Problem 24 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{7 x-3}{x^{3}+2 x^{2}-3 x} $$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

The denominator of the given rational function is \(x^3 + 2x^2 - 3x\). We start by factoring it.First, factor out a common factor of \(x\):\[x(x^2 + 2x - 3)\]Next, factor the quadratic \(x^2 + 2x - 3\), which can be factored as \((x+3)(x-1)\). Therefore, the denominator is:\[x(x+3)(x-1)\]

Step 2: Set Up the Partial Fraction Decomposition

Now that the denominator is factored, set up the partial fraction decomposition. Since the denominator is \(x(x+3)(x-1)\), the partial fraction decomposition can be expressed as:\[\frac{A}{x} + \frac{B}{x+3} + \frac{C}{x-1}\]

Step 3: Solve for the Coefficients A, B, and C

We rewrite the original rational function with the partial fractions:\[\frac{7x-3}{x(x+3)(x-1)} = \frac{A}{x} + \frac{B}{x+3} + \frac{C}{x-1}\]Multiply through by the common denominator \(x(x+3)(x-1)\) to eliminate the denominators:\[7x - 3 = A(x+3)(x-1) + B(x)(x-1) + C(x)(x+3)\]Now, expand and collect like terms for each product:- \(A(x+3)(x-1) = A(x^2 + 2x - 3)\) expands to \(Ax^2 + 2Ax - 3A\)- \(B(x)(x-1) = B(x^2 - x)\) expands to \(Bx^2 - Bx\)- \(C(x)(x+3) = C(x^2 + 3x)\) expands to \(Cx^2 + 3Cx\)Combine these:\[(A + B + C)x^2 + (2A - B + 3C)x - 3A\]Equate coefficients with \(7x - 3\):1. \(A + B + C = 0\)2. \(2A - B + 3C = 7\)3. \(-3A = -3\)From equation (3), solve for \(A\):\(A = 1\)

Step 4: Substitute and Solve for B and C

Substitute \(A = 1\) into equations (1) and (2):1. \(1 + B + C = 0\) gives \(B + C = -1\)2. \(2(1) - B + 3C = 7\) simplifies to \(2 - B + 3C = 7\) or \(-B + 3C = 5\)We now have a system of equations:- \(B + C = -1\)- \(-B + 3C = 5\)Add the equations to eliminate \(B\):\(B + C - B + 3C = -1 + 5\)\(4C = 4\)Thus, \(C = 1\).Substituting \(C = 1\) back into \(B + C = -1\):\(B + 1 = -1\)\(B = -2\).

Step 5: Write the Partial Fraction Decomposition

Now that we have \(A = 1\), \(B = -2\), and \(C = 1\), substitute back into the partial fractions:\[\frac{7x-3}{x(x+3)(x-1)} = \frac{1}{x} - \frac{2}{x+3} + \frac{1}{x-1}\]This is the partial fraction decomposition of the rational function.

Key Concepts

Factoring Polynomials

Factoring polynomials is a critical skill, especially when dealing with rational functions. The first step is to simplify the denominator by breaking it into factors. In the given exercise, the polynomial in the denominator is of the form \(x^3 + 2x^2 - 3x\). The first step in factoring such an expression is to find a common factor. Here, we have a common factor of \(x\), which gives us an initial factorization:\[x(x^2 + 2x - 3)\]Next, we need to focus on factoring the quadratic expression \(x^2 + 2x - 3\). Quadratics often factor into the product of two binomials. A helpful method is to look for two numbers that multiply to give the constant term, \(-3\), and add to give the middle term, \(2\). These two numbers are \(3\) and \(-1\), leading to the factorization of:\[(x+3)(x-1)\]Put together, the completely factored form of the denominator is \(x(x+3)(x-1)\). This process simplifies further calculations in partial fraction decomposition.

Solving Equations

Once the polynomial is factored, we move on to solving for unknown coefficients in the partial fraction decomposition. Here, the original rational function \(\frac{7x-3}{x(x+3)(x-1)}\) is rewritten as a sum of simpler fractions:

• \(\frac{A}{x}\)
• \(\frac{B}{x+3}\)
• \(\frac{C}{x-1}\)

To find \(A\), \(B\), and \(C\), equate the expanded equation to the original polynomial. First, clear the denominators by multiplying through by \(x(x+3)(x-1)\). The key to solving these equations lies in matching the coefficients from both sides. For example, from the expression:\[(A + B + C)x^2 + (2A - B + 3C)x - 3A\]we know the left side, \(7x - 3\), has coefficients \(0\), \(7\), and \(-3\) respectively. This gives us a system of equations:\[\begin{align*}A + B + C &= 0, \2A - B + 3C &= 7, \-3A &= -3\end{align*}\]Solving these, we find the coefficients. Start with equation \(-3A = -3\) to find that \(A = 1\). Substitute \(A = 1\) into the other two equations to solve for \(B\) and \(C\) using methods such as substitution or elimination.

Rational Functions

Rational functions are expressions that represent the ratio of two polynomials. In cases where the degree of the numerator is less than the degree of the denominator, the rational function can be decomposed into partial fractions. This makes complex expressions easier to integrate or differentiate. For the expression \(\frac{7x-3}{x^3 + 2x^2 - 3x}\), after factoring the denominator, we rewrite this rational function as a sum of simpler rational expressions:

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

These individual fractions can be separately integrated or used in other calculations, each depending only on a single variable factor. Understanding this decomposition is crucial, as it turns a complex function into a series of manageable parts. This method also highlights the underlying relationships between polynomial functions and helps solve integrals and differential equations efficiently.