Question

fva 25 - 26: algebra ii q3
adding and subtracting rational expressions
\\(\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 + \boxed{-4})(x - \boxed{3})\\)

Explanation:

Step1: Identify factors of denominators

First denominator: \(x^2 - 3x - 4=(x - 4)(x + 1)\) (given as correct).
Second denominator: \(x^2 - 6x + 8=(x - 4)(x - 2)\) (given as correct).

Step2: Determine least common denominator (LCD)

The LCD is the product of the highest power of each distinct factor.
Distinct factors: \((x - 4)\), \((x + 1)\), \((x - 2)\).
So, LCD \(=(x - 4)(x + 1)(x - 2)\).

Answer:

The least common denominator is \((x - 4)(x + 1)(x - 2)\). So the missing values are \(1\) (in \(x + 1\)) and \(2\) (in \(x - 2\)), meaning the correct boxes should be \(1\) (for \(x + \boldsymbol{1}\)) and \(2\) (for \(x - \boldsymbol{2}\)) instead of \(-4\) and \(3\).