Algebra Examples

16 x^{2} y

$16 x^{2} y$ ,

32 x

$32 x$

Step 1

Since

16 x^{2} y, 32 x

$16 x^{2} y, 32 x$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part

16, 32

$16, 32$ then find LCM for the variable part

x^{2}, y^{1}, x^{1}

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

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

The prime factors for

16

$16$ are

2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2$ .

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Step 3.1

16

$16$ has factors of

2

$2$ and

8

$8$ .

2 \cdot 8

$2 \cdot 8$

Step 3.2

8

$8$ has factors of

2

$2$ and

4

$4$ .

2 \cdot 2 \cdot 4

$2 \cdot 2 \cdot 4$

Step 3.3

4

$4$ has factors of

2

$2$ and

2

$2$ .

2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2$

2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2$

Step 4

The prime factors for

32

$32$ are

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ .

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Step 4.1

32

$32$ has factors of

2

$2$ and

16

$16$ .

2 \cdot 16

$2 \cdot 16$

Step 4.2

16

$16$ has factors of

2

$2$ and

8

$8$ .

2 \cdot 2 \cdot 8

$2 \cdot 2 \cdot 8$

Step 4.3

8

$8$ has factors of

2

$2$ and

4

$4$ .

2 \cdot 2 \cdot 2 \cdot 4

$2 \cdot 2 \cdot 2 \cdot 4$

Step 4.4

4

$4$ has factors of

2

$2$ and

2

$2$ .

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

Step 5

Multiply

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ .

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Step 5.1

Multiply

2

$2$ by

2

$2$ .

4 \cdot 2 \cdot 2 \cdot 2

$4 \cdot 2 \cdot 2 \cdot 2$

Step 5.2

Multiply

4

$4$ by

2

$2$ .

8 \cdot 2 \cdot 2

$8 \cdot 2 \cdot 2$

Step 5.3

Multiply

8

$8$ by

2

$2$ .

16 \cdot 2

$16 \cdot 2$

Step 5.4

Multiply

16

$16$ by

2

$2$ .

32

$32$

32

$32$

Step 6

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 7

The factor for

y^{1}

$y^{1}$ is

y

$y$ itself.

y^{1} = y

$y^{1} = y$

y

$y$ occurs

1

$1$ time.

Step 8

The factor for

x^{1}

$x^{1}$ is

x

$x$ itself.

x^{1} = x

$x^{1} = x$

x

$x$ occurs

1

$1$ time.

Step 9

The LCM of

x^{2}, y^{1}, x^{1}

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

x \cdot x \cdot y

$x \cdot x \cdot y$

Step 10

Multiply

x

$x$ by

x

$x$ .

x^{2} y

$x^{2} y$

Step 11

The LCM for

16 x^{2} y, 32 x

$16 x^{2} y, 32 x$ is the numeric part

32

$32$ multiplied by the variable part.

32 x^{2} y

$32 x^{2} y$