Find the Excluded Values ((-4x-12)/(x-7))÷((x^2-12x+36)/(2x-14))

$\frac{- 4 x - 12}{x - 7} \div \frac{x^{2} - 12 x + 36}{2 x - 14}$

Step 1

Set the denominator in $\frac{- 4 x - 12}{x - 7}$ equal to $0$ to find where the expression is undefined.

$x - 7 = 0$

Step 2

Add $7$ to both sides of the equation.

$x = 7$

Step 3

Set the denominator in $\frac{x^{2} - 12 x + 36}{2 x - 14}$ equal to $0$ to find where the expression is undefined.

$2 x - 14 = 0$

Step 4

Solve for $x$ .

Add $14$ to both sides of the equation.

$2 x = 14$

Divide each term in $2 x = 14$ by $2$ and simplify.

Divide each term in $2 x = 14$ by $2$ .

$\frac{2 x}{2} = \frac{14}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{14}{2}$

Divide $x$ by $1$ .

$x = \frac{14}{2}$

Simplify the right side.

Divide $14$ by $2$ .

$x = 7$

Step 5

Set the denominator in $\frac{\frac{- 4 x - 12}{x - 7}}{\frac{x^{2} - 12 x + 36}{2 x - 14}}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} - 12 x + 36}{2 x - 14} = 0$

Step 6

Solve for $x$ .

Set the numerator equal to zero.

$x^{2} - 12 x + 36 = 0$

Solve the equation for $x$ .

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$x^{2} - 12 x + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 6$ .

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

Set the $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 7

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 $0$ .

$x = 6, x = 7$

Step 8

Find the Excluded Values ((-4x-12)/(x-7))÷((x^2-12x+36)/(2x-14))

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