Simplify (-8x^2-48xy-72y^2)/(2x^2-18y^2)

$\frac{- 8 x^{2} - 48 x y - 72 y^{2}}{2 x^{2} - 18 y^{2}}$

Step 1

Cancel the common factor of $- 8 x^{2} - 48 x y - 72 y^{2}$ and $2 x^{2} - 18 y^{2}$ .

Factor $2$ out of $- 8 x^{2}$ .

$\frac{2 (- 4 x^{2}) - 48 x y - 72 y^{2}}{2 x^{2} - 18 y^{2}}$

Factor $2$ out of $- 48 x y$ .

$\frac{2 (- 4 x^{2}) + 2 (- 24 x y) - 72 y^{2}}{2 x^{2} - 18 y^{2}}$

Factor $2$ out of $2 (- 4 x^{2}) + 2 (- 24 x y)$ .

$\frac{2 (- 4 x^{2} - 24 x y) - 72 y^{2}}{2 x^{2} - 18 y^{2}}$

Factor $2$ out of $- 72 y^{2}$ .

$\frac{2 (- 4 x^{2} - 24 x y) + 2 (- 36 y^{2})}{2 x^{2} - 18 y^{2}}$

Factor $2$ out of $2 (- 4 x^{2} - 24 x y) + 2 (- 36 y^{2})$ .

$\frac{2 (- 4 x^{2} - 24 x y - 36 y^{2})}{2 x^{2} - 18 y^{2}}$

Cancel the common factors.

Factor $2$ out of $2 x^{2}$ .

$\frac{2 (- 4 x^{2} - 24 x y - 36 y^{2})}{2 (x^{2}) - 18 y^{2}}$

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

$\frac{2 (- 4 x^{2} - 24 x y - 36 y^{2})}{2 (x^{2}) + 2 (- 9 y^{2})}$

Factor $2$ out of $2 (x^{2}) + 2 (- 9 y^{2})$ .

$\frac{2 (- 4 x^{2} - 24 x y - 36 y^{2})}{2 (x^{2} - 9 y^{2})}$

Cancel the common factor.

$\frac{2 (- 4 x^{2} - 24 x y - 36 y^{2})}{2 (x^{2} - 9 y^{2})}$

Rewrite the expression.

$\frac{- 4 x^{2} - 24 x y - 36 y^{2}}{x^{2} - 9 y^{2}}$

Step 2

Simplify the numerator.

Factor $4$ out of $- 4 x^{2} - 24 x y - 36 y^{2}$ .

Factor $4$ out of $- 4 x^{2}$ .

$\frac{4 (- x^{2}) - 24 x y - 36 y^{2}}{x^{2} - 9 y^{2}}$

Factor $4$ out of $- 24 x y$ .

$\frac{4 (- x^{2}) + 4 (- 6 x y) - 36 y^{2}}{x^{2} - 9 y^{2}}$

Factor $4$ out of $- 36 y^{2}$ .

$\frac{4 (- x^{2}) + 4 (- 6 x y) + 4 (- 9 y^{2})}{x^{2} - 9 y^{2}}$

Factor $4$ out of $4 (- x^{2}) + 4 (- 6 x y)$ .

$\frac{4 (- x^{2} - 6 x y) + 4 (- 9 y^{2})}{x^{2} - 9 y^{2}}$

Factor $4$ out of $4 (- x^{2} - 6 x y) + 4 (- 9 y^{2})$ .

$\frac{4 (- x^{2} - 6 x y - 9 y^{2})}{x^{2} - 9 y^{2}}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 9 = 9$ and whose sum is $b = - 6$ .

Reorder terms.

$\frac{4 (- x^{2} - 9 y^{2} - 6 x y)}{x^{2} - 9 y^{2}}$

Reorder $- 9 y^{2}$ and $- 6 x y$ .

$\frac{4 (- x^{2} - 6 x y - 9 y^{2})}{x^{2} - 9 y^{2}}$

Factor $- 6$ out of $- 6 x y$ .

$\frac{4 (- x^{2} - 6 (x y) - 9 y^{2})}{x^{2} - 9 y^{2}}$

Rewrite $- 6$ as $- 3$ plus $- 3$

$\frac{4 (- x^{2} + (- 3 - 3) (x y) - 9 y^{2})}{x^{2} - 9 y^{2}}$

Apply the distributive property.

$\frac{4 (- x^{2} - 3 (x y) - 3 (x y) - 9 y^{2})}{x^{2} - 9 y^{2}}$

Move parentheses.

$\frac{4 (- x^{2} - 3 x y - 3 x y - 9 y^{2})}{x^{2} - 9 y^{2}}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{4 ((- x^{2} - 3 x y) - 3 x y - 9 y^{2})}{x^{2} - 9 y^{2}}$

Factor out the greatest common factor (GCF) from each group.

$\frac{4 (x (- x - 3 y) + 3 y (- x - 3 y))}{x^{2} - 9 y^{2}}$

Factor the polynomial by factoring out the greatest common factor, $- x - 3 y$ .

$\frac{4 ((- x - 3 y) (x + 3 y))}{x^{2} - 9 y^{2}}$

$\frac{4 (- x - 3 y) (x + 3 y)}{x^{2} - 9 y^{2}}$

Combine exponents.

Factor $- 1$ out of $- x$ .

$\frac{4 (- (x) - 3 y) (x + 3 y)}{x^{2} - 9 y^{2}}$

Factor $- 1$ out of $- 3 y$ .

$\frac{4 (- (x) - (3 y)) (x + 3 y)}{x^{2} - 9 y^{2}}$

Factor $- 1$ out of $- (x) - (3 y)$ .

$\frac{4 (- (x + 3 y)) (x + 3 y)}{x^{2} - 9 y^{2}}$

Rewrite $- (x + 3 y)$ as $- 1 (x + 3 y)$ .

$\frac{4 (- 1 (x + 3 y)) (x + 3 y)}{x^{2} - 9 y^{2}}$

Raise $x + 3 y$ to the power of $1$ .

$\frac{4 \cdot - 1 ({(x + 3 y)}^{1} (x + 3 y))}{x^{2} - 9 y^{2}}$

Raise $x + 3 y$ to the power of $1$ .

$\frac{4 \cdot - 1 ({(x + 3 y)}^{1} {(x + 3 y)}^{1})}{x^{2} - 9 y^{2}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{4 \cdot - 1 {(x + 3 y)}^{1 + 1}}{x^{2} - 9 y^{2}}$

Add $1$ and $1$ .

$\frac{4 \cdot - 1 {(x + 3 y)}^{2}}{x^{2} - 9 y^{2}}$

Multiply $4$ by $- 1$ .

$\frac{- 4 {(x + 3 y)}^{2}}{x^{2} - 9 y^{2}}$

Step 3

Simplify the denominator.

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

$\frac{- 4 {(x + 3 y)}^{2}}{x^{2} - {(3 y)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3 y$ .

$\frac{- 4 {(x + 3 y)}^{2}}{(x + 3 y) (x - (3 y))}$

Multiply $3$ by $- 1$ .

$\frac{- 4 {(x + 3 y)}^{2}}{(x + 3 y) (x - 3 y)}$

Step 4

Reduce the expression by cancelling the common factors.

Cancel the common factor of ${(x + 3 y)}^{2}$ and $x + 3 y$ .

Factor $x + 3 y$ out of $- 4 {(x + 3 y)}^{2}$ .

$\frac{(x + 3 y) (- 4 (x + 3 y))}{(x + 3 y) (x - 3 y)}$

Cancel the common factors.

Cancel the common factor.

$\frac{(x + 3 y) (- 4 (x + 3 y))}{(x + 3 y) (x - 3 y)}$

Rewrite the expression.

$\frac{- 4 (x + 3 y)}{x - 3 y}$

Move the negative in front of the fraction.

$- \frac{4 (x + 3 y)}{x - 3 y}$

Simplify (-8x^2-48xy-72y^2)/(2x^2-18y^2)

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