Find the LCD (4x+12)/(x^2+5x+6) and (5x+15)/(10x+20)

$\frac{4 x + 12}{x^{2} + 5 x + 6}$ and $\frac{5 x + 15}{10 x + 20}$

Step 1

Simplify each polynomial.

Factor $4$ out of $4 x + 12$ .

Factor $4$ out of $4 x$ .

$\frac{4 (x) + 12}{x^{2} + 5 x + 6}, \frac{5 x + 15}{10 x + 20}$

Factor $4$ out of $12$ .

$\frac{4 x + 4 \cdot 3}{x^{2} + 5 x + 6}, \frac{5 x + 15}{10 x + 20}$

Factor $4$ out of $4 x + 4 \cdot 3$ .

$\frac{4 (x + 3)}{x^{2} + 5 x + 6}, \frac{5 x + 15}{10 x + 20}$

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

$\frac{4 (x + 3)}{(x + 2) (x + 3)}, \frac{5 x + 15}{10 x + 20}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$\frac{4 (x + 3)}{(x + 2) (x + 3)}, \frac{5 x + 15}{10 x + 20}$

Rewrite the expression.

$\frac{4}{x + 2}, \frac{5 x + 15}{10 x + 20}$

Cancel the common factor of $5 x + 15$ and $10 x + 20$ .

Factor $5$ out of $5 x$ .

$\frac{4}{x + 2}, \frac{5 (x) + 15}{10 x + 20}$

Factor $5$ out of $15$ .

$\frac{4}{x + 2}, \frac{5 x + 5 \cdot 3}{10 x + 20}$

Factor $5$ out of $5 x + 5 \cdot 3$ .

$\frac{4}{x + 2}, \frac{5 (x + 3)}{10 x + 20}$

Cancel the common factors.

Factor $5$ out of $10 x$ .

$\frac{4}{x + 2}, \frac{5 (x + 3)}{5 (2 x) + 20}$

Factor $5$ out of $20$ .

$\frac{4}{x + 2}, \frac{5 (x + 3)}{5 (2 x) + 5 (4)}$

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

$\frac{4}{x + 2}, \frac{5 (x + 3)}{5 (2 x + 4)}$

Cancel the common factor.

$\frac{4}{x + 2}, \frac{5 (x + 3)}{5 (2 x + 4)}$

Rewrite the expression.

$\frac{4}{x + 2}, \frac{x + 3}{2 x + 4}$

Factor $2$ out of $2 x + 4$ .

Factor $2$ out of $2 x$ .

$\frac{4}{x + 2}, \frac{x + 3}{2 (x) + 4}$

Factor $2$ out of $4$ .

$\frac{4}{x + 2}, \frac{x + 3}{2 x + 2 \cdot 2}$

Factor $2$ out of $2 x + 2 \cdot 2$ .

$\frac{4}{x + 2}, \frac{x + 3}{2 (x + 2)}$

Step 2

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

$x + 2, 2 (x + 2)$

Step 3

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 4

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

Not prime

Step 5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 6

The LCM of $1, 2$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

Step 7

The factor for $x + 2$ is $x + 2$ itself.

$(x + 2) = x + 2$

$(x + 2)$ occurs $1$ time.

Step 8

The LCM of $x + 2, x + 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x + 2$

Step 9

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

$2 (x + 2)$

Find the LCD (4x+12)/(x^2+5x+6) and (5x+15)/(10x+20)

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