Question

what are the excluded values of x for \\(\frac{x + 4}{-3x^2 + 12x + 36}\\)?
\\(\bigcirc\\) \\(x = -6, x = 2\\)
\\(\bigcirc\\) \\(x = -6, x = -3, x = 2\\)
\\(\bigcirc\\) \\(x = -2, x = 6\\)
\\(\bigcirc\\) \\(x = -2, x = 3, x = 6\\)

Explanation:

Step1: Find denominator's zero

Set denominator \(-3x^2 + 12x + 36 = 0\). Divide by \(-3\): \(x^2 - 4x - 12 = 0\).

Step2: Factor quadratic

Factor \(x^2 - 4x - 12\): \((x - 6)(x + 2) = 0\).

Step3: Solve for x

Set factors to zero: \(x - 6 = 0 \Rightarrow x = 6\); \(x + 2 = 0 \Rightarrow x = -2\).

Answer:

\(x = -2, x = 6\) (corresponding to the option: \(x = -2, x = 6\))