Problem 34 - College Algebra

Question

For the following exercises, 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

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

Step 1: Identify the Denominator Structure

The denominator of the given function is \((x+1)(x^2 + 5x - 2)\). This consists of a linear factor \((x+1)\) and an irreducible quadratic factor \((x^2 + 5x - 2)\). We will decompose the fraction based on these terms.

Step 2: Set Up the Partial Fraction Form

Since the denominator includes both a linear and an irreducible quadratic factor, we set up the partial fraction as follows:\[\frac{x^2 + 3x + 1}{(x+1)(x^2 + 5x - 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 5x - 2}\] where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Multiply to Clear the Denominator

Multiply both sides by the denominator \((x+1)(x^2 + 5x - 2)\) to eliminate the fractions:\[x^2 + 3x + 1 = A(x^2 + 5x - 2) + (Bx + C)(x+1)\]

Step 4: Expand and Simplify

Expand the right side to combine like terms:\[A(x^2 + 5x - 2) = Ax^2 + 5Ax - 2A\]\[(Bx + C)(x+1) = Bx^2 + Bx + Cx + C = Bx^2 + (B + C)x + C\]Combine:\[x^2 + 3x + 1 = (A + B)x^2 + (5A + B + C)x + (-2A + C)\]

Step 5: Equate Coefficients

Compare coefficients of both sides to set up equations:- For \(x^2\): \(A + B = 1\)- For \(x\): \(5A + B + C = 3\)- Constant term: \(-2A + C = 1\)

Step 6: Solve the System of Equations

Solve these equations:1. \(A + B = 1\)2. \(5A + B + C = 3\)3. \(-2A + C = 1\)From (1): \(B = 1 - A\).Substitute \(B\) in (2):\[5A + (1 - A) + C = 3 \4A + C = 2 \C = 2 - 4A\]Substitute \(C\) in (3):\[-2A + (2 - 4A) = 1 \-6A = -1 \A = \frac{1}{6}\]Now, find \(B\):\[B = 1 - \frac{1}{6} = \frac{5}{6}\]Finally, \(C\):\[C = 2 - 4\left(\frac{1}{6}\right) = 2 - \frac{2}{3} = \frac{4}{3}\]

Step 7: Write the Decomposed Partial Fractions

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

Key Concepts

Irreducible Quadratic Factor

In partial fraction decomposition, you might face polynomials that do not factor into real linear factors. These polynomials are known as irreducible quadratic factors. They have the form \(ax^2 + bx + c\) and there are no real numbers that will factor them into simpler polynomials. To work with these, you need to recognize them in the denominator of the function. For example, in the expression \(\frac{x^2 + 3x + 1}{(x+1)(x^2 + 5x - 2)}\), \(x^2 + 5x - 2\) is the irreducible quadratic factor.

When setting up partial fractions for these cases, ensure to use a linear term (such as \(Bx + C\)) over the quadratic factor.

Linear Factor

A linear factor is a simple polynomial of the first degree, such as \(x + 1\). These are the simplest type of factors you'll manage in partial fraction decomposition. When faced with a rational expression, identify linear factors easily because they take the form \(x + a\). For the expression in our exercise, \(x + 1\) is the linear factor of the denominator.

When you decompose your expression into partial fractions, you'll set up a form like \(\frac{A}{x + 1}\) for each linear factor. "A" will be a constant you'll need to determine through solving.

System of Equations

Solving for the constants in partial fraction decomposition often requires setting up and solving a system of equations. Once you've expressed your partial fraction into a sum based on the factors, multiplying through by the original denominator clears the fractions and leaves you with a polynomial equation.

In our exercise, multiplying gives us the equation:
\(x^2 + 3x + 1 = A(x^2 + 5x - 2) + (Bx + C)(x+1)\).
This is expanded and simplified to obtain a system of equations:

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

These equations represent the coefficients of corresponding terms on either side of the equation. Solving them gives the values for \(A\), \(B\), and \(C\).

Coefficient Comparison

The technique of coefficient comparison is crucial in partial fraction decomposition. After aligning terms during expansion, both sides of the equation should look similar. You then compare like terms (coefficients) to form equations.

For example, considering the polynomial equation \(x^2 + 3x + 1 = (A + B)x^2 + (5A + B + C)x + (-2A + C)\), we set up equations based on:

• \(x^2\) term: \(A + B = 1\)
• \(x\) term: \(5A + B + C = 3\)
• Constant term: \(-2A + C = 1\)

By solving these, you find the necessary constant values. Coefficient comparison is a powerful method to equate and determine coefficients systematically, leading to finding the partial fraction components.