Question

simplify the expression completely if possible.\\(\frac{x^2 - 16}{x^2 + 3x - 28}\\)

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 16\) as a difference of squares: \(x^2 - 16=(x - 4)(x + 4)\).
Factor the quadratic in the denominator \(x^2+3x - 28\). We need two numbers that multiply to \(- 28\) and add to \(3\). Those numbers are \(7\) and \(-4\), so \(x^2+3x - 28=(x + 7)(x - 4)\).

Step2: Cancel common factors

The expression becomes \(\frac{(x - 4)(x + 4)}{(x + 7)(x - 4)}\). Cancel the common factor \((x - 4)\) (assuming \(x
eq4\) to avoid division by zero).

Answer:

\(\frac{x + 4}{x + 7}\) (for \(x
eq4\) and \(x
eq - 7\))