Question

multiply the rational expression and write the product in simplest form: \\(\frac{x^2 + 2x - 3}{x^2 + 3x - 4} cdot \frac{x^2 + 6x + 8}{x^2 - 9}\\)\
(1 point)\
\\(-\frac{2}{3}\\)\
\\(\frac{x + 2}{x - 3}\\)\
\\(\frac{(x + 4)(x + 2)}{(x + 3)(x - 3)}\\)\
\\(\frac{(x + 2)}{x + 3}\\)

Explanation:

Step1: Factor each quadratic

Factor \(x^2 + 2x - 3\): \(x^2 + 2x - 3=(x + 3)(x - 1)\)
Factor \(x^2 + 3x - 4\): \(x^2 + 3x - 4=(x + 4)(x - 1)\)
Factor \(x^2 + 6x + 8\): \(x^2 + 6x + 8=(x + 2)(x + 4)\)
Factor \(x^2 - 9\): \(x^2 - 9=(x + 3)(x - 3)\)

Step2: Substitute factored forms into the expression

The original expression becomes:
\(\frac{(x + 3)(x - 1)}{(x + 4)(x - 1)} \cdot \frac{(x + 2)(x + 4)}{(x + 3)(x - 3)}\)

Step3: Cancel common factors

Cancel \((x + 3)\), \((x - 1)\), and \((x + 4)\) from numerator and denominator:
After canceling, we get \(\frac{x + 2}{x - 3}\)

Answer:

\(\boldsymbol{\frac{x + 2}{x - 3}}\) (corresponding to the option with this expression)