Simplify (6y^2+18y-60)/(3y^2-12y)*(y^2-16)/(y^2+2y-8)

$\frac{6 y^{2} + 18 y - 60}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Simplify the numerator.

Factor $6$ out of $6 y^{2} + 18 y - 60$ .

Factor $6$ out of $6 y^{2}$ .

$\frac{6 (y^{2}) + 18 y - 60}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $6$ out of $18 y$ .

$\frac{6 (y^{2}) + 6 (3 y) - 60}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $6$ out of $- 60$ .

$\frac{6 y^{2} + 6 (3 y) + 6 \cdot - 10}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $6$ out of $6 y^{2} + 6 (3 y)$ .

$\frac{6 (y^{2} + 3 y) + 6 \cdot - 10}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $6$ out of $6 (y^{2} + 3 y) + 6 \cdot - 10$ .

$\frac{6 (y^{2} + 3 y - 10)}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $y^{2} + 3 y - 10$ 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 $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

$\frac{6 ((y - 2) (y + 5))}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

$\frac{6 (y - 2) (y + 5)}{3 y^{2} - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $3 y$ out of $3 y^{2} - 12 y$ .

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

$\frac{6 (y - 2) (y + 5)}{3 y (y) - 12 y} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $3 y$ out of $- 12 y$ .

$\frac{6 (y - 2) (y + 5)}{3 y (y) + 3 y (- 4)} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Factor $3 y$ out of $3 y (y) + 3 y (- 4)$ .

$\frac{6 (y - 2) (y + 5)}{3 y (y - 4)} \cdot \frac{y^{2} - 16}{y^{2} + 2 y - 8}$

Simplify the numerator.

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

$\frac{6 (y - 2) (y + 5)}{3 y (y - 4)} \cdot \frac{y^{2} - 4^{2}}{y^{2} + 2 y - 8}$

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 $b = 4$ .

$\frac{6 (y - 2) (y + 5)}{3 y (y - 4)} \cdot \frac{(y + 4) (y - 4)}{y^{2} + 2 y - 8}$

Factor $y^{2} + 2 y - 8$ 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 $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$\frac{6 (y - 2) (y + 5)}{3 y (y - 4)} \cdot \frac{(y + 4) (y - 4)}{(y - 2) (y + 4)}$

Simplify terms.

Cancel the common factor of $y - 2$ .

Factor $y - 2$ out of $6 (y - 2) (y + 5)$ .

$\frac{(y - 2) (6 (y + 5))}{3 y (y - 4)} \cdot \frac{(y + 4) (y - 4)}{(y - 2) (y + 4)}$

Cancel the common factor.

$\frac{(y - 2) (6 (y + 5))}{3 y (y - 4)} \cdot \frac{(y + 4) (y - 4)}{(y - 2) (y + 4)}$

Rewrite the expression.

$\frac{6 (y + 5)}{3 y (y - 4)} \cdot \frac{(y + 4) (y - 4)}{y + 4}$

Cancel the common factor of $y - 4$ .

Factor $y - 4$ out of $3 y (y - 4)$ .

$\frac{6 (y + 5)}{(y - 4) (3 y)} \cdot \frac{(y + 4) (y - 4)}{y + 4}$

Factor $y - 4$ out of $(y + 4) (y - 4)$ .

$\frac{6 (y + 5)}{(y - 4) (3 y)} \cdot \frac{(y - 4) (y + 4)}{y + 4}$

Cancel the common factor.

$\frac{6 (y + 5)}{(y - 4) (3 y)} \cdot \frac{(y - 4) (y + 4)}{y + 4}$

Rewrite the expression.

$\frac{6 (y + 5)}{3 y} \cdot \frac{y + 4}{y + 4}$

Multiply $\frac{6 (y + 5)}{3 y}$ by $\frac{y + 4}{y + 4}$ .

$\frac{6 (y + 5) (y + 4)}{3 y (y + 4)}$

Cancel the common factor of $6$ and $3$ .

Factor $3$ out of $6 (y + 5) (y + 4)$ .

$\frac{3 (2 (y + 5) (y + 4))}{3 y (y + 4)}$

Cancel the common factors.

Factor $3$ out of $3 y (y + 4)$ .

$\frac{3 (2 (y + 5) (y + 4))}{3 (y (y + 4))}$

Cancel the common factor.

$\frac{3 (2 (y + 5) (y + 4))}{3 (y (y + 4))}$

Rewrite the expression.

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

Cancel the common factor of $y + 4$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 (y + 5)}{y}$

Simplify (6y^2+18y-60)/(3y^2-12y)*(y^2-16)/(y^2+2y-8)

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