Simplify ((2y^2-7y-15)/(3y^2-8y-3)*(9y^2-1)/(4y^2-9))÷((y^2+3y-10)/(2y^2-9y+9))

$\frac{2 y^{2} - 7 y - 15}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \div \frac{y^{2} + 3 y - 10}{2 y^{2} - 9 y + 9}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{2 y^{2} - 7 y - 15}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Step 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 $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 y$ .

$\frac{2 y^{2} - 7 (y) - 15}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Rewrite $- 7$ as $3$ plus $- 10$

$\frac{2 y^{2} + (3 - 10) y - 15}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Apply the distributive property.

$\frac{2 y^{2} + 3 y - 10 y - 15}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 y^{2} + 3 y) - 10 y - 15}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

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

$\frac{y (2 y + 3) - 5 (2 y + 3)}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Factor the polynomial by factoring out the greatest common factor, $2 y + 3$ .

$\frac{(2 y + 3) (y - 5)}{3 y^{2} - 8 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Step 3

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 $a \cdot c = 3 \cdot - 3 = - 9$ and whose sum is $b = - 8$ .

Factor $- 8$ out of $- 8 y$ .

$\frac{(2 y + 3) (y - 5)}{3 y^{2} - 8 (y) - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Rewrite $- 8$ as $1$ plus $- 9$

$\frac{(2 y + 3) (y - 5)}{3 y^{2} + (1 - 9) y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Apply the distributive property.

$\frac{(2 y + 3) (y - 5)}{3 y^{2} + 1 y - 9 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Multiply $y$ by $1$ .

$\frac{(2 y + 3) (y - 5)}{3 y^{2} + y - 9 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 y + 3) (y - 5)}{(3 y^{2} + y) - 9 y - 3} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

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

$\frac{(2 y + 3) (y - 5)}{y (3 y + 1) - 3 (3 y + 1)} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Factor the polynomial by factoring out the greatest common factor, $3 y + 1$ .

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{9 y^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Step 4

Simplify the numerator.

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

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{{(3 y)}^{2} - 1}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Rewrite $1$ as $1^{2}$ .

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{{(3 y)}^{2} - 1^{2}}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 y$ and $b = 1$ .

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{4 y^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Step 5

Simplify the denominator.

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

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{{(2 y)}^{2} - 9} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Rewrite $9$ as $3^{2}$ .

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{{(2 y)}^{2} - 3^{2}} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

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

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{(2 y + 3) (2 y - 3)} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Step 6

Simplify terms.

Cancel the common factor of $2 y + 3$ .

Cancel the common factor.

$\frac{(2 y + 3) (y - 5)}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{(2 y + 3) (2 y - 3)} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Rewrite the expression.

$\frac{y - 5}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{2 y - 3} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Cancel the common factor of $3 y + 1$ .

Cancel the common factor.

$\frac{y - 5}{(3 y + 1) (y - 3)} \cdot \frac{(3 y + 1) (3 y - 1)}{2 y - 3} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Rewrite the expression.

$\frac{y - 5}{y - 3} \cdot \frac{3 y - 1}{2 y - 3} \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Multiply $\frac{y - 5}{y - 3}$ by $\frac{3 y - 1}{2 y - 3}$ .

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{2 y^{2} - 9 y + 9}{y^{2} + 3 y - 10}$

Step 7

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 $a \cdot c = 2 \cdot 9 = 18$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 y$ .

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{2 y^{2} - 9 (y) + 9}{y^{2} + 3 y - 10}$

Rewrite $- 9$ as $- 3$ plus $- 6$

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{2 y^{2} + (- 3 - 6) y + 9}{y^{2} + 3 y - 10}$

Apply the distributive property.

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{2 y^{2} - 3 y - 6 y + 9}{y^{2} + 3 y - 10}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{(2 y^{2} - 3 y) - 6 y + 9}{y^{2} + 3 y - 10}$

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

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{y (2 y - 3) - 3 (2 y - 3)}{y^{2} + 3 y - 10}$

Factor the polynomial by factoring out the greatest common factor, $2 y - 3$ .

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{(2 y - 3) (y - 3)}{y^{2} + 3 y - 10}$

Step 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.

$\frac{(y - 5) (3 y - 1)}{(y - 3) (2 y - 3)} \cdot \frac{(2 y - 3) (y - 3)}{(y - 2) (y + 5)}$

Step 9

Cancel the common factor of $(2 y - 3) (y - 3)$ .

Factor $(2 y - 3) (y - 3)$ out of $(y - 3) (2 y - 3)$ .

$\frac{(y - 5) (3 y - 1)}{(2 y - 3) (y - 3) (1)} \cdot \frac{(2 y - 3) (y - 3)}{(y - 2) (y + 5)}$

Cancel the common factor.

$\frac{(y - 5) (3 y - 1)}{(2 y - 3) (y - 3) \cdot 1} \cdot \frac{(2 y - 3) (y - 3)}{(y - 2) (y + 5)}$

Rewrite the expression.

$(y - 5) (3 y - 1) \frac{1}{(y - 2) (y + 5)}$

Step 10

Expand $(y - 5) (3 y - 1)$ using the FOIL Method.

Apply the distributive property.

$(y (3 y - 1) - 5 (3 y - 1)) \frac{1}{(y - 2) (y + 5)}$

Apply the distributive property.

$(y (3 y) + y \cdot - 1 - 5 (3 y - 1)) \frac{1}{(y - 2) (y + 5)}$

Apply the distributive property.

$(y (3 y) + y \cdot - 1 - 5 (3 y) - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Step 11

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$(3 y \cdot y + y \cdot - 1 - 5 (3 y) - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$(3 (y \cdot y) + y \cdot - 1 - 5 (3 y) - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Multiply $y$ by $y$ .

$(3 y^{2} + y \cdot - 1 - 5 (3 y) - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Move $- 1$ to the left of $y$ .

$(3 y^{2} - 1 \cdot y - 5 (3 y) - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Rewrite $- 1 y$ as $- y$ .

$(3 y^{2} - y - 5 (3 y) - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Multiply $3$ by $- 5$ .

$(3 y^{2} - y - 15 y - 5 \cdot - 1) \frac{1}{(y - 2) (y + 5)}$

Multiply $- 5$ by $- 1$ .

$(3 y^{2} - y - 15 y + 5) \frac{1}{(y - 2) (y + 5)}$

Subtract $15 y$ from $- y$ .

$(3 y^{2} - 16 y + 5) \frac{1}{(y - 2) (y + 5)}$

Step 12

Multiply $3 y^{2} - 16 y + 5$ by $\frac{1}{(y - 2) (y + 5)}$ .

$\frac{3 y^{2} - 16 y + 5}{(y - 2) (y + 5)}$

Step 13

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 $a \cdot c = 3 \cdot 5 = 15$ and whose sum is $b = - 16$ .

Factor $- 16$ out of $- 16 y$ .

$\frac{3 y^{2} - 16 (y) + 5}{(y - 2) (y + 5)}$

Rewrite $- 16$ as $- 1$ plus $- 15$

$\frac{3 y^{2} + (- 1 - 15) y + 5}{(y - 2) (y + 5)}$

Apply the distributive property.

$\frac{3 y^{2} - 1 y - 15 y + 5}{(y - 2) (y + 5)}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(3 y^{2} - 1 y) - 15 y + 5}{(y - 2) (y + 5)}$

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

$\frac{y (3 y - 1) - 5 (3 y - 1)}{(y - 2) (y + 5)}$

Factor the polynomial by factoring out the greatest common factor, $3 y - 1$ .

$\frac{(3 y - 1) (y - 5)}{(y - 2) (y + 5)}$

Simplify ((2y^2-7y-15)/(3y^2-8y-3)*(9y^2-1)/(4y^2-9))÷((y^2+3y-10)/(2y^2-9y+9))

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