Find the Domain (k^3+k)/(k^2+k-42)

$\frac{k^{3} + k}{k^{2} + k - 42}$

Step 1

Set the denominator in $\frac{k^{3} + k}{k^{2} + k - 42}$ equal to $0$ to find where the expression is undefined.

$k^{2} + k - 42 = 0$

Step 2

Solve for $k$ .

Factor $k^{2} + k - 42$ 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 $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

$(k - 6) (k + 7) = 0$

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

$k - 6 = 0$

$k + 7 = 0$

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

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

$k - 6 = 0$

Add $6$ to both sides of the equation.

$k = 6$

Set $k + 7$ equal to $0$ and solve for $k$ .

Set $k + 7$ equal to $0$ .

$k + 7 = 0$

Subtract $7$ from both sides of the equation.

$k = - 7$

The final solution is all the values that make $(k - 6) (k + 7) = 0$ true.

$k = 6, - 7$

Step 3

The domain is all values of $k$ that make the expression defined.

Interval Notation:

$(- \infty, - 7) \cup (- 7, 6) \cup (6, \infty)$

Set-Builder Notation:

${k | k \neq 6, - 7}$

Step 4

Find the Domain (k^3+k)/(k^2+k-42)

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