Find the LCD (5x+2)/(4x^2-1) , 3/x , (9x)/(2x+1)

$\frac{5 x + 2}{4 x^{2} - 1}$ , $\frac{3}{x}$ , $\frac{9 x}{2 x + 1}$

Step 1

Simplify the denominator.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{5 x + 2}{{(2 x)}^{2} - 1}, \frac{3}{x}, \frac{9 x}{2 x + 1}$

Rewrite $1$ as $1^{2}$ .

$\frac{5 x + 2}{{(2 x)}^{2} - 1^{2}}, \frac{3}{x}, \frac{9 x}{2 x + 1}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$\frac{5 x + 2}{(2 x + 1) (2 x - 1)}, \frac{3}{x}, \frac{9 x}{2 x + 1}$

Step 2

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

$(2 x + 1) (2 x - 1), x, 2 x + 1$

Step 3

Since $(2 x + 1) (2 x - 1), x, 2 x + 1$ contains both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for $(2 x + 1) (2 x - 1), x, 2 x + 1$ are:

1. Find the LCM for the numeric part $1, 1, 1$ .

2. Find the LCM for the variable part $x^{1}$ .

3. Find the LCM for the compound variable part $2 x + 1, 2 x - 1, 2 x + 1$ .

4. Multiply each LCM together.

Step 4

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 5

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

Not prime

Step 6

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

$1$

Step 7

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 8

The LCM of $x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x$

Step 9

The factor for $2 x + 1$ is $2 x + 1$ itself.

$(2 x + 1) = 2 x + 1$

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

Step 10

The factor for $2 x - 1$ is $2 x - 1$ itself.

$(2 x - 1) = 2 x - 1$

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

Step 11

The factor for $2 x + 1$ is $2 x + 1$ itself.

$(2 x + 1) = 2 x + 1$

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

Step 12

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

$(2 x + 1) (2 x - 1)$

Step 13

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

$x (2 x + 1) (2 x - 1)$

Find the LCD (5x+2)/(4x^2-1) , 3/x , (9x)/(2x+1)

Related Questions

Support LCMGCF.com by disabling your adblocker.