Find Where Undefined/Discontinuous ((x^2-16)/(4x-4))÷((x^2+8x+16)/(x^2+3x-4))

$\frac{x^{2} - 16}{4 x - 4} \div \frac{x^{2} + 8 x + 16}{x^{2} + 3 x - 4}$

Step 1

Set the denominator in $\frac{x^{2} - 16}{4 x - 4}$ equal to $0$ to find where the expression is undefined.

$4 x - 4 = 0$

Step 2

Solve for $x$ .

Add $4$ to both sides of the equation.

$4 x = 4$

Divide each term in $4 x = 4$ by $4$ and simplify.

Divide each term in $4 x = 4$ by $4$ .

$\frac{4 x}{4} = \frac{4}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{4}{4}$

Divide $x$ by $1$ .

$x = \frac{4}{4}$

Simplify the right side.

Divide $4$ by $4$ .

$x = 1$

Step 3

Set the denominator in $\frac{x^{2} + 8 x + 16}{x^{2} + 3 x - 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 3 x - 4 = 0$

Step 4

Solve for $x$ .

Factor $x^{2} + 3 x - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$(x - 1) (x + 4) = 0$

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

$x - 1 = 0$

$x + 4 = 0$

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 $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 1) (x + 4) = 0$ true.

$x = 1, - 4$

Step 5

Set the denominator in $\frac{\frac{x^{2} - 16}{4 x - 4}}{\frac{x^{2} + 8 x + 16}{x^{2} + 3 x - 4}}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} + 8 x + 16}{x^{2} + 3 x - 4} = 0$

Step 6

Solve for $x$ .

Set the numerator equal to zero.

$x^{2} + 8 x + 16 = 0$

Solve the equation for $x$ .

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$x^{2} + 8 x + 4^{2} = 0$

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 4$ .

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

Set the $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Exclude the solutions that do not make $\frac{x^{2} + 8 x + 16}{x^{2} + 3 x - 4} = 0$ 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 $0$ .

$x = - 4, x = 1$

Step 8

Find Where Undefined/Discontinuous ((x^2-16)/(4x-4))÷((x^2+8x+16)/(x^2+3x-4))

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