Question

complete the following steps to find the lcd and write the sum of the numerators for the given problem:
\\(\frac{2}{x^2 - 3x - 4} + \frac{3}{x^2 - 6x + 8}\\)
factor each denominator:
\\(x^2 - 3x - 4 = \\) a \\((x - 4)(x + 1)\\)
\\(x^2 - 6x + 8 = \\) b \\((x - 4)(x - 2)\\)
the least common denominator is:
\\((x - 4)(x + \square)(x - \square)\\)

Explanation:

Step1: Identify the denominators

The denominators are \(x^2 - 3x - 4\) and \(x^2 - 6x + 8\). We already factored them as \((x - 4)(x + 1)\) (from \(x^2 - 3x - 4\)) and \((x - 4)(x - 2)\) (from \(x^2 - 6x + 8\)).

Step2: Determine the least common denominator (LCD)

To find the LCD, we take the product of the unique factors with the highest power each factor appears. The factors are \((x - 4)\), \((x + 1)\), and \((x - 2)\). So the LCD is \((x - 4)(x + 1)(x - 2)\).

Answer:

The least common denominator is \((x - 4)(x + 1)(x - 2)\), so the first blank is \(1\) (for \((x + 1)\)) and the second blank is \(2\) (for \((x - 2)\)).