Find Where Undefined/Discontinuous (2x(x-3))/((x+4)(x^2-9))

$\frac{2 x (x - 3)}{(x + 4) (x^{2} - 9)}$

Step 1

Set the denominator in $\frac{2 x (x - 3)}{(x + 4) (x^{2} - 9)}$ equal to $0$ to find where the expression is undefined.

$(x + 4) (x^{2} - 9) = 0$

Step 2

Solve for $x$ .

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

$x + 4 = 0$

$x^{2} - 9 = 0$

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 $x^{2} - 9$ equal to $0$ and solve for $x$ .

Set $x^{2} - 9$ equal to $0$ .

$x^{2} - 9 = 0$

Solve $x^{2} - 9 = 0$ for $x$ .

Add $9$ to both sides of the equation.

$x^{2} = 9$

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

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

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

$x = \pm 3$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

The final solution is all the values that make $(x + 4) (x^{2} - 9) = 0$ true.

$x = - 4, 3, - 3$

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 $0$ .

$x = - 4, x = - 3, x = 3$

Step 4

Find Where Undefined/Discontinuous (2x(x-3))/((x+4)(x^2-9))

Related Questions

Support LCMGCF.com by disabling your adblocker.