Simplify (y^2-36)/(8y^2+96y+288)

$\frac{y^{2} - 36}{8 y^{2} + 96 y + 288}$

Simplify the numerator.

Rewrite $36$ as $6^{2}$ .

$\frac{y^{2} - 6^{2}}{8 y^{2} + 96 y + 288}$

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 $b = 6$ .

$\frac{(y + 6) (y - 6)}{8 y^{2} + 96 y + 288}$

Simplify the denominator.

Factor $8$ out of $8 y^{2} + 96 y + 288$ .

Factor $8$ out of $8 y^{2}$ .

$\frac{(y + 6) (y - 6)}{8 (y^{2}) + 96 y + 288}$

Factor $8$ out of $96 y$ .

$\frac{(y + 6) (y - 6)}{8 (y^{2}) + 8 (12 y) + 288}$

Factor $8$ out of $288$ .

$\frac{(y + 6) (y - 6)}{8 y^{2} + 8 (12 y) + 8 \cdot 36}$

Factor $8$ out of $8 y^{2} + 8 (12 y)$ .

$\frac{(y + 6) (y - 6)}{8 (y^{2} + 12 y) + 8 \cdot 36}$

Factor $8$ out of $8 (y^{2} + 12 y) + 8 \cdot 36$ .

$\frac{(y + 6) (y - 6)}{8 (y^{2} + 12 y + 36)}$

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$\frac{(y + 6) (y - 6)}{8 (y^{2} + 12 y + 6^{2})}$

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

$12 y = 2 \cdot y \cdot 6$

Rewrite the polynomial.

$\frac{(y + 6) (y - 6)}{8 (y^{2} + 2 \cdot y \cdot 6 + 6^{2})}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 6$ .

$\frac{(y + 6) (y - 6)}{8 {(y + 6)}^{2}}$

Cancel the common factors.

Factor $y + 6$ out of $8 {(y + 6)}^{2}$ .

$\frac{(y + 6) (y - 6)}{(y + 6) (8 (y + 6))}$

Cancel the common factor.

$\frac{(y + 6) (y - 6)}{(y + 6) (8 (y + 6))}$

Rewrite the expression.

$\frac{y - 6}{8 (y + 6)}$