Problem 44 - College Algebra

Question

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. $$\frac{3 x^{3}+2 x^{2}+14 x+15}{\left(x^{2}+4\right)^{2}}$$

Step-by-Step Solution

Answer

\(\frac{3x+2}{x^2+4} + \frac{2x+7}{(x^2+4)^2}\)

Step 1: Identify the Denominator

The denominator is \((x^2 + 4)^2\). This is a repeated irreducible quadratic factor.

Step 2: Set Up the Partial Fraction Decomposition

For an irreducible repeated quadratic factor \((x^2 + 4)^2\), the decomposition form will be \(\frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2}\).

Step 3: Express the Numerator in Terms of the Partial Fractions

Write \(3x^3 + 2x^2 + 14x + 15\) as \((Ax + B)(x^2 + 4) + (Cx + D)\) and equate the coefficients to form an equation.

Step 4: Expand the Equation

Expand the equation to get \(Ax^3 + Bx^2 + 4Ax + 4B + Cx + D\). Simplify it to \(Ax^3 + Bx^2 + (4A + C)x + (4B + D)\).

Step 5: Match Coefficients

Equate the coefficients on both sides: - For \(x^3\): \(A = 3\)- For \(x^2\): \(B = 2\)- For \(x\): \(4A + C = 14\)- For constant term: \(4B + D = 15\).

Step 6: Solve the System of Equations

Solve the equations sequentially:- From \(A = 3\), we substitute into \(4A + C = 14\) to solve for \(C\). - \(12 + C = 14\) - \(C = 2\)- From \(B = 2\), we substitute into \(4B + D = 15\) to solve for \(D\). - \(8 + D = 15\) - \(D = 7\).

Step 7: Write the Final Partial Fraction Decomposition

Using the solved coefficients \(A = 3\), \(B = 2\), \(C = 2\), \(D = 7\), write the partial fraction decomposition as:\[\frac{3x + 2}{x^2 + 4} + \frac{2x + 7}{(x^2 + 4)^2}\]

Key Concepts

Irreducible Quadratic Factor

An irreducible quadratic factor is a quadratic polynomial that cannot be factored further over the real numbers. In the context of real coefficients, this means the polynomial has no real roots, and you cannot break it down into two linear factors. For example, in the denominator \( (x^2 + 4)^2 \), \( x^2 + 4 \) is irreducible because it does not have any real roots (since the equation \( x^2 = -4 \) has no solution in real numbers).
This concept is important in partial fraction decomposition because irreducible quadratic factors in the denominator need to be handled differently than linear factors. They require us to set up the partial fraction with both a linear term (\( Ax + B \) in the numerator) over each irreducible factor. This ensures that we're accounting for all possible polynomial terms that could exist in the numerator.

Repeating Factors

Repeating factors are factors that appear more than once in the denominator of a fraction. In our problem, the denominator is \( (x^2 + 4)^2 \), where \( x^2 + 4 \) is repeated. When dealing with repeating irreducible quadratic factors, we must include several terms in the partial fraction decomposition, each with increasing powers of these factors.
So, for \( (x^2 + 4)^2 \), we must include decomposed terms for both \( \frac{Ax + B}{x^2 + 4} \) and \( \frac{Cx + D}{(x^2 + 4)^2} \). This inclusion ensures all possible ways the polynomial in the numerator could interact with the repeated denominator powers are considered.
This extra step of adding terms for each power of the repeating factor is crucial in capturing the full behavior of the rational expression.

Polynomial Division

Polynomial division is a method used to divide one polynomial by another, leading to a quotient and a remainder. Although not directly used in the provided problem, understanding it is essential when performing partial fraction decomposition, especially when the degree of the numerator is higher than that of the denominator.
If you encounter a fraction with a higher-degree numerator, you would perform polynomial division first to simplify the expression and reduce the degree of the numerator, which can then be handled by partial fraction decomposition.
In our example, the degree of the numerator (3) is less than the degree of the denominator (4), so polynomial division wasn't necessary. However, remembering this step ensures you're prepared for any scenario where it might be required.

Coefficient Comparison

Coefficient comparison is a technique used to solve for unknowns in partial fraction decomposition. Once you've set up your partial fraction form, you expand the expression and then write it such that it matches the original polynomial (numerator).
In our solution, after setting the partial fractions, we expanded both sides, resulting in an equation where each term on the left must be equal to its corresponding term on the right. For example, when expanding \[ (Ax + B)(x^2 + 4) + (Cx + D) \]we equate coefficients of like terms in \( 3x^3 + 2x^2 + 14x + 15 \).
- For \( x^3 \), the coefficient \( A \) must equal 3.- For \( x^2 \), the coefficient \( B \) equals 2.- For \( x \), solve \( 4A + C = 14 \).- For the constant term, solve \( 4B + D = 15 \).Using coefficient comparison allows you to systematically solve for unknowns, ensuring that every aspect of the original expression is matched.