Problem 53 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. $$\frac{2 x-9}{\left(x^{2}-x\right)^{2}}$$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

First, factor the denominator \((x^2 - x)^2\). We can factor it as \(x^2 - x = x(x-1)\). So, the denominator can be rewritten as \((x(x-1))^2\).

Step 2: Setup the Partial Fraction Structure

Since the denominator includes \((x(x-1))^2\), which is a square-term, we'll set up the partial fraction decomposition as:\[ \frac{2x-9}{(x(x-1))^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2}. \]

Step 3: Clear the Fractions by Multiplying the Denominator

Multiply through by the common denominator \((x^2 - x)^2 = (x(x-1))^2\) to eliminate the fractions:\[ 2x - 9 = A(x(x-1))^2/x + B(x(x-1))^2/x^2 + C(x(x-1))^2/(x-1) + D(x(x-1))^2/(x-1)^2. \]

Step 4: Simplify and Expand

Simplify the terms:\[ 2x - 9 = A(x(x-1)) + B(x-1)^2 + Cx^2 + Dx. \]Now, expand and simplify both sides for easier comparison.

Step 5: Equate and Solve for Coefficients

Equate the coefficients from both sides of the expanded equation for \(x^2\), \(x\), and the constant terms:- Coefficient of \(x^2\): \(A + C = 0\)- Coefficient of \(x\): \(D - A + 2B = 2\)- Constant term: \(-B - C = -9\)Solve these simultaneous equations to find the values of \(A\), \(B\), \(C\), and \(D\).

Step 6: Write the Decomposed Partial Fractions

Substitute the found values back into the partial fraction structure to write the decomposed partial fraction expression.

Key Concepts

Irreducible Quadratic Factor

When we discuss partial fraction decomposition, an irreducible quadratic factor is a polynomial of degree 2 that cannot be factored further over the real numbers. It is important to identify whether a quadratic factor is irreducible as it influences the setup of the partial fraction. In our exercise, the expression \((x^2 - x)^2\) can be factored into linear terms due to the presence of a common factor: \(x(x - 1)\). This means it isn't strictly an irreducible quadratic, but rather a product of linear factors. Understanding this helps in constructing the appropriate form for partial fraction decomposition.
In the case of irreducible quadratics, if we had a term like \(x^2 + 1\), which cannot be simplified over real numbers, it would be represented in its partial fraction form as \(\frac{Ax + B}{x^2 + 1}\).
Recognizing whether a quadratic factor can be simplified is crucial in formulating the right decomposition.

Factor the Denominator

Factoring the denominator is the first step in partial fraction decomposition. For this problem, the expression \((x^2 - x)^2\) needed simplification. By factoring, we found \(x^2 - x = x(x-1)\). So, the square of this product becomes \((x(x-1))^2\). This lays the groundwork for breaking the expression into simpler parts.
Why factor first? Because it shows the fundamental pieces that we will later assign variables (like A, B, C, D) in the partial fraction setup. Each unique factor contributes to a term in the final decomposition. Without factoring, we cannot accurately represent the original expression's structure.
Identifying such factors is a core skill. In some cases, you may encounter factors that require quadratic formula or other techniques to simplify, thereby increasing complexity.

Solve Simultaneous Equations

After clearing the fractions by multiplying through by the denominator, the next task is to solve for unknown constants. In our setup:\[2x - 9 = A(x^2 - x) + B(x - 1)^2 + Cx^2 + Dx\]We use simultaneous equations based on coefficients comparison to find A, B, C, and D. By expanding the terms, we gather equations for the coefficients of \(x^2\), \(x\), and constant terms:

• Coefficient of \(x^2\): \(A + C = 0\)
• Coefficient of \(x\): \(D - A + 2B = 2\)
• Constant term: \(-B - C = -9\)

Solve these equations together, often leveraging methods like substitution or elimination, to pinpoint each constant. This is crucial because each constant helps reconstruct the original expression in its decomposed form.
Learning to solve these can significantly aid in handling similar algebraic challenges effectively.

Equating Coefficients

Equating coefficients is a tactical approach used to identify the unknown variables in a partial fraction decomposition. Once we expand the newly expressed equation from the partial fraction setup, we compare each term's coefficients.This process involves lining up terms of similar degree on both sides of the equation, such as:

• Matching \(x^2\) terms on both sides gives \(A + C\). Here, these two should equal the coefficient of \(x^2\) on the left-hand side, which is 0.
• For \(x\) terms, determine \(D - A + 2B\) and set it equal to the \(x\) coefficient, in this case, 2.
• Finally, for constant terms, \(-B - C\) should equal the constant number in the expression, which corresponds to -9.

Once you find these relationships, solving them simultaneously gives values for A, B, C, and D. Once solved, these values can be plugged back into the expression to achieve successful partial fraction decomposition. Understanding and practicing equating coefficients can build a strong foundation for complex polynomial and algebra problems.