Problem 43 - College Algebra

Question

Perform each long division and write the partial fraction decomposition of the remainder term. $$\frac{x^{5}+2}{x^{2}-1}$$

Step-by-Step Solution

Answer

The division quotient is \(x^3\) and the partial fraction decomposition of the remainder is \(\frac{A}{x - 1} + \frac{B}{x + 1}\), with \(A\) and \(B\) being computed as described in the steps above.

Step 1: Perform Long Division

To perform the long division, \(x^{5}\) is divided by \(x^{2}\), resulting in \(x^{3}\). This means \(x^{3}\) is the first term of the quotient. Subtract \(x^{3}\) multiplied by \(x^{2} - 1\) from the dividend to find the remainder. Repeat this process until the degree of the remainder is less than the degree of the divisor. The final quotient is the result of the division, and the remainder will be decomposed into partial fractions in the next step.

Step 2: Decompose the Remainder into Partial Fractions

Given that the divisor \(x^{2} -1\) can be factored as \(x - 1\) and \(x + 1\), the partial fraction decomposition of the remainder \(2x^{2} + x + 2\) can be written as \(\frac{A}{x - 1} + \frac{B}{x + 1}\), where \(A\) and \(B\) are constants to be determined. To find these constants, multiply both sides of the equation by the common denominator \(x^2 - 1\), followed by equating coefficients and solving the system of equations to find the values of \(A\) and \(B\).

Step 3: Conclusion

Finally, combine the quotient from the long division and the partial fraction decomposition of the remainder to form the complete solution.

Key Concepts

Long Division in Algebra

Long division in algebra is a crucial technique used to divide polynomials. The process is similar to the traditional long division method used with numbers. The main goal is to simplify complex expressions by breaking them down into simpler components.

• Start by dividing the first term of the dividend by the first term of the divisor.
• The result is the first term of the quotient.
• Multiply the entire divisor by this term and subtract it from the dividend.
• Repeat the process with the new polynomial formed by your remainder.

This method continues until the degree of the remainder is less than the degree of the divisor. At that point, you have your final quotient and remainder. The remainder can then be expressed as part of a partial fraction decomposition. This step is particularly helpful in calculus and complex algebra to solve equations easily.

Polynomial Division

Polynomial division is the process of dividing two polynomials. Like numerical division, it involves finding how many times one polynomial can "fit into" another.

• First, align the dividend and divisor, ensuring that each term's degree is in descending order.
• Begin the division by dividing the highest degree term in the dividend by the highest degree term in the divisor.
• The result is the next term in the quotient, helping to gradually simplify the dividend.
• For example, if dividing \(x^5 + 2\) by \(x^2 - 1\), we start by dividing \(x^5\) by \(x^2\), resulting in \(x^3\).

This systematic breakdown helps manage more complicated polynomials and assists in later steps such as solving algebraic equations.

Factoring Quadratics

Factoring quadratics is the step of breaking down a quadratic equation into simpler parts, usually products of linear equations. This is essential for partial fraction decomposition, especially when dealing with remainders from polynomial division.

• A quadratic equation, in standard form \(ax^2 + bx + c\), can often be factored into two binomials: \((x - p)(x - q)\).
• The factors are values for which the quadratic equals zero.
• Sometimes factoring is not straightforward — in these cases, use methods like the quadratic formula, completing the square, or inspection.

In the exercise, the divisor \(x^2 - 1\) is factored into \(x - 1\) and \(x + 1\). This step is vital to set up partial fractions accurately.

Equating Coefficients

Equating coefficients is a technique used after setting up an equation that involves polynomial expressions on both sides. It helps find unknown variables by matching coefficients of corresponding terms.

• When decomposing the remainder into partial fractions, set up an identity where the expression equals the sum of the fractions.
• Multiply through by the least common denominator to eliminate fractions.
• Equate the coefficients of corresponding terms on both sides of the equation.
• Solve the resulting linear equations for the unknowns.

For example, when breaking down the remainder \(2x^2 + x + 2\) into \(\frac{A}{x - 1} + \frac{B}{x + 1}\), multiply by \(x^2 - 1\) and equate coefficients. Solving these equations gives the values of \(A\) and \(B\), finalizing the partial fraction decomposition.