Simplify (2x^2-7x-15)/(x^2-25)*(x^2+x)/(2x^2+5x+3)

$\frac{2 x^{2} - 7 x - 15}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

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

Factor $- 7$ out of $- 7 x$ .

$\frac{2 x^{2} - 7 (x) - 15}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

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

$\frac{2 x^{2} + (3 - 10) x - 15}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

Apply the distributive property.

$\frac{2 x^{2} + 3 x - 10 x - 15}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{2} + 3 x) - 10 x - 15}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

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

$\frac{x (2 x + 3) - 5 (2 x + 3)}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

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

$\frac{(2 x + 3) (x - 5)}{x^{2} - 25} \cdot \frac{x^{2} + x}{2 x^{2} + 5 x + 3}$

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

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

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 $b = 5$ .

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

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

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

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

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

Factor $x$ out of $x \cdot x + x \cdot 1$ .

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

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ as $2$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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

Factor $2 x + 3$ out of $(x + 1) (2 x + 3)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 5}{(x + 5) (x - 5)} \cdot \frac{x (x + 1)}{x + 1}$

Multiply $\frac{x - 5}{(x + 5) (x - 5)}$ by $\frac{x (x + 1)}{x + 1}$ .

$\frac{(x - 5) (x (x + 1))}{(x + 5) (x - 5) (x + 1)}$

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$\frac{(x - 5) (x (x + 1))}{(x + 5) (x - 5) (x + 1)}$

Rewrite the expression.

$\frac{x (x + 1)}{(x + 5) (x + 1)}$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{x (x + 1)}{(x + 5) (x + 1)}$

Rewrite the expression.

$\frac{x}{x + 5}$

Simplify (2x^2-7x-15)/(x^2-25)*(x^2+x)/(2x^2+5x+3)

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