Find the Excluded Values (x+4)/(-3x^2+12x+36)

$\frac{x + 4}{- 3 x^{2} + 12 x + 36}$

Set the denominator in $\frac{x + 4}{- 3 x^{2} + 12 x + 36}$ equal to $0$ to find where the expression is undefined.

$- 3 x^{2} + 12 x + 36 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} + 12 x + 36$ .

Factor $- 3$ out of $- 3 x^{2}$ .

$- 3 x^{2} + 12 x + 36 = 0$

Factor $- 3$ out of $12 x$ .

$- 3 x^{2} - 3 (- 4 x) + 36 = 0$

Factor $- 3$ out of $36$ .

$- 3 x^{2} - 3 (- 4 x) - 3 \cdot - 12 = 0$

Factor $- 3$ out of $- 3 (x^{2}) - 3 (- 4 x)$ .

$- 3 (x^{2} - 4 x) - 3 \cdot - 12 = 0$

Factor $- 3$ out of $- 3 (x^{2} - 4 x) - 3 (- 12)$ .

$- 3 (x^{2} - 4 x - 12) = 0$

Factor.

Factor $x^{2} - 4 x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$- 3 ((x - 6) (x + 2)) = 0$

Remove unnecessary parentheses.

$- 3 (x - 6) (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 6 = 0$

$x + 2 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $- 3 (x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 2, x = 6$

Find the Excluded Values (x+4)/(-3x^2+12x+36)

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