Find Where Undefined/Discontinuous ((6-x)/(x^2+2x-3))÷((x^2-4x-12)/(x^2+4x+3))

$\frac{6 - x}{x^{2} + 2 x - 3} \div \frac{x^{2} - 4 x - 12}{x^{2} + 4 x + 3}$

Step 1

Set the denominator in $\frac{6 - x}{x^{2} + 2 x - 3}$ equal to $0$ to find where the expression is undefined.

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

Step 2

Solve for $x$ .

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

$- 1, 3$

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

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

$x - 1 = 0$

$x + 3 = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Step 3

Set the denominator in $\frac{x^{2} - 4 x - 12}{x^{2} + 4 x + 3}$ equal to $0$ to find where the expression is undefined.

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

Step 4

Solve for $x$ .

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

$1, 3$

Write the factored form using these integers.

$(x + 1) (x + 3) = 0$

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

$x + 1 = 0$

$x + 3 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

Step 5

Set the denominator in $\frac{\frac{6 - x}{x^{2} + 2 x - 3}}{\frac{x^{2} - 4 x - 12}{x^{2} + 4 x + 3}}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} - 4 x - 12}{x^{2} + 4 x + 3} = 0$

Step 6

Solve for $x$ .

Set the numerator equal to zero.

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

Solve the equation for $x$ .

Factor $x^{2} - 4 x - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$(x - 6) (x + 2) = 0$

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

$x - 6 = 0$

$x + 2 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

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 $0$ .

$x = - 3, x = - 2, x = - 1, x = 1, x = 6$

Step 8

Find Where Undefined/Discontinuous ((6-x)/(x^2+2x-3))÷((x^2-4x-12)/(x^2+4x+3))

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