Simplify ((5y)/(y^2+4y+4))÷((20y^2)/(y^2-4))

$\frac{5 y}{y^{2} + 4 y + 4} \div \frac{20 y^{2}}{y^{2} - 4}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{5 y}{y^{2} + 4 y + 4} \cdot \frac{y^{2} - 4}{20 y^{2}}$

Step 2

Simplify terms.

Cancel the common factor of $5 y$ .

Factor $5 y$ out of $20 y^{2}$ .

$\frac{5 y}{y^{2} + 4 y + 4} \cdot \frac{y^{2} - 4}{5 y (4 y)}$

Cancel the common factor.

$\frac{5 y}{y^{2} + 4 y + 4} \cdot \frac{y^{2} - 4}{5 y (4 y)}$

Rewrite the expression.

$\frac{1}{y^{2} + 4 y + 4} \cdot \frac{y^{2} - 4}{4 y}$

Multiply $\frac{1}{y^{2} + 4 y + 4}$ by $\frac{y^{2} - 4}{4 y}$ .

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

Step 3

Simplify the numerator.

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

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

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

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

Step 4

Factor using the perfect square rule.

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

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

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

$4 y = 2 \cdot y \cdot 2$

Rewrite the polynomial.

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

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

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

Step 5

Cancel the common factors.

Factor $y + 2$ out of ${(y + 2)}^{2} \cdot 4 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

Move $4$ to the left of $y + 2$ .

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

Step 7

Reorder factors in $\frac{y - 2}{4 (y + 2) y}$ .

$\frac{y - 2}{4 y (y + 2)}$