Problem 53 - College Algebra

Question

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

Decompose \(\frac{2x-9}{(x(x-1))^2}\) into partial fractions using coefficients \(A, B, C, D, E, F\).

Step 1: Factor the Denominator

The given denominator is \((x^2 - x)^2\). First, factor out the common term in the expression \(x^2 - x\) to get \(x(x-1)\). So the denominator becomes \((x(x-1))^2\) or \(x^2(x-1)^2\).

Step 2: Set Up Partial Fraction Decomposition

Since the denominator contains a repeated quadratic factor, we set up the partial fraction decomposition as follows:\[\frac{2x-9}{(x^2(x-1)^2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{(x-1)} + \frac{Ex+F}{(x-1)^2}\] where \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) are constants to be determined.

Step 3: Clear the Fractions

Multiply both sides of the equation by the common denominator \(x^2(x-1)^2\) to clear the fractions:\[2x - 9 = A(x(x-1)^2) + B((x-1)^2) + (Cx+D)x(x-1) + (Ex+F)x^2\]

Step 4: Expand and Collect Terms

Expand the right-hand side of the equation:- For \(A(x(x-1)^2)\), expand and simplify.- For \(B((x-1)^2)\), expand and simplify.- For \((Cx+D)x(x-1)\), expand and simplify.- For \((Ex+F)x^2\), expand and simplify.Collect similar terms to form a polynomial expression in terms of powers of \(x\).

Step 5: Solve for Coefficients

Comparing the coefficients of the polynomial on both sides of the equation, solve for the unknown constants \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\). This involves solving a system of equations for each coefficient of \(x^i\) for distinct powers of \(x\).

Step 6: Substitute Values Back

Once you have calculated the values for \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\), substitute them back into the partial fraction decomposition to express the fraction in its decomposed form.

Key Concepts

Repeated Quadratic Factor

When dealing with partial fraction decomposition, a repeated quadratic factor in the denominator can make things more challenging.
In simpler terms, using repeated quadratic factors requires considering multiple fractions for each repeating factor. In this scenario, \(x^2(x-1)^2\) is our denominator, comprising the repeated quadratic factor \(x-1\) squared.
For each quadratic factor that appears r times, we highlight these points:

• If a quadratic \(ax^2+bx+c\) appears 2 times (as in this case with \(x-1\)), you set up two different terms for the decomposition.
• The terms will be in the form of \(\frac{Cx+D}{ax^2+bx+c}\) and \(\frac{Ex+F}{(ax^2+bx+c)^2}\).
• These extra terms ensure that the decomposition accounts for all parts of the repeated factor.

Understanding this concept is vital as it ensures the decomposition captures the structure of the polynomial, allowing for easier integration or simplification later on.

Polynomial Expansion

Expanding polynomials is the process of multiplying terms together to get them into a standard polynomial form.
During partial fraction decomposition, once you've set up your proper fractions, expanding these is crucial for finding common terms.
Some essential points about polynomial expansion:

• Every term in the polynomial must be expanded correctly to ensure that when summed, they reconstruct the original polynomial.
• Since you work on multiple terms such as \(A(x(x-1)^2)\), \(B((x-1)^2)\), and so forth, it becomes imperative to handle each expansion meticulously.
• The primary purpose is to simplify and then compare with the numerator of the given term to establish equality in the system of equations you'll solve.

Accurately expanded polynomials make it straightforward to identify coefficients, leading to the solution for unknowns needed in decomposition.

Coefficients in Polynomials

Coefficients play a critical role in obtaining the final decomposition form in partial fraction decomposition.
They are the numerical factors attached to terms in a polynomial that determine the polynomial's shape and value for given inputs.
Key concepts include:

• Once polynomials are expanded, the problem shifts to matching coefficients from both sides of the equation.
• For instance, if your original expression's numerator was \(2x - 9\), after expanding both sides, you'll set up equations such that the sum of terms involving \(x^i\) equals 2 or \-9\ depending on the power of \(x\).
• This process might involve solving for multiple unknowns (A, B, C, etc.), each representing coefficients that you equate across the original and expanded forms.

Understanding how coefficients interact in different polynomial terms is essential, allowing us to solve the system of equations effectively, thereby achieving a successful decomposition.