Problem 7 - Algebra for College Students

Question

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{x^{2}-18 x+5}{(x-1)(x+2)(x-3)}\)

Step-by-Step Solution

Answer

\( \frac{-2}{x-1} + \frac{-3}{x+2} + \frac{6}{x-3} \)

Step 1: Identify Form of Partial Fractions

Given the rational expression \( \frac{x^2-18x+5}{(x-1)(x+2)(x-3)} \), first note that the denominator \((x-1)(x+2)(x-3)\) is a product of linear factors. So, we express it as a sum of partial fractions: \( \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3} \) where \( A, B, \) and \( C \) are constants to be determined.

Step 2: Set up the Equation

Multiply every term by the denominator \((x-1)(x+2)(x-3)\) to eliminate the denominators, obtaining:\[ x^2 - 18x + 5 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) \] now expand each product on the right-hand side.

Step 3: Expand and Simplify

Expand each term:- For \(A(x+2)(x-3)\), expand to \(A(x^2 - x - 6) = Ax^2 - Ax - 6A\).- For \(B(x-1)(x-3)\), expand to \(B(x^2 - 4x + 3) = Bx^2 - 4Bx + 3B\).- For \(C(x-1)(x+2)\), expand to \(C(x^2 + x - 2) = Cx^2 + Cx - 2C\).The equation becomes: \[ x^2 - 18x + 5 = (A+B+C)x^2 + (-A-4B+C)x - 6A + 3B - 2C \]

Step 4: Equate Coefficients

Match the coefficients of \(x^2\), \(x\), and the constant term from both sides to form a system of equations:1. \(A + B + C = 1\) for the \(x^2\) terms.2. \(-A - 4B + C = -18\) for the \(x\) terms.3. \(-6A + 3B - 2C = 5\) for the constant terms.

Step 5: Solve the System of Equations

Solve the system:From 1: \(A + B + C = 1\).From 2: \( -A - 4B + C = -18\).From 3: \( -6A + 3B - 2C = 5\).Using substitution or elimination methods, solve this system:- First, solve Equation 1 for \(A\): \(A = 1 - B - C \).- Substitute \(A\) in the other two equations and simplify to find the values of \(B\) and \(C\).- With \(B\) and \(C\) found, use Equation 1 again to find \(A\).The solution is: \( A = -2 \), \( B = -3 \), \( C = 6 \).

Step 6: Write the Partial Fraction Decomposition

Substitute \( A = -2 \), \( B = -3 \), and \( C = 6 \) back into the partial fractions:\[ \frac{x^2 - 18x + 5}{(x-1)(x+2)(x-3)} = \frac{-2}{x-1} + \frac{-3}{x+2} + \frac{6}{x-3} \]

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are both polynomials. In these expressions, the key is that you cannot divide by zero, so the values that make the denominator zero are not in the domain of the expression.
To find the partial fraction decomposition of a rational expression, the degree of the numerator must be less than the degree of the denominator. If this is not the case, polynomial long division can be used to reduce the degree of the numerator before starting the decomposition process.
Partial fraction decomposition involves breaking down a complex rational expression into simpler fractions that can be more easily manipulated or integrated. This is particularly useful in calculus, where integration of simpler expressions is much more straightforward.
In our exercise, the rational expression \(\frac{x^2-18x+5}{(x-1)(x+2)(x-3)}\) is already set up for decomposition, as the degree of the numerator (2) is less than the degree of the denominator (3). By expressing it in terms of partial fractions, we aim to write it as a sum of simpler fractions.

Linear Factors

Linear factors are components of a polynomial that are of the first degree, usually taking the form \(ax + b\). If a polynomial is fully factored into linear factors, it is expressed as a product of terms like \((x - r)\), where \(r\) is a root of the polynomial.
For partial fraction decomposition, decomposing the denominator into linear factors is crucial. This simplification allows the division of the original rational expression into simpler fractions with these factors as the denominators.
In our problem, the denominator \((x-1)(x+2)(x-3)\) is the product of three linear factors: \((x-1)\), \((x+2)\), and \((x-3)\). The presence of linear factors implies that the corresponding partial fractions will have numerators that are constants \(A\), \(B\), and \(C\). This turns the complex rational expression into a form that is easier to work with for integration or further algebraic manipulation.
It's important to note that if any of the factors were repeated (e.g., \((x-1)^2)\), the decomposition would include terms like \(\frac{D}{(x-1)^2}\). However, in this exercise, there are no repeated factors, simplifying our task.

System of Equations

A system of equations is a set of equations with multiple variables that you solve together to find a common solution for the variables. In the context of partial fraction decomposition, once we have set up our expression with constants \(A\), \(B\), and \(C\), we obtain a system of equations by equating coefficients from both sides of the equation.
For the given expression, expanding the right-hand side and matching with the original numerator gives us a system containing:

• \(A + B + C = 1\) for the \(x^2\) terms
• \(-A - 4B + C = -18\) for the \(x\) terms
• \(-6A + 3B - 2C = 5\) for the constant terms

By solving this system using substitution or elimination, we find the specific values of \(A\), \(B\), and \(C\) that fit the original expression.
Solving a system of equations is a common process in algebra that engages logic and procedural knowledge, helping to solidify understanding of linear relationships and variables. By solving step by step, we obtain \(A = -2\), \(B = -3\), and \(C = 6\), providing us with the complete decomposition of the original rational expression.