Find Where Undefined/Discontinuous ((x^2-36)/(6x-6))÷((x^2+12x+36)/(x^2+5x-6))

$\frac{x^{2} - 36}{6 x - 6} \div \frac{x^{2} + 12 x + 36}{x^{2} + 5 x - 6}$

Step 1

Set the denominator in $\frac{x^{2} - 36}{6 x - 6}$ equal to $0$ to find where the expression is undefined.

$6 x - 6 = 0$

Step 2

Solve for $x$ .

Add $6$ to both sides of the equation.

$6 x = 6$

Divide each term in $6 x = 6$ by $6$ and simplify.

Divide each term in $6 x = 6$ by $6$ .

$\frac{6 x}{6} = \frac{6}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{6}{6}$

Divide $x$ by $1$ .

$x = \frac{6}{6}$

Simplify the right side.

Divide $6$ by $6$ .

$x = 1$

Step 3

Set the denominator in $\frac{x^{2} + 12 x + 36}{x^{2} + 5 x - 6}$ equal to $0$ to find where the expression is undefined.

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

Step 4

Solve for $x$ .

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

$- 1, 6$

Write the factored form using these integers.

$(x - 1) (x + 6) = 0$

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

$x - 1 = 0$

$x + 6 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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 $(x - 1) (x + 6) = 0$ true.

$x = 1, - 6$

Step 5

Set the denominator in $\frac{\frac{x^{2} - 36}{6 x - 6}}{\frac{x^{2} + 12 x + 36}{x^{2} + 5 x - 6}}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} + 12 x + 36}{x^{2} + 5 x - 6} = 0$

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 $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Exclude the solutions that do not make $\frac{x^{2} + 12 x + 36}{x^{2} + 5 x - 6} = 0$ true.

$No solution$

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 $0$ .

$x = - 6, x = 1$

Step 8

Find Where Undefined/Discontinuous ((x^2-36)/(6x-6))÷((x^2+12x+36)/(x^2+5x-6))

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