91Ó°ÊÓ

Understanding rational expressions is crucial for dealing with the concept of partial fraction decomposition. A rational expression is simply a fraction where both the numerator and the denominator are polynomials. For example, in the given problem, the expression \( \frac{x^2 - 3x + 14}{(x+3)(x-1)^2} \) is a rational expression. The polynomial in the numerator is \( x^2 - 3x + 14 \), and the polynomial in the denominator is \( (x+3)(x-1)^2 \). The whole idea of partial fraction decomposition is to break down this complicated rational expression into simpler fractions that are easier to work with.

To start with partial fraction decomposition, we first need to identify the linear and quadratic factors in the denominator. In this case, the factors are \( x+3 \) and \( (x-1)^2 \). Instead of grappling with the more complex expression, we aim to convert it into simpler fractions, each having these factors as their denominators.

Combining fractions involves finding a common denominator. This step is essential to simplify the fractions during partial fraction decomposition. Initially, we write our equation as follows:
\( \frac{x^2 - 3x + 14}{(x+3)(x-1)^2} = \frac{A}{x+3} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \).

Here, \( A, B, \) and \( C \) are constants, the values we need to determine. Next, to combine these partial fractions over a single common denominator, we need to express them all over \( (x+3)(x-1)^2 \). This process results in:
\( \frac{A}{x+3} + \frac{B}{x-1} + \frac{C}{(x-1)^2} = \frac{A(x-1)^2 + B(x+3)(x-1) + C(x+3)}{(x+3)(x-1)^2} \).

We ensure to align these fractions under a common denominator, so we can equate and match the numerators, making it easier to handle complex fractions.

Matching coefficients is a foundational step in simplifying and solving polynomial equations. Once we have a consolidated fraction, as seen in \( \frac{A(x-1)^2 + B(x+3)(x-1) + C(x+3)}{(x+3)(x-1)^2} \), we equate the numerators on both sides:
\( x^2 - 3x + 14 = A(x-1)^2 + B(x+3)(x-1) + C(x+3) \).

Next, we expand and simplify the right-hand side, which provides:
\( Ax^2 - 2Ax + A + Bx^2 + 2Bx - 3B + Cx + 3C \). By combining like terms, it becomes:
\( (A + B)x^2 + (-2A + 2B + C)x + (A - 3B + 3C) \).

We specifically match coefficients of like terms with the left-hand side \( x^2 - 3x + 14 \). This yields:

• For the \( x^2 \) term: \( A + B = 1 \).
• For the \( x \) term: \( -2A + 2B + C = -3 \).
• For the constant term: \( A - 3B + 3C = 14 \).

Matching these coefficients steps us closer to solving for \( A, B, \) and \( C \).

When coefficients are matched in partial fraction decomposition, we end up with a system of equations. For our problem, these are:
1) \( A + B = 1 \)
2) \( -2A + 2B + C = -3 \)
3) \( A - 3B + 3C = 14 \).

We first isolate \( A \) from the equation \( A + B = 1 \), giving us \( A = 1 - B \). Next, substituting \( A = 1 - B \) into the other two equations simplifies them:

• The second equation becomes: \( -2(1 - B) + 2B + C = -3 \), which simplifies to \( -2 + 2B + 2B + C = -3 \).
• The third equation transforms into: \( (1 - B) - 3B + 3C = 14 \), which simplifies to \( 1 - B - 3B + 3C = 14 \).

By solving these equations, we find the precise values for \( A, B, \) and \( C \). These values are then substituted back into the partial fractions, completing the decomposition.