Divide (6/(k-4))÷((k+6)/(k^2+2k-24))

$\frac{6}{k - 4} \div \frac{k + 6}{k^{2} + 2 k - 24}$

To divide by a fraction, multiply by its reciprocal.

$\frac{6}{k - 4} \cdot \frac{k^{2} + 2 k - 24}{k + 6}$

Factor $k^{2} + 2 k - 24$ 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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Write the factored form using these integers.

$\frac{6}{k - 4} \cdot \frac{(k - 4) (k + 6)}{k + 6}$

Cancel the common factor of $k - 4$ .

Cancel the common factor.

$\frac{6}{k - 4} \cdot \frac{(k - 4) (k + 6)}{k + 6}$

Rewrite the expression.

$6 \frac{k + 6}{k + 6}$

Combine $6$ and $\frac{k + 6}{k + 6}$ .

$\frac{6 (k + 6)}{k + 6}$

Cancel the common factor of $k + 6$ .

Cancel the common factor.

$\frac{6 (k + 6)}{k + 6}$

Divide $6$ by $1$ .

$6$

Divide (6/(k-4))÷((k+6)/(k^2+2k-24))

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