Find the LCD (x^2+8x+15)/((x^2+6)(7x+1))-(x^3-64)/(7x^2-49x-28)

$\frac{x^{2} + 8 x + 15}{(x^{2} + 6) (7 x + 1)} - \frac{x^{3} - 64}{7 x^{2} - 49 x - 28}$

Step 1

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.

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{x^{3} - 64}{7 x^{2} - 49 x - 28}$

Step 2

Simplify the numerator.

Rewrite $64$ as $4^{3}$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{x^{3} - 4^{3}}{7 x^{2} - 49 x - 28}$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 4$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + x \cdot 4 + 4^{2})}{7 x^{2} - 49 x - 28}$

Simplify.

Move $4$ to the left of $x$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 \cdot x + 4^{2})}{7 x^{2} - 49 x - 28}$

Raise $4$ to the power of $2$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 x + 16)}{7 x^{2} - 49 x - 28}$

Step 3

Factor $7$ out of $7 x^{2} - 49 x - 28$ .

Factor $7$ out of $7 x^{2}$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 x + 16)}{7 (x^{2}) - 49 x - 28}$

Factor $7$ out of $- 49 x$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 x + 16)}{7 (x^{2}) + 7 (- 7 x) - 28}$

Factor $7$ out of $- 28$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 x + 16)}{7 x^{2} + 7 (- 7 x) + 7 \cdot - 4}$

Factor $7$ out of $7 x^{2} + 7 (- 7 x)$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 x + 16)}{7 (x^{2} - 7 x) + 7 \cdot - 4}$

Factor $7$ out of $7 (x^{2} - 7 x) + 7 \cdot - 4$ .

$\frac{(x + 3) (x + 5)}{(x^{2} + 6) (7 x + 1)} - \frac{(x - 4) (x^{2} + 4 x + 16)}{7 (x^{2} - 7 x - 4)}$

Step 4

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$(x^{2} + 6) (7 x + 1), 7 (x^{2} - 7 x - 4)$

Step 5

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 6

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 7

Since $7$ has no factors besides $1$ and $7$ .

$7$ is a prime number

Step 8

The LCM of $1, 7$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$7$

Step 9

The factor for $x^{2} + 6$ is $x^{2} + 6$ itself.

$(x^{2} + 6) = x^{2} + 6$

$(x^{2} + 6)$ occurs $1$ time.

Step 10

The factor for $7 x + 1$ is $7 x + 1$ itself.

$(7 x + 1) = 7 x + 1$

$(7 x + 1)$ occurs $1$ time.

Step 11

The factor for $x^{2} - 7 x - 4$ is $x^{2} - 7 x - 4$ itself.

$(x^{2} - 7 x - 4) = x^{2} - 7 x - 4$

$(x^{2} - 7 x - 4)$ occurs $1$ time.

Step 12

The LCM of $x^{2} + 6, 7 x + 1, x^{2} - 7 x - 4$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x^{2} + 6) (7 x + 1) (x^{2} - 7 x - 4)$

Step 13

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$7 (x^{2} + 6) (7 x + 1) (x^{2} - 7 x - 4)$

Find the LCD (x^2+8x+15)/((x^2+6)(7x+1))-(x^3-64)/(7x^2-49x-28)

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