Find the Excluded Values (v^2+1)/(v^2-v-6)

$\frac{v^{2} + 1}{v^{2} - v - 6}$

Step 1

Set the denominator in $\frac{v^{2} + 1}{v^{2} - v - 6}$ equal to $0$ to find where the expression is undefined.

$v^{2} - v - 6 = 0$

Step 2

Solve for $v$ .

Factor $v^{2} - v - 6$ 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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(v - 3) (v + 2) = 0$

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

$v - 3 = 0$

$v + 2 = 0$

Set $v - 3$ equal to $0$ and solve for $v$ .

Set $v - 3$ equal to $0$ .

$v - 3 = 0$

Add $3$ to both sides of the equation.

$v = 3$

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

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

$v + 2 = 0$

Subtract $2$ from both sides of the equation.

$v = - 2$

The final solution is all the values that make $(v - 3) (v + 2) = 0$ true.

$v = 3, - 2$

Step 3

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 $0$ .

$v = - 2, v = 3$

Step 4

Find the Excluded Values (v^2+1)/(v^2-v-6)

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