Find Where Undefined/Discontinuous (x^2+15x+54)/(x^3+2x^2-24x)

$\frac{x^{2} + 15 x + 54}{x^{3} + 2 x^{2} - 24 x}$

Step 1

Set the denominator in $\frac{x^{2} + 15 x + 54}{x^{3} + 2 x^{2} - 24 x}$ equal to $0$ to find where the expression is undefined.

$x^{3} + 2 x^{2} - 24 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 2 x^{2} - 24 x$ .

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

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

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

$x \cdot x^{2} + x (2 x) - 24 x = 0$

Factor $x$ out of $- 24 x$ .

$x \cdot x^{2} + x (2 x) + x \cdot - 24 = 0$

Factor $x$ out of $x \cdot x^{2} + x (2 x)$ .

$x (x^{2} + 2 x) + x \cdot - 24 = 0$

Factor $x$ out of $x (x^{2} + 2 x) + x \cdot - 24$ .

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

Factor.

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.

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

Remove unnecessary parentheses.

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

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

$x = 0$

$x - 4 = 0$

$x + 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

$x = 0, 4, - 6$

Step 3

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 $0$ .

$x = - 6, x = 0, x = 4$

Step 4

Find Where Undefined/Discontinuous (x^2+15x+54)/(x^3+2x^2-24x)

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