problem: \( \frac{2}{x^{2}-3 x-4}+\frac{3}{x^{2}-6 x+8} \) Factor each denominator: \( x^{2}-3 x-4=\square \) \( x^{2}-6 x+8=\square \)

1. Factor \( x^2 - 3x - 4 \): - We need two numbers that multiply to \(-4\) and add to \(-3\). - The numbers \(-4\) and \(1\) work because \((-4) \times 1 = -4\) and \((-4) + 1 = -3\). - Thus, \[ x^2 - 3x - 4 = (x - 4)(x + 1). \] 2. Factor \( x^2 - 6x + 8 \): - We need two numbers that multiply to \(8\) and add to \(-6\). - The numbers \(-2\) and \(-4\) work because \((-2) \times (-4) = 8\) and \((-2) + (-4) = -6\). - Thus, \[ x^2 - 6x + 8 = (x - 2)(x - 4). \] Final answers: \[ x^2 - 3x - 4 = (x - 4)(x + 1) \] \[ x^2 - 6x + 8 = (x - 2)(x - 4) \]