Simplify ((6u+72)/(2u+24))÷((u^2+21u+108)/(u^2+20u+96))

$\frac{6 u + 72}{2 u + 24} \div \frac{u^{2} + 21 u + 108}{u^{2} + 20 u + 96}$

To divide by a fraction, multiply by its reciprocal.

$\frac{6 u + 72}{2 u + 24} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Simplify terms.

Cancel the common factor of $6 u + 72$ and $2 u + 24$ .

Factor $2$ out of $6 u$ .

$\frac{2 (3 u) + 72}{2 u + 24} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $2$ out of $72$ .

$\frac{2 (3 u) + 2 \cdot 36}{2 u + 24} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $2$ out of $2 (3 u) + 2 (36)$ .

$\frac{2 (3 u + 36)}{2 u + 24} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Cancel the common factors.

Factor $2$ out of $2 u$ .

$\frac{2 (3 u + 36)}{2 (u) + 24} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $2$ out of $24$ .

$\frac{2 (3 u + 36)}{2 u + 2 \cdot 12} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $2$ out of $2 u + 2 \cdot 12$ .

$\frac{2 (3 u + 36)}{2 (u + 12)} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Cancel the common factor.

$\frac{2 (3 u + 36)}{2 (u + 12)} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Rewrite the expression.

$\frac{3 u + 36}{u + 12} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $3$ out of $3 u + 36$ .

Factor $3$ out of $3 u$ .

$\frac{3 (u) + 36}{u + 12} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $3$ out of $36$ .

$\frac{3 u + 3 \cdot 12}{u + 12} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $3$ out of $3 u + 3 \cdot 12$ .

$\frac{3 (u + 12)}{u + 12} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Cancel the common factor of $u + 12$ .

Cancel the common factor.

$\frac{3 (u + 12)}{u + 12} \cdot \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Divide $3$ by $1$ .

$3 \frac{u^{2} + 20 u + 96}{u^{2} + 21 u + 108}$

Factor $u^{2} + 20 u + 96$ 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 $96$ and whose sum is $20$ .

$8, 12$

Write the factored form using these integers.

$3 \frac{(u + 8) (u + 12)}{u^{2} + 21 u + 108}$

Factor $u^{2} + 21 u + 108$ 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 $108$ and whose sum is $21$ .

$9, 12$

Write the factored form using these integers.

$3 \frac{(u + 8) (u + 12)}{(u + 9) (u + 12)}$

Simplify terms.

Cancel the common factor of $u + 12$ .

Cancel the common factor.

$3 \frac{(u + 8) (u + 12)}{(u + 9) (u + 12)}$

Rewrite the expression.

$3 \frac{u + 8}{u + 9}$

Combine $3$ and $\frac{u + 8}{u + 9}$ .

$\frac{3 (u + 8)}{u + 9}$

Simplify ((6u+72)/(2u+24))÷((u^2+21u+108)/(u^2+20u+96))

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