Finite Math Examples

- 5 x^{2}

$- 5 x^{2}$ ,

4 x

$4 x$

Step 1

Since

- 5 x^{2}, 4 x

$- 5 x^{2}, 4 x$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part

- 5, 4

$- 5, 4$ then find LCM for the variable part

x^{2}, x^{1}

$x^{2}, x^{1}$ .

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

Since the LCM is the smallest positive number,

LCM (- 5, 4) = LCM (5, 4)

$LCM (- 5, 4) = LCM (5, 4)$

Step 4

Since

5

$5$ has no factors besides

1

$1$ and

5

$5$ .

5

$5$ is a prime number

Step 5

4

$4$ has factors of

2

$2$ and

2

$2$ .

2 \cdot 2

$2 \cdot 2$

Step 6

The LCM of

5, 4

$5, 4$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

2 \cdot 2 \cdot 5

$2 \cdot 2 \cdot 5$

Step 7

Multiply

2 \cdot 2 \cdot 5

$2 \cdot 2 \cdot 5$ .

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Step 7.1

Multiply

2

$2$ by

2

$2$ .

4 \cdot 5

$4 \cdot 5$

Step 7.2

Multiply

4

$4$ by

5

$5$ .

20

$20$

20

$20$

Step 8

The factors for

x^{2}

$x^{2}$ are

x \cdot x

$x \cdot x$ , which is

x

$x$ multiplied by each other

2

$2$ times.

x^{2} = x \cdot x

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

x

$x$ occurs

2

$2$ times.

Step 9

The factor for

x^{1}

$x^{1}$ is

x

$x$ itself.

x^{1} = x

$x^{1} = x$

x

$x$ occurs

1

$1$ time.

Step 10

The LCM of

x^{2}, x^{1}

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

x \cdot x

$x \cdot x$

Step 11

Multiply

x

$x$ by

x

$x$ .

x^{2}

$x^{2}$

Step 12

The LCM for

- 5 x^{2}, 4 x

$- 5 x^{2}, 4 x$ is the numeric part

20

$20$ multiplied by the variable part.

20 x^{2}

$20 x^{2}$