Problem 32 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. $$\frac{4 x^{2}+9 x+23}{(x-1)\left(x^{2}+6 x+11\right)}$$

Step-by-Step Solution

Answer

Partial fraction decomposition is \(\frac{2}{x-1} + \frac{2x - 1}{x^2 + 6x + 11}\).

Step 1: Denote Partial Fraction Decomposition

We are given: \( \frac{4x^2 + 9x + 23}{(x-1)(x^2 + 6x + 11)} \). To decompose into partial fractions, assume:\[\frac{4x^2 + 9x + 23}{(x-1)(x^2 + 6x + 11)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 6x + 11}\]where \(A\), \(B\), and \(C\) are constants to be determined.

Step 2: Multiply Both Sides by the Denominator

Multiply both sides by the denominator \((x-1)(x^2 + 6x + 11)\) to clear the fractions:\[4x^2 + 9x + 23 = A(x^2 + 6x + 11) + (Bx + C)(x-1)\]

Step 3: Expand the Equation

Expand both terms on the right side:- \(A(x^2 + 6x + 11) = Ax^2 + 6Ax + 11A\)- \((Bx + C)(x - 1) = Bx^2 + (C - B)x - C\)Thus, the equation is:\[4x^2 + 9x + 23 = Ax^2 + 6Ax + 11A + Bx^2 + (C - B)x - C\]

Step 4: Combine Like Terms

Combine like terms on the right side:\[(A + B)x^2 + (6A + C - B)x + (11A - C)\]

Step 5: Compare Coefficients

Compare the coefficients of corresponding terms on both sides:1. For \(x^2\): \(A + B = 4\)2. For \(x\): \(6A + C - B = 9\)3. For the constant term: \( 11A - C = 23\)

Step 6: Solve the System of Equations

Solve the system of linear equations obtained in Step 5:1. \(A + B = 4\)2. \(6A + C - B = 9\)3. \(11A - C = 23\)By substitution or elimination, find values of \(A, B,\) and \(C\):- From (3), \(C = 11A - 23\)- Substitute \(C\) in (2): \[6A + (11A - 23) - B = 9\] \[17A - B = 32\]Using (1) and the above:- \(A + B = 4\)- \(B = 17A - 32\)Substitute into the 1st equation:- \(A + 17A - 32 = 4\)- \(18A = 36\)- \(A = 2\)Using \(A = 2\) to find \(B\) and \(C\):- \(B = 17(2) - 32 = 2\)- \(C = 11(2) - 23 = -1\)

Step 7: Write the Decomposition

Substitute the values found back into the decomposition:\[\frac{4x^2 + 9x + 23}{(x-1)(x^2 + 6x + 11)} = \frac{2}{x-1} + \frac{2x - 1}{x^2 + 6x + 11}\]

Key Concepts

Irreducible Quadratic Factors

In partial fraction decomposition, one important concept is identifying irreducible quadratic factors. But what does irreducible mean? In the simplest terms, an irreducible quadratic factor cannot be factorized further into real linear factors. For example, given a quadratic expression like \(x^2 + 6x + 11\), you cannot break it down further using real numbers.
This happens because the discriminant \(b^2 - 4ac\) in the quadratic formula is negative. Here, for \(x^2 + 6x + 11\), the discriminant is \(6^2 - 4 \cdot 1 \cdot 11 = 36 - 44 = -8\), meaning there are no real roots. Thus, it's "irreducible" over the real numbers.
Understanding irreducible quadratic factors is crucial because they determine how we set up our partial fractions. For an irreducible factor, we use a linear expression in the numerator, such as \(Bx + C\), rather than a single constant \(A\) as you would with linear factors like \(x-1\). This setup allows us to correctly decompose the fraction and solve for unknown constants.

System of Equations

During partial fraction decomposition, determining the constants requires solving a system of equations. This involves setting up equations based on the coefficients of the expanded polynomial. Let's dive into how this works.
When we expand the equation, we equate the coefficients of the polynomial on both sides, namely the given polynomial's coefficients and those coming from our assumed decomposition. This gives us a set of equations to solve.
For instance, in the expression \(4x^2 + 9x + 23 = (A + B)x^2 + (6A + C - B)x + (11A - C)\), comparing like terms provides us with these equations:

• \(A + B = 4\)
• \(6A + C - B = 9\)
• \(11A - C = 23\)

These equations form a system of linear equations, which we can solve using various methods such as substitution or elimination. Solving these allows us to find the values of \(A\), \(B\), and \(C\), which are crucial for accurately forming the decomposed expression.

Coefficients Comparison

Coefficients comparison is an essential step for successfully decomposing a fraction into partial fractions. It involves lining up corresponding powers and constants from both the original polynomial and the expanded form of our assumed partial fractions. This helps us find equations that link unknowns to known coefficients.
For a fraction like \(\frac{4x^2 + 9x + 23}{(x-1)(x^2 + 6x + 11)}\), after assuming \(\frac{A}{x-1} + \frac{Bx + C}{x^2 + 6x + 11}\), we expanded and organized terms to compare coefficients of terms like \(x^2\), \(x\), and the constant.
Consider these specific comparisons:
- For \(x^2\) terms: comparing gives \(A + B = 4\).- For \(x\) terms: leads to \(6A + C - B = 9\).- For constants: \(11A - C = 23\).
With these equations, we can see how the original function's structure must match our partially decomposed version in each degree and constant term, offering a direct path to solve \(A\), \(B\), and \(C\). This powerful method makes coefficients comparison a cornerstone of breaking down complex fractions.