Problem 35 - College Algebra with Corequisite Support

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor. $$ \frac{4 x^{2}+17 x-1}{(x+3)\left(x^{2}+6 x+1\right)} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{2}{x+3} + \frac{2x-1}{x^2+6x+1}\).

Step 1: Identify the Factors

Identify the factors of the denominator. In this case, the denominator is \((x+3)(x^2+6x+1)\), which includes a linear factor \((x+3)\) and an irreducible quadratic factor \((x^2+6x+1)\).

Step 2: Set Up the Partial Fraction Form

Write the partial fraction decomposition form: \[\frac{4x^2 + 17x - 1}{(x+3)(x^2+6x+1)} = \frac{A}{x+3} + \frac{Bx + C}{x^2+6x+1}\]where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides by the denominator \((x+3)(x^2+6x+1)\) to clear the fractions:\[4x^2 + 17x - 1 = A(x^2 + 6x + 1) + (Bx + C)(x + 3)\]

Step 4: Expand and Simplify

Expand the right-hand side of the equation:\[A(x^2 + 6x + 1) = Ax^2 + 6Ax + A\]\[(Bx + C)(x + 3) = Bx^2 + 3Bx + Cx + 3C\]Combine like terms:\[= (A + B)x^2 + (6A + 3B + C)x + (A + 3C)\]

Step 5: Match Coefficients

Equate the coefficients of \(x^2\), \(x\), and the constant term from both sides:1. For \(x^2\): \(A + B = 4\)2. For \(x\): \(6A + 3B + C = 17\)3. For the constant: \(A + 3C = -1\)

Step 6: Solve the System of Equations

From the system of equations obtained in Step 5:1. \(A + B = 4\)2. \(6A + 3B + C = 17\)3. \(A + 3C = -1\)Solve for \(A\), \(B\), and \(C\) using substitution or elimination methods.

Step 7: Calculate and Verify Solutions

Solve the equations:1. From \(A + B = 4\), express \(B = 4 - A\).2. Substitute into \(6A + 3B + C = 17\): \(6A + 3(4 - A) + C = 17\) \(3A + C = 5\)3. From \(A + 3C = -1\), solve for \(C\): Subtract the two equations (3 - step): \(3A + C = 5\) \(A + 3C = -1\) Solving these gives \(A = 2\), \(B = 2\), \(C = -1\).Verify by substituting back to check if all initial conditions match.

Key Concepts

Irreducible Quadratic Factor

When dealing with partial fraction decomposition, one crucial concept is the irreducible quadratic factor. An irreducible quadratic factor is a quadratic expression that cannot be factored over the real numbers into linear terms, meaning it has no real roots. In our example, the term \(x^2 + 6x + 1\) is quadratic; however, upon examining its discriminant (found using \(b^2 - 4ac\)), you will see it cannot produce real roots because this value is negative. Thus, it remains in its quadratic form.

Irreducible quadratics in the partial fraction decomposition method compel us to pair them with linear numerators, such as \(Bx + C\) in our solution. These numerators are flexible enough to account for all linear scenarios within the quadratic's presence in the larger polynomial. This setup helps analyze complex rational expressions by breaking them down into simpler, more manageable components.

System of Equations

In partial fraction decomposition, setting up a system of equations is key to finding the coefficients for each decomposed term. For our specific exercise, once you have the expanded form of the partial fraction, collecting terms with similar powers of \(x\) allows you to create individual equations by matching coefficients.

For instance, through our decomposition process, we derived three equations based on the coefficients:

• \(A + B = 4\) derived from the coefficient of \(x^2\)
• \(6A + 3B + C = 17\) from the coefficient of \(x\)
• \(A + 3C = -1\) pertaining to the constant term

These equations represent a system of linear equations that can be solved using substitution or elimination methods, guiding you to determine the values of \(A\), \(B\), and \(C\). The consistency of these values ensures the integrity of the decomposed fractions, aligning them with the original expression.

Polynomial Long Division

Polynomial long division is often used before applying partial fraction decomposition, especially when the numerator's degree is equal to or higher than the denominator's. However, in this problem, our preliminary factorization allows us directly to set up partial fractions without immediate division.

When a polynomial has already been factored into identifiable parts, partial fraction decomposition skips polynomial long division. However, understanding polynomial long division remains useful, particularly when dealing with more complex numerators or when simplifying higher degree rational expressions. This division helps reduce complicated polynomials into simpler terms, allowing for an easier application of partial fraction methods. Such simplification ensures our decomposed fractions maintain manageable and solvable forms, ultimately aligning with our mathematical goals and providing clarity throughout the process.