Simplify (10x+2x^2)/(25-x^2)*(x^2-2x-15)/(3x^2+12x+9)

$\frac{10 x + 2 x^{2}}{25 - x^{2}} \cdot \frac{x^{2} - 2 x - 15}{3 x^{2} + 12 x + 9}$

Step 1

Factor $2 x$ out of $10 x + 2 x^{2}$ .

Factor $2 x$ out of $10 x$ .

$\frac{2 x (5) + 2 x^{2}}{25 - x^{2}} \cdot \frac{x^{2} - 2 x - 15}{3 x^{2} + 12 x + 9}$

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

$\frac{2 x (5) + 2 x (x)}{25 - x^{2}} \cdot \frac{x^{2} - 2 x - 15}{3 x^{2} + 12 x + 9}$

Factor $2 x$ out of $2 x (5) + 2 x (x)$ .

$\frac{2 x (5 + x)}{25 - x^{2}} \cdot \frac{x^{2} - 2 x - 15}{3 x^{2} + 12 x + 9}$

Step 2

Simplify the denominator.

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

$\frac{2 x (5 + x)}{5^{2} - x^{2}} \cdot \frac{x^{2} - 2 x - 15}{3 x^{2} + 12 x + 9}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$\frac{2 x (5 + x)}{(5 + x) (5 - x)} \cdot \frac{x^{2} - 2 x - 15}{3 x^{2} + 12 x + 9}$

Step 3

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$\frac{2 x (5 + x)}{(5 + x) (5 - x)} \cdot \frac{(x - 5) (x + 3)}{3 x^{2} + 12 x + 9}$

Step 4

Simplify the denominator.

Factor $3$ out of $3 x^{2} + 12 x + 9$ .

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

$\frac{2 x (5 + x)}{(5 + x) (5 - x)} \cdot \frac{(x - 5) (x + 3)}{3 (x^{2}) + 12 x + 9}$

Factor $3$ out of $12 x$ .

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

Factor $3$ out of $9$ .

$\frac{2 x (5 + x)}{(5 + x) (5 - x)} \cdot \frac{(x - 5) (x + 3)}{3 x^{2} + 3 (4 x) + 3 \cdot 3}$

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

$\frac{2 x (5 + x)}{(5 + x) (5 - x)} \cdot \frac{(x - 5) (x + 3)}{3 (x^{2} + 4 x) + 3 \cdot 3}$

Factor $3$ out of $3 (x^{2} + 4 x) + 3 \cdot 3$ .

$\frac{2 x (5 + x)}{(5 + x) (5 - x)} \cdot \frac{(x - 5) (x + 3)}{3 (x^{2} + 4 x + 3)}$

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

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

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

Step 5

Simplify terms.

Cancel the common factor of $5 + x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $x - 5$ and $5 - x$ .

Factor $- 1$ out of $x$ .

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

Rewrite $- 5$ as $- 1 (5)$ .

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

Factor $- 1$ out of $- 1 (- x) - 1 (5)$ .

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

Reorder terms.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the expression.

Multiply $- 1$ by $2$ .

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

Move the negative in front of the fraction.

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

Simplify (10x+2x^2)/(25-x^2)*(x^2-2x-15)/(3x^2+12x+9)

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