Problem 34 - College Algebra

Question

Find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. \(\frac{x^{2}+3 x+1}{(x+1)\left(x^{2}+5 x-2\right)}\)

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{1}{6(x+1)} + \frac{5x/6 + 4/3}{x^2+5x-2}\).

Step 1: Identify the Factors of the Denominator

The given function is \( \frac{x^{2}+3x+1}{(x+1)(x^{2}+5x-2)} \). Here, the denominator is already factored into \((x+1)(x^{2}+5x-2)\). The first factor \((x+1)\) is linear, and the second factor \((x^{2}+5x-2)\) is an irreducible quadratic.

Step 2: Set Up the Partial Fraction Decomposition

Since we have a linear factor \((x+1)\) and an irreducible quadratic factor \((x^{2}+5x-2)\), the partial fraction decomposition will be of the form:\[\frac{x^{2}+3x+1}{(x+1)(x^{2}+5x-2)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+5x-2}.\]Here, \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Combine the Fractions

To solve for \(A\), \(B\), and \(C\), combine the right side into a single fraction:\[\frac{A(x^{2}+5x-2) + (Bx+C)(x+1)}{(x+1)(x^{2}+5x-2)}.\]The numerators must be equal, so we equate:\[A(x^{2}+5x-2) + (Bx+C)(x+1) = x^{2}+3x+1.\]

Step 4: Expand and Collect Like Terms

Expand the equation:\[A(x^{2}+5x-2) + (Bx+C)(x+1) = Ax^{2} + 5Ax - 2A + Bx^{2} + Bx + Cx + C.\]Combine like terms:\[(A + B)x^{2} + (5A + B + C)x + (-2A + C).\]

Step 5: Set Up System of Equations

Equate the coefficients from the expanded polynomial and the numerator \(x^{2}+3x+1\):1. \(A + B = 1\)2. \(5A + B + C = 3\)3. \(-2A + C = 1\)

Step 6: Solve the System of Equations

From equation 1: \(A + B = 1\), so \(B = 1 - A\).Substitute \(B = 1 - A\) into equation 2:\[5A + (1 - A) + C = 3\] gives \[4A + C = 2\].Now address equation 3: \(-2A + C = 1\).Solve simultaneously:1. \(4A + C = 2\)2. \(-2A + C = 1\)Subtract equation 2 from 1: \(6A = 1\) so \(A = \frac{1}{6}\).Substitute \(A\) back into \(B = 1 - A\) gives \(B = \frac{5}{6}\).Substitute \(A\) into \(-2A + C = 1\): \(-\frac{2}{6} + C = 1\) gives \(C = \frac{8}{6} = \frac{4}{3}\).

Step 7: Write the Final Partial Fraction Decomposition

Substitute \(A = \frac{1}{6}\), \(B = \frac{5}{6}\), and \(C = \frac{4}{3}\) back into the decomposition:\[\frac{x^{2}+3x+1}{(x+1)(x^{2}+5x-2)} = \frac{1/6}{x+1} + \frac{(5/6)x + 4/3}{x^{2}+5x-2}.\]

Key Concepts

Quadratic Factor

A quadratic factor is a polynomial of degree two, which can often be expressed in the form \(ax^2 + bx + c\). In the context of partial fraction decomposition, it is essential to know whether such a factor can be further factored into linear terms or not. If not, it is termed as irreducible. In the problem, the quadratic factor \(x^2 + 5x - 2\) is irreducible within the real numbers, because it cannot be factored into simpler polynomials with real coefficients.
In partial fraction decomposition, the presence of a quadratic factor affects how we structure the decomposition itself. It prompts us to include terms of the form \(\frac{Bx+C}{ax^2+bx+c}\), where \(B\) and \(C\) are constants to be determined. This differs from linear factors which only require a single constant in the numerator.

Irreducible Quadratic

An irreducible quadratic is a polynomial of degree two that cannot be factored into linear terms over the real number set. For example, \(x^2 + 5x - 2\) is considered irreducible because there's no way to write it as a product of two first-degree polynomials with real coefficients. This property is significant when setting up partial fraction decomposition because it dictates how we represent fractions involving these factors.
The partial fraction form for an irreducible quadratic \(ax^2 + bx + c\) is \(\frac{Bx+C}{ax^2+bx+c}\). It's crucial to have both a linear term and a constant term \((Bx+C)\) in the numerator, because the quadratic factor allows these. To determine \(B\) and \(C\), we later use algebraic steps such as expanding the polynomial and comparing coefficients.

System of Equations

When performing partial fraction decomposition, particularly when more than one term requires a numerator with multiple coefficients, we end up with a system of equations. This happens because we expand the fractions and equate them to the original polynomial to balance both sides of the equation.
In this exercise, setting up the system meant equating coefficients from both the numerators. We had:

• \(A + B = 1\)
• \(5A + B + C = 3\)
• \(-2A + C = 1\)

The solution to this system involves expressing one of the variables in terms of another and substituting it into the next equation. This process continues until all variables are determined. Solving such systems is crucial to correctly carrying out partial fraction decomposition with irreducible quadratics and ensures that the transformation is accurately adjusted to reflect the original polynomial.

Linear Factor

A linear factor in the context of polynomials is an expression of the form \(ax + b\) where the highest power of the variable \(x\) is one. In our problem, the term \(x+1\) behaves as the linear factor in the denominator. Linear factors are the simplest kind to deal with in partial fraction decomposition. Unlike quadratic factors, they are represented by a single constant term in the numerator—for instance, \(\frac{A}{x+1}\).
To find \(A\), we first set up the partial fraction decomposition and then solve using algebraic methods, such as combining fractions and comparing terms. Handling linear factors becomes particularly straightforward, as they involve solving simple equations that directly yield the constant coefficients required for the decomposition.