Simplify ((x^2+12x+27)/(42x-6x^2)*(49-x^2)/(x^2+16x+63))÷((x^2+8x+15)/(100-4x^2))

$\frac{x^{2} + 12 x + 27}{42 x - 6 x^{2}} \cdot \frac{49 - x^{2}}{x^{2} + 16 x + 63} \div \frac{x^{2} + 8 x + 15}{100 - 4 x^{2}}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} + 12 x + 27}{42 x - 6 x^{2}} \cdot \frac{49 - x^{2}}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Step 2

Factor $x^{2} + 12 x + 27$ 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 $27$ and whose sum is $12$ .

$3, 9$

Write the factored form using these integers.

$\frac{(x + 3) (x + 9)}{42 x - 6 x^{2}} \cdot \frac{49 - x^{2}}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Step 3

Factor $6 x$ out of $42 x - 6 x^{2}$ .

Factor $6 x$ out of $42 x$ .

$\frac{(x + 3) (x + 9)}{6 x (7) - 6 x^{2}} \cdot \frac{49 - x^{2}}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Factor $6 x$ out of $- 6 x^{2}$ .

$\frac{(x + 3) (x + 9)}{6 x (7) + 6 x (- x)} \cdot \frac{49 - x^{2}}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Factor $6 x$ out of $6 x (7) + 6 x (- x)$ .

$\frac{(x + 3) (x + 9)}{6 x (7 - x)} \cdot \frac{49 - x^{2}}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Step 4

Simplify the numerator.

Rewrite $49$ as $7^{2}$ .

$\frac{(x + 3) (x + 9)}{6 x (7 - x)} \cdot \frac{7^{2} - x^{2}}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = x$ .

$\frac{(x + 3) (x + 9)}{6 x (7 - x)} \cdot \frac{(7 + x) (7 - x)}{x^{2} + 16 x + 63} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Step 5

Factor $x^{2} + 16 x + 63$ 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 $63$ and whose sum is $16$ .

$7, 9$

Write the factored form using these integers.

$\frac{(x + 3) (x + 9)}{6 x (7 - x)} \cdot \frac{(7 + x) (7 - x)}{(x + 7) (x + 9)} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Step 6

Simplify terms.

Cancel the common factor of $x + 9$ .

Factor $x + 9$ out of $(x + 3) (x + 9)$ .

$\frac{(x + 9) (x + 3)}{6 x (7 - x)} \cdot \frac{(7 + x) (7 - x)}{(x + 7) (x + 9)} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Factor $x + 9$ out of $(x + 7) (x + 9)$ .

$\frac{(x + 9) (x + 3)}{6 x (7 - x)} \cdot \frac{(7 + x) (7 - x)}{(x + 9) (x + 7)} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Cancel the common factor.

$\frac{(x + 9) (x + 3)}{6 x (7 - x)} \cdot \frac{(7 + x) (7 - x)}{(x + 9) (x + 7)} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Rewrite the expression.

$\frac{x + 3}{6 x (7 - x)} \cdot \frac{(7 + x) (7 - x)}{x + 7} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Cancel the common factor of $7 - x$ .

Factor $7 - x$ out of $6 x (7 - x)$ .

$\frac{x + 3}{(7 - x) (6 x)} \cdot \frac{(7 + x) (7 - x)}{x + 7} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Factor $7 - x$ out of $(7 + x) (7 - x)$ .

$\frac{x + 3}{(7 - x) (6 x)} \cdot \frac{(7 - x) (7 + x)}{x + 7} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Cancel the common factor.

$\frac{x + 3}{(7 - x) (6 x)} \cdot \frac{(7 - x) (7 + x)}{x + 7} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Rewrite the expression.

$\frac{x + 3}{6 x} \cdot \frac{7 + x}{x + 7} \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Multiply $\frac{x + 3}{6 x}$ by $\frac{7 + x}{x + 7}$ .

$\frac{(x + 3) (7 + x)}{6 x (x + 7)} \cdot \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Cancel the common factor of $7 + x$ and $x + 7$ .

Reorder terms.

$\frac{(x + 3) (x + 7)}{6 x (x + 7)} \cdot \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Cancel the common factor.

$\frac{(x + 3) (x + 7)}{6 x (x + 7)} \cdot \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Rewrite the expression.

$\frac{x + 3}{6 x} \cdot \frac{100 - 4 x^{2}}{x^{2} + 8 x + 15}$

Step 7

Simplify the numerator.

Factor $4$ out of $100 - 4 x^{2}$ .

Factor $4$ out of $100$ .

$\frac{x + 3}{6 x} \cdot \frac{4 (25) - 4 x^{2}}{x^{2} + 8 x + 15}$

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

$\frac{x + 3}{6 x} \cdot \frac{4 (25) + 4 (- x^{2})}{x^{2} + 8 x + 15}$

Factor $4$ out of $4 (25) + 4 (- x^{2})$ .

$\frac{x + 3}{6 x} \cdot \frac{4 (25 - x^{2})}{x^{2} + 8 x + 15}$

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

$\frac{x + 3}{6 x} \cdot \frac{4 (5^{2} - x^{2})}{x^{2} + 8 x + 15}$

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 $b = x$ .

$\frac{x + 3}{6 x} \cdot \frac{4 (5 + x) (5 - x)}{x^{2} + 8 x + 15}$

Step 8

Factor $x^{2} + 8 x + 15$ 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 $8$ .

$3, 5$

Write the factored form using these integers.

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

Step 9

Simplify terms.

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{6 x} \cdot \frac{4 (5 + x) (5 - x)}{x + 5}$

Cancel the common factor of $2$ .

Factor $2$ out of $6 x$ .

$\frac{1}{2 (3 x)} \cdot \frac{4 (5 + x) (5 - x)}{x + 5}$

Factor $2$ out of $4 (5 + x) (5 - x)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Reorder terms.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify ((x^2+12x+27)/(42x-6x^2)*(49-x^2)/(x^2+16x+63))÷((x^2+8x+15)/(100-4x^2))

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