Question

Decompose the expression \(x^{2} /\left(x^{3}-2 x^{2}-5 x+6\right)\) into partial functions.

Short answer

The decomposed expression of \(\frac{x^2}{(x^3 - 2x^2 - 5x + 6)}\) into partial functions is: \( \frac{-\frac{1}{2}}{x - 1} + \frac{\frac{9}{10}}{x - 3} + \frac{-\frac{4}{5}}{x + 2} \)

Step-by-step solution

Factor the denominator

We begin by factoring the denominator \(x^3 - 2x^2 - 5x + 6\). By trial and error, we find that \(x = 1\) is a root of the denominator. Therefore, we can perform synthetic division or long division to find the remaining quadratic factor: \( (x^3 - 2x^2 - 5x + 6) = (x - 1)(x^2 - x - 6) \) Now, we continue by factoring the quadratic part: \( x^2 - x - 6 = (x - 3)(x + 2) \) So, the fully factored denominator is: \( (x - 1)(x - 3)(x + 2) \)

Decompose the expression using partial fraction decomposition

Now we decompose the given expression using partial fraction decomposition: \( \frac{x^2}{(x - 1)(x - 3)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x - 3} + \frac{C}{x + 2} \) We need to find the values of A, B, and C by eliminating the denominators and then solving the resulting system of linear equations. So, first, we multiply both sides of the equation by the denominator \((x - 1)(x - 3)(x + 2)\): \( x^2 = A(x - 3)(x + 2) + B(x - 1)(x + 2) + C(x - 1)(x - 3) \) Now, we'll find the values of A, B, and C by plugging in suitable values for x: 1. Let x = 1: \( (1)^2 = A(1 - 3)(1 + 2) + B(1 - 1)(1 + 2) + C(1 - 1)(1 - 3) \) \( 1 = -2A \Rightarrow A = -\frac{1}{2} \) 2. Let x = 3: \( (3)^2 = A(3 - 3)(3 + 2) + B(3 - 1)(3 + 2) + C(3 - 1)(3 - 3) \) \( 9 = 10B \Rightarrow B = \frac{9}{10} \) 3. Let x = -2: \( (-2)^2 = A(-2 - 3)(-2 + 2) + B(-2 - 1)(-2 + 2) + C(-2 - 1)(-2 - 3) \) \( 4 = -5C \Rightarrow C = -\frac{4}{5} \)

Write the expression as the sum of partial functions

Now that we have found the values of A, B, and C, we can write the expression as the sum of partial functions: \( \frac{x^2}{(x - 1)(x - 3)(x + 2)} = \frac{-\frac{1}{2}}{x - 1} + \frac{\frac{9}{10}}{x - 3} + \frac{-\frac{4}{5}}{x + 2} \) So, the decomposed expression is: \( \frac{-\frac{1}{2}}{x - 1} + \frac{\frac{9}{10}}{x - 3} + \frac{-\frac{4}{5}}{x + 2} \)

Key concepts

Factoring Polynomials

Factoring polynomials is akin to breaking down a complex expression into simpler, more manageable multipliers. It forms the cornerstone of algebra and is widely used in calculus, especially in tasks such as partial fraction decomposition. A polynomial is an expression consisting of variables and coefficients, involving operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

For example, the polynomial in the given exercise is factored by first identifying an obvious root. Here, the root is discovered through trial and error or by using the Rational Root Theorem, which provides a list of potential rational roots based on the factors of the constant term and the leading coefficient. Once a root is identified, methods like synthetic division or long division are employed to further simplify the polynomial into its component linear factors.

Synthetic Division

Synthetic division is a shorthand, efficient way of dividing polynomials by binomials of the form \(x - c\) and obtaining the quotient and the remainder quickly. This method is particularly useful in the case of our exercise for factoring the denominator polynomial after identifying a root.

How to Perform Synthetic Division

• Write down the coefficients of the polynomial.
• Bring down the leading coefficient to the bottom row.
• Multiply by the root identified and place this value under the next coefficient.
• Continue the process until all coefficients have been worked through.

The result of synthetic division gives you the coefficients of the quotient polynomial, which are used in subsequent factoring steps. In the context of our exercise, synthetic division facilitated factoring into the quadratic \(x^2 - x - 6\) after identifying \(x = 1\) as a root.

Solving Systems of Linear Equations

Solving systems of linear equations is an essential skill for various mathematical applications, such as during the partial fraction decomposition stage of our main problem. Once the expression is set up with variables representing the unknown coefficients, the task becomes to find values for these variables that satisfy all equations simultaneously.

In the context of partial fraction decomposition, after determining the partial fractions, you create a system of equations by equating coefficients from both sides of an equation. In our problem, we cleverly chose values for \(x\) to simplify the system into individual equations that could be solved immediately. Outside of such straightforward cases, more complex systems might require methods like substitution, elimination, or matrix operations (using augmented matrices and row operations) to find the solution.