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

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

Step 1

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

$x^{2} - x - 12 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} - x - 12$ 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 $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$(x - 4) (x + 3) = 0$

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

$x - 4 = 0$

$x + 3 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Step 3

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

$x - 4 = 0$

Step 4

Add $4$ to both sides of the equation.

$x = 4$

Step 5

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 $0$ .

$x = - 3, x = 4$

Step 6

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

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