Simplify ((q^2+2q)/(6q))÷((q^2-4)/(3q^2))

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

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{q^{2} + 2 q}{6 q} \cdot \frac{3 q^{2}}{q^{2} - 4}$

Step 2

Simplify terms.

Cancel the common factor of $3 q$ .

Factor $3 q$ out of $6 q$ .

$\frac{q^{2} + 2 q}{3 q (2)} \cdot \frac{3 q^{2}}{q^{2} - 4}$

Factor $3 q$ out of $3 q^{2}$ .

$\frac{q^{2} + 2 q}{3 q (2)} \cdot \frac{3 q (q)}{q^{2} - 4}$

Cancel the common factor.

$\frac{q^{2} + 2 q}{3 q \cdot 2} \cdot \frac{3 q \cdot q}{q^{2} - 4}$

Rewrite the expression.

$\frac{q^{2} + 2 q}{2} \cdot \frac{q}{q^{2} - 4}$

Multiply $\frac{q^{2} + 2 q}{2}$ by $\frac{q}{q^{2} - 4}$ .

$\frac{(q^{2} + 2 q) q}{2 (q^{2} - 4)}$

Step 3

Simplify the numerator.

Factor $q$ out of $q^{2} + 2 q$ .

Factor $q$ out of $q^{2}$ .

$\frac{(q \cdot q + 2 q) q}{2 (q^{2} - 4)}$

Factor $q$ out of $2 q$ .

$\frac{(q \cdot q + q \cdot 2) q}{2 (q^{2} - 4)}$

Factor $q$ out of $q \cdot q + q \cdot 2$ .

$\frac{q (q + 2) q}{2 (q^{2} - 4)}$

Combine exponents.

Raise $q$ to the power of $1$ .

$\frac{q^{1} q (q + 2)}{2 (q^{2} - 4)}$

Raise $q$ to the power of $1$ .

$\frac{q^{1} q^{1} (q + 2)}{2 (q^{2} - 4)}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{q^{1 + 1} (q + 2)}{2 (q^{2} - 4)}$

Add $1$ and $1$ .

$\frac{q^{2} (q + 2)}{2 (q^{2} - 4)}$

Step 4

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$\frac{q^{2} (q + 2)}{2 (q^{2} - 2^{2})}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = q$ and $b = 2$ .

$\frac{q^{2} (q + 2)}{2 (q + 2) (q - 2)}$

Step 5

Cancel the common factor of $q + 2$ .

Cancel the common factor.

$\frac{q^{2} (q + 2)}{2 (q + 2) (q - 2)}$

Rewrite the expression.

$\frac{q^{2}}{2 (q - 2)}$

Simplify ((q^2+2q)/(6q))÷((q^2-4)/(3q^2))

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