Problem 35 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}$$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

In the given rational expression, the denominator \(x^{3}+x^{2}+x+1\) can be factored by grouping. \[x^{3}+x^{2}+x+1 = x^{2}(x+1) + 1(x+1) = (x^{2}+1)(x+1)\]

Step 2: Set Up the Partial Fractions

Set up the partial fractions with the factored denominators. Let the given fraction be equal to a sum of two simpler fractions with denominators as the factors of the denominator of the original fraction and coefficients A and B respectively. Thus we have, \[\frac{6x^{2} - x + 1}{x^{3}+x^{2}+x+1} = \frac{A}{x+1} + \frac{B}{x^{2}+1}\]

Step 3: Find the Coefficients

Multiply each side of the equation by the common denominator of the fractions on the right side (which is actually the denominator of the original fraction) and simplify to create a polynomial equation. By equating the coefficients of the polynomial to match the numerator of the original fraction, we can find the values of A and B.\[6x^{2} - x + 1 = A(x^{2}+1) + B(x + 1)\]\nThat simplifies to: \[6x^{2} - x + 1 = Ax^{2} + A + Bx + B\]\nTo get the coefficients, equate the expressions on both sides of the equation for similar powers of x: \[For\, x^{2}:\, A=6\] \[For\, x:\, B=-1\] \[For\, constant\, terms:\, A + B = 1\]\nThus, A = 6 and B = -5.

Step 4: Write Down the Final Partial fractions

Replace the values of A and B in the partial fractions obtained in step 2: \[\frac{6x^{2} - x + 1}{x^{3}+x^{2}+x+1} = \frac{6}{x+1} - \frac{5}{x^{2}+1}\]

Key Concepts

Rational Expression

A rational expression is a fraction where the numerator and the denominator are both polynomials. The example in the exercise, \( \frac{6x^{2} - x + 1}{x^{3} + x^{2} + x + 1} \), illustrates a complex rational expression that can be simplified for better understanding and operation using partial fraction decomposition. This process involves breaking down the complicated fraction into simpler fractions that are much more manageable.

Understanding rational expressions is crucial as they appear frequently in calculus, especially when integrating functions. Grasping how to handle these expressions will build a strong foundation for tackling more complex mathematical concepts. When attempting to simplify a rational expression, ensure that the expression is first simplified by factoring where possible and that all extraneous solutions (such as those that might make the denominator zero) are considered and excluded from the solution set.

Factoring Polynomials

Factoring polynomials is a vital skill in simplifying rational expressions and solving polynomial equations. A polynomial is factored when it is written as a product of its factors, much like breaking down a number into a product of primes. In the exercise, the denominator \(x^{3} + x^{2} + x + 1\) is factored into \(x^{2} + 1\) and \(x + 1\) using a technique called grouping.

Steps for Factoring

• Look for a common factor in all terms of the polynomial.
• If no common factor is found, group terms in a strategic way, seeking to factor by grouping.
• Use formulas for special products or patterns like differences of squares or cubic equations when applicable.

By recognizing common factoring techniques such as the difference of squares, perfect square trinomials, and others, one can simplify complex polynomials that at first glance seem unfactorable. Mastery of factoring techniques will not only aid in decomposition but also in finding zeros and graphing polynomial functions.

Polynomial Equation

A polynomial equation is an equality involving a polynomial on one side of the equation, typically set to zero. Solving these equations usually involves factoring the polynomial and using the Zero Product Property to find the roots. However, when it comes to partial fraction decomposition, equating coefficients is the tool of choice.

In the process of decomposing a rational expression, after factoring the denominator, coefficient matching is used to create a system of equations. These systems arise by multiplying the entire equation by the common denominator to eliminate the fraction, leading to an equation that is purely polynomial. The example from the exercise transitions into the polynomial equation \(6x^{2} - x + 1 = A(x^{2}+1) + B(x + 1)\), which is then solved for the coefficients A and B. This method requires careful solving to ensure that the polynomial on both sides of the equation is balanced with respect to each degree of x.