Find Where Undefined/Discontinuous (x^2-4)/(4x^2-12x)*(2x+4)/(-x^2+5x-6)

$\frac{x^{2} - 4}{4 x^{2} - 12 x} \cdot \frac{2 x + 4}{- x^{2} + 5 x - 6}$

Step 1

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

$4 x^{2} - 12 x = 0$

Step 2

Solve for $x$ .

Factor $4 x$ out of $4 x^{2} - 12 x$ .

Factor $4 x$ out of $4 x^{2}$ .

$4 x (x) - 12 x = 0$

Factor $4 x$ out of $- 12 x$ .

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

Factor $4 x$ out of $4 x (x) + 4 x (- 3)$ .

$4 x (x - 3) = 0$

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

$x = 0$

$x - 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 0, 3$

Step 3

Set the denominator in $\frac{2 x + 4}{- x^{2} + 5 x - 6}$ equal to $0$ to find where the expression is undefined.

$- x^{2} + 5 x - 6 = 0$

Step 4

Solve for $x$ .

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 5 x - 6$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 5 x - 6 = 0$

Factor $- 1$ out of $5 x$ .

$- (x^{2}) - (- 5 x) - 6 = 0$

Rewrite $- 6$ as $- 1 (6)$ .

$- (x^{2}) - (- 5 x) - 1 \cdot 6 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 5 x)$ .

$- (x^{2} - 5 x) - 1 \cdot 6 = 0$

Factor $- 1$ out of $- (x^{2} - 5 x) - 1 (6)$ .

$- (x^{2} - 5 x + 6) = 0$

Factor.

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$- ((x - 3) (x - 2)) = 0$

Remove unnecessary parentheses.

$- (x - 3) (x - 2) = 0$

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

$x - 3 = 0$

$x - 2 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $- (x - 3) (x - 2) = 0$ true.

$x = 3, 2$

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 $0$ .

$x = 0, x = 2, x = 3$

Step 6

Find Where Undefined/Discontinuous (x^2-4)/(4x^2-12x)*(2x+4)/(-x^2+5x-6)

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