Problem 64 - Calculus Single Variable

Question

Find the partial fraction decomposition of \(\frac{1}{x^{3}-3 x^{2}+2}\).

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{-1/3}{x-1} + \frac{(1/3)x - 1/3}{x^2 - 2x - 2}\).

Step 1: Factor the Denominator

First, we need to factor the cubic polynomial in the denominator: \(x^3 - 3x^2 + 2\). The goal is to express it as a product of simpler polynomials. We begin by finding the roots. Testing possible roots using the Rational Root Theorem, we find that \(x=1\) is a root. Performing synthetic division on \(x^3 - 3x^2 + 2\) by \(x-1\), we obtain \(x^3 - 3x^2 + 2 = (x-1)(x^2 - 2x - 2)\). We then further factor \(x^2 - 2x - 2\), obtaining \((x-1)(x-rac{1+ ext{ or }- ext{ root of }5}{2})\).

Step 2: Set Up Partial Fraction Form

With the denominator factored, we can express the fraction as a sum of partial fractions. The partial fraction decomposition of \(\frac{1}{(x-1)(x^2 - 2x - 2)}\) takes the form: \(\frac{A}{x-1} + \frac{Bx + C}{x^2 - 2x - 2}\). Our task is to determine the coefficients \(A\), \(B\), and \(C\).

Step 3: Clear the Denominator

Multiply through by the common denominator \((x-1)(x^2 - 2x - 2)\) to eliminate the fractions: \[1 = A(x^2 - 2x - 2) + (Bx + C)(x - 1)\]. Expanding both terms gives: \[1 = Ax^2 - 2Ax - 2A + Bx^2 -Bx + Cx - C\].

Step 4: Equate Coefficients

Group like terms and equate coefficients with the left-hand side. This gives us three equations: \(A + B = 0\) (for \(x^2\)), \(-2A - B + C = 0\) (for \(x\)), \(-2A - C = 1\) (for constant term).

Step 5: Solve the System of Equations

Solve the system of equations from the previous step: First, from \(A + B = 0\), we have \(B = -A\). Substituting \(B = -A\) into \(-2A - B + C = 0\), we get: \(-2A - (-A) + C = 0 \Rightarrow -A + C = 0 \Rightarrow C = A\). From \(-2A - C = 1\), substitute \(C = A\): \(-2A - A = 1 \Rightarrow -3A = 1 \Rightarrow A = -\frac{1}{3}\).Substitute \(A = -\frac{1}{3}\) into \(B = -A\) and \(C = A\): \(B = \frac{1}{3}\) and \(C = -\frac{1}{3}\).

Step 6: Write the Decomposition

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

Key Concepts

Polynomial Factorization

Polynomial factorization is a critical concept in algebra that involves decomposing a polynomial into a product of simpler polynomials. For example, when we have a polynomial like \(x^3 - 3x^2 + 2\), the goal is to express it as a product of linear or quadratic factors to simplify operations like solving equations or partial fraction decomposition.

The first step in factoring is to look for roots, which are values of \(x\) that make the polynomial equal zero. The Rational Root Theorem can help identify potential rational roots by considering factors of the constant term and the leading coefficient. Once a root is found, tools like synthetic division can simplify the polynomial further.

In our exercise, synthetic division was used to divide the cubic polynomial by \((x-1)\), simplifying it to \((x-1)(x^2 - 2x - 2)\). Further factorization identifies the quadratic expression \(x^2 - 2x - 2\), making it easier to express the original polynomial as a product of its roots.

Synthetic Division

Synthetic division is an efficient method for dividing a polynomial by a linear binomial of the form \((x-c)\), where \(c\) is a constant. This technique is particularly useful when evaluating polynomials at specific points or simplifying them during the factorization process.

To perform synthetic division, set up the process by writing the coefficients of the polynomial in sequence. Next, use the known root to perform the division and determine the quotient and remainder, if any. This method eliminates the variables in calculations, making it faster and easier compared to traditional long division.

For the polynomial \(x^3 - 3x^2 + 2\), synthetic division reveals that \((x-1)\) is a factor, leaving us with the quadratic \(x^2 - 2x - 2\). This helps in further breaking down the polynomial, crucial for partial fraction decomposition.

Rational Root Theorem

The Rational Root Theorem is a powerful tool for finding potential rational roots of a polynomial equation of the form \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0\). It suggests that any potential rational root, expressed as \(\frac{p}{q}\), must be a factor of the constant term \(a_0\) divided by a factor of the leading coefficient \(a_n\).

This theorem narrows down the list of possible rational roots, making it easier to test and find actual roots. Once a root is confirmed, it can be used to simplify the polynomial further, often by using synthetic division.

In our example, by applying the Rational Root Theorem, we find that \(x=1\) is a root of the polynomial \(x^3 - 3x^2 + 2\). Knowing this, we can apply synthetic division to factor the polynomial efficiently.

System of Equations

Solving a system of equations involves finding the values of variables that satisfy all given equations simultaneously. In the context of partial fraction decomposition, we often end up with a system of linear equations derived from equating coefficients.

For instance, in our equation: \[1 = A(x^2 - 2x - 2) + (Bx + C)(x - 1)\]Expanding and simplifying gives different systems of equations like: - \(A + B = 0\) (for \(x^2\))- \(-2A - B + C = 0\) (for \(x\))- \(-2A - C = 1\)\ (for the constant term).

This system can be solved step-by-step, finding values like \(A = -\frac{1}{3}\), \(B = \frac{1}{3}\), and \(C = -\frac{1}{3}\), which are then used to express the partial fractions. By systematically solving each equation, we find the coefficients needed for the partial fraction decomposition.