Find Where Undefined/Discontinuous (3x^2-48)/(6x^2)*(2x^3)/(x^3+10x^2+24x)

$\frac{3 x^{2} - 48}{6 x^{2}} \cdot \frac{2 x^{3}}{x^{3} + 10 x^{2} + 24 x}$

Step 1

Set the denominator in $\frac{3 x^{2} - 48}{6 x^{2}}$ equal to $0$ to find where the expression is undefined.

$6 x^{2} = 0$

Step 2

Solve for $x$ .

Divide each term in $6 x^{2} = 0$ by $6$ and simplify.

Divide each term in $6 x^{2} = 0$ by $6$ .

$\frac{6 x^{2}}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x^{2}}{6} = \frac{0}{6}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x^{2} = 0$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 3

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

$x^{3} + 10 x^{2} + 24 x = 0$

Step 4

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 10 x^{2} + 24 x$ .

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

$x \cdot x^{2} + 10 x^{2} + 24 x = 0$

Factor $x$ out of $10 x^{2}$ .

$x \cdot x^{2} + x (10 x) + 24 x = 0$

Factor $x$ out of $24 x$ .

$x \cdot x^{2} + x (10 x) + x \cdot 24 = 0$

Factor $x$ out of $x \cdot x^{2} + x (10 x)$ .

$x (x^{2} + 10 x) + x \cdot 24 = 0$

Factor $x$ out of $x (x^{2} + 10 x) + x \cdot 24$ .

$x (x^{2} + 10 x + 24) = 0$

Factor.

Factor $x^{2} + 10 x + 24$ 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 $10$ .

$4, 6$

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 $0$ .

$x + 4 = 0$

Subtract $4$ from 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 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 $0$ .

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

Step 6

Find Where Undefined/Discontinuous (3x^2-48)/(6x^2)*(2x^3)/(x^3+10x^2+24x)

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