Find Where Undefined/Discontinuous (x^2+9x+20)/(x+3)*(x^2-x-12)/(x^2+3x-4)

$\frac{x^{2} + 9 x + 20}{x + 3} \cdot \frac{x^{2} - x - 12}{x^{2} + 3 x - 4}$

Step 1

Set the denominator in $\frac{x^{2} + 9 x + 20}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Step 2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3

Set the denominator in $\frac{x^{2} - x - 12}{x^{2} + 3 x - 4}$ 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

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 $0$ .

$x = - 4, x = - 3, x = 1$

Step 6

Find Where Undefined/Discontinuous (x^2+9x+20)/(x+3)*(x^2-x-12)/(x^2+3x-4)

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