Simplify ((a^2+4a+4)/(16-b^4))÷((4-a^2)/(4+b^2))

$\frac{a^{2} + 4 a + 4}{16 - b^{4}} \div \frac{4 - a^{2}}{4 + b^{2}}$

To divide by a fraction, multiply by its reciprocal.

$\frac{a^{2} + 4 a + 4}{16 - b^{4}} \cdot \frac{4 + b^{2}}{4 - a^{2}}$

Factor using the perfect square rule.

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

$\frac{a^{2} + 4 a + 2^{2}}{16 - b^{4}} \cdot \frac{4 + b^{2}}{4 - a^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 a = 2 \cdot a \cdot 2$

Rewrite the polynomial.

$\frac{a^{2} + 2 \cdot a \cdot 2 + 2^{2}}{16 - b^{4}} \cdot \frac{4 + b^{2}}{4 - a^{2}}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = a$ and $b = 2$ .

$\frac{{(a + 2)}^{2}}{16 - b^{4}} \cdot \frac{4 + b^{2}}{4 - a^{2}}$

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

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

Rewrite $b^{4}$ as ${(b^{2})}^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = b^{2}$ .

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

Simplify.

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

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

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

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

Simplify terms.

Cancel the common factor of $4 + b^{2}$ .

Factor $4 + b^{2}$ out of $(4 + b^{2}) (2 + b) (2 - b)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{{(a + 2)}^{2}}{(2 + b) (2 - b)} \cdot \frac{1}{4 - a^{2}}$

Multiply $\frac{{(a + 2)}^{2}}{(2 + b) (2 - b)}$ by $\frac{1}{4 - a^{2}}$ .

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

Simplify the denominator.

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

$\frac{{(a + 2)}^{2}}{(2 + b) (2 - b) (2^{2} - a^{2})}$

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

$\frac{{(a + 2)}^{2}}{(2 + b) (2 - b) (2 + a) (2 - a)}$

Cancel the common factor of ${(a + 2)}^{2}$ and $2 + a$ .

Reorder terms.

$\frac{{(a + 2)}^{2}}{(2 + b) (2 - b) (a + 2) (2 - a)}$

Factor $a + 2$ out of ${(a + 2)}^{2}$ .

$\frac{(a + 2) (a + 2)}{(2 + b) (2 - b) (a + 2) (2 - a)}$

Cancel the common factors.

Factor $a + 2$ out of $(2 + b) (2 - b) (a + 2) (2 - a)$ .

$\frac{(a + 2) (a + 2)}{(a + 2) ((2 + b) (2 - b) (2 - a))}$

Cancel the common factor.

$\frac{(a + 2) (a + 2)}{(a + 2) ((2 + b) (2 - b) (2 - a))}$

Rewrite the expression.

$\frac{a + 2}{(2 + b) (2 - b) (2 - a)}$

Simplify ((a^2+4a+4)/(16-b^4))÷((4-a^2)/(4+b^2))

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