Problem 14 - Precalculus

Question

(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{x^{2}-3 x-1}{x^{3}-x^{2}+2 x-2}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is: \(-\frac{1}{x-1} + \frac{2x - 1}{x^2 + 2}\).

Step 1: Factor the Denominator

First, we need to factor the cubic polynomial in the denominator: \(x^3 - x^2 + 2x - 2\). To do this, attempt polynomial division by searching for rational roots using the Rational Root Theorem, which suggests testing \(\pm 1\) and \(\pm 2\) as potential roots.Upon testing, we find that \(x = 1\) is a root (as plugging in gives zero). Hence, divide the polynomial by \((x - 1)\) to further factor.Divide:\[x^3 - x^2 + 2x - 2 \div (x - 1) = x^2 + 2\]The complete factorization of the denominator is: \[(x-1)(x^2 + 2)\].

Step 2: Set Up the Partial Fraction Decomposition

Next, we set up the partial fraction decomposition using the factors found in Step 1. The rational expression can be represented as:\[\frac{x^2 - 3x - 1}{(x-1)(x^2 + 2)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 2}\]This form includes one linear term \(A\) over the \((x-1)\) and a linear numerator \(Bx + C\) over the quadratic \((x^2+2)\).

Step 3: Clear the Denominators

Multiply through by the common denominator \((x-1)(x^2+2)\) to clear the fractions:\[x^2 - 3x - 1 = A(x^2 + 2) + (Bx + C)(x - 1)\]Expand the equation:\[x^2 - 3x - 1 = Ax^2 + 2A + Bx^2 - Bx + Cx - C\]

Step 4: Group and Solve for Constants

Combine like terms:\[x^2 - 3x - 1 = (A + B)x^2 + (-B + C)x + (2A - C)\]Equate coefficients:- \(A + B = 1\) for \(x^2\) terms- \(-B + C = -3\) for \(x\) terms- \(2A - C = -1\) for constant termsFrom these equations, we solve the system for \(A\), \(B\), and \(C\):1. From \(A + B = 1\) : \(B = 1 - A\)2. Substitute \(B = 1 - A\) into \(-B + C = -3\), leading to: \[ -(1 - A) + C = -3 \Rightarrow C = -3 + 1 - A = -2 - A \]3. Substitute \(C = -2 - A\) into \(2A - C = -1\): \[ 2A - (-2 - A) = -1 \Rightarrow 2A + 2 + A = -1 \Rightarrow 3A = -3 \Rightarrow A = -1 \]4. Plug \(A = -1\) back into equations for \(B\) and \(C\): - \(B = 1 - (-1) = 2\) - \(C = -2 - (-1) = -1\)

Step 5: Verify the Solution

With our constants, the decomposition becomes:\[\frac{x^2 - 3x - 1}{x^3 - x^2 + 2x - 2} = \frac{-1}{x-1} + \frac{2x - 1}{x^2 + 2}\]Verify by back substitution into the expression to ensure the coefficients match, which they do as confirmed previously.

Key Concepts

Polynomial Division

Polynomial division is a method that allows us to break down complex polynomial expressions into simpler ones. It is similar to long division with numbers, where you divide one polynomial (the dividend) by another (the divisor). This technique is essential when simplifying rational expressions or factoring polynomials.
For example, in the given problem, the denominator is a cubic polynomial:

• Polynomial: \(x^3 - x^2 + 2x - 2\)

We used polynomial division to find a way to factor this polynomial by dividing it by one of its factors, which was established by the Rational Root Theorem. Once we determined that \(x - 1\) was a factor (since \(x = 1\) is a root), we divided to factor the cubic polynomial completely into \((x-1)(x^2 + 2)\). Division helped simplify the polynomial making it easier to work with in further steps.

Rational Root Theorem

The Rational Root Theorem is a useful tool for finding potential rational roots of polynomial equations. It states that any rational root, expressed in the form \(\frac{p}{q}\), of a polynomial equation with integer coefficients is such that \(p\) divides the constant term and \(q\) divides the leading coefficient.
In the problem at hand, the polynomial in question is:

• Polynomial: \(x^3 - x^2 + 2x - 2\)

Here, the constant term is \(-2\) and the leading coefficient is \(1\), leading us to test potential roots \(\pm 1\) and \(\pm 2\). By plugging these values into the polynomial, we found that \(x = 1\) is indeed a root, helping in the factorization of the polynomial. By identifying these roots with the Rational Root Theorem, we can systematically find factors of polynomials that may seem complex at first glance.

Coefficient Comparison

Coefficient comparison is a technique that involves equating coefficients of similar terms on both sides of an equation. This method is particularly handy when performing partial fraction decomposition, as it allows you to establish a system of equations for the unknowns.
In the given exercise, after setting up the decomposition, we expanded and grouped terms:

• Original: \(x^2 - 3x - 1\)
• Partial: \((A + B)x^2 + (-B + C)x + (2A - C)\)

By matching coefficients from both sides:

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

Using these equations, you can solve for \(A\), \(B\), and \(C\). Coefficient comparison simplifies the process of determining unknown constants in partial fraction forms.

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving these systems involves finding common variable values that satisfy all equations simultaneously.
In partial fraction decomposition, systems of equations arise naturally when comparing coefficients. Using our equations:

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

With these conditions, we solve the system step-by-step:- **Substitute**: Substitute \(B = 1 - A\) into the second equation to express \(C\) in terms of \(A\).- **Solve Sequentially**: Substitute \(C\) into the third equation to find \(A\).- **Find All**: Back-substitute to find precise values for \(B\) and \(C\).
This stepwise manipulation allows us to find integer values for \(A\), \(B\), and \(C\), which are necessary in constructing the final decomposition. System of equations methods ensures that solutions are exact and consistent with all given conditions.