Pre-Algebra Examples

$12 x$ , $39$ , $4 x^{2}$

Step 1

Since $12 x, 39, 4 x^{2}$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $12, 39, 4$ then find LCM for the variable part $x^{1}, x^{2}$ .

Step 2

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 3

The prime factors for $12$ are $2 \cdot 2 \cdot 3$ .

Tap for more steps...

Step 3.1

$12$ has factors of $2$ and $6$ .

$2 \cdot 6$

Step 3.2

$6$ has factors of $2$ and $3$ .

$2 \cdot 2 \cdot 3$

Step 4

$39$ has factors of $3$ and $13$ .

$3 \cdot 13$

Step 5

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Step 6

The LCM of $12, 39, 4$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 2 \cdot 3 \cdot 13$

Step 7

Multiply $2 \cdot 2 \cdot 3 \cdot 13$ .

Tap for more steps...

Step 7.1

Multiply $2$ by $2$ .

$4 \cdot 3 \cdot 13$

Step 7.2

Multiply $4$ by $3$ .

$12 \cdot 13$

Step 7.3

Multiply $12$ by $13$ .

$156$

Step 8

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

$x^{1} = x$

$x$ occurs $1$ time.

Step 9

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

Step 10

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

$x \cdot x$

Step 11

Multiply $x$ by $x$ .

$x^{2}$

Step 12

The LCM for $12 x, 39, 4 x^{2}$ is the numeric part $156$ multiplied by the variable part.

$156 x^{2}$