Find Where Undefined/Discontinuous (x^2+9x+18)/(x+2)*(x^2-3x-10)/(x^2+2x-24)

$\frac{x^{2} + 9 x + 18}{x + 2} \cdot \frac{x^{2} - 3 x - 10}{x^{2} + 2 x - 24}$

Step 1

Set the denominator in $\frac{x^{2} + 9 x + 18}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Step 2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3

Set the denominator in $\frac{x^{2} - 3 x - 10}{x^{2} + 2 x - 24}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 2 x - 24 = 0$

Step 4

Solve for $x$ .

Factor $x^{2} + 2 x - 24$ 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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Write the factored form using these integers.

$(x - 4) (x + 6) = 0$

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

$x - 4 = 0$

$x + 6 = 0$

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 $x + 6$ equal to $0$ and solve for $x$ .

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x - 4) (x + 6) = 0$ true.

$x = 4, - 6$

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 $0$ .

$x = - 6, x = - 2, x = 4$

Step 6

Find Where Undefined/Discontinuous (x^2+9x+18)/(x+2)*(x^2-3x-10)/(x^2+2x-24)

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