division gives us the reduced polynomial quotient, which can be further factored if necessary.

This method is appreciated for its efficiency, saving time and reducing chances for errors compared to the more verbose polynomial long division. Synthetic division is a must-know tool for algebra students tackling polynomial division and partial fraction decomposition.