91Ó°ÊÓ

Factoring polynomials is one of the most important steps in partial fraction decomposition. It simplifies the rational expression and makes the problem more manageable.
In our exercise, we started by factoring the denominator. The given polynomial was: \[ x^3 - 2x^2 - 3x \] First, we factored out the common term \[ x \] giving: \[ x(x^2 - 2x - 3) \] Next, we need to factor the quadratic \[ x^2 - 2x - 3 \]. To do this, we look for two numbers that multiply to \[ -3 \] (the constant term) and add up to \[ -2 \] (the coefficient of the linear term). These numbers are \[ -3 \] and \[ 1 \]
Thus, the factoring is: \[ x^2 - 2x - 3 = (x - 3)(x + 1) \], which leads us to the fully factored form: \[ x(x - 3)(x + 1) \]
This transformation is key for setting up partial fractions as it reveals the simpler components of the polynomial.

Rational expressions are fractions where both the numerator and the denominator are polynomials. In this exercise, the rational expression is: \[ \frac{7x + 3}{x(x - 3)(x + 1)} \]
To decompose this rational expression into partial fractions, we express it as a sum of simpler fractions. For our problem, it looks like: \[ \frac{7x + 3}{x(x - 3)(x + 1)} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{x + 1} \]
Here, \[ A \], \[ B \], and \[ C \] are constants that need to be determined. Each term has a polynomial in the denominator that matches one of the factors of the original denominator. This breakdown simplifies the rational expression into elements that can each be easily integrated or otherwise manipulated.

To find the constants \[ A \], \[ B \], and \[ C \] for the partial fractions, we need to set up a system of equations. We get these equations by clearing the denominator and expanding the equation \[ 7x + 3 = A(x - 3)(x + 1) + B(x)(x + 1) + C(x)(x - 3) \]
This results in: \[ 7x + 3 = A(x^2 - 2x - 3) + B(x^2 + x) + C(x^2 - 3x) \]
Combine like terms to get: \[ 7x + 3 = (A + B + C)x^2 + (-2A + B - 3C)x - 3A \] By matching the coefficients of \[ x^2 \], \[ x \], and the constant term, we derive the following system of equations:

• \[ A + B + C = 0 \]
• \[ -2A + B - 3C = 7 \]
• \[ -3A = 3 \]

We solve this system step-by-step to find the values of \[ A \], \[ B \], and \[ C \], which are critical for completing the partial fraction decomposition.

Algebraic manipulation involves rearranging and simplifying equations to get to the solution. In this exercise, after setting up our system of equations, we used substitution and addition methods to solve for \[ A \], \[ B \], and \[ C \]. First, from the equation: \[ -3A = 3 \] we found \[ A = -1 \].
Next, by substituting \[ A = -1 \] into the other equations: \[ -1 + B + C = 0 \] (which simplifies to \[ B + C = 1 \]) and \[ 2 + B - 3C = 7 \] (which simplifies to \[ B - 3C = 5 \]), we obtained another set of equations:

• \[ B + C = 1 \]
• \[ B - 3C = 5 \]

By solving these, we found \[ B = 2 \] and \[ C = -1 \]. This algebraic process is essential for breaking down complex relationships into simpler elements, allowing us to solve for unknowns systematically.