Algebra Examples

$x^{2} - 64$ , $8 x + 64$

Step 1

Factor $x^{2} - 64$ .

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Step 1.1

Rewrite $64$ as $8^{2}$ .

$x^{2} - 8^{2}$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 8$ .

$(x + 8) (x - 8)$

Step 2

Factor $8$ out of $8 x + 64$ .

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Step 2.1

Factor $8$ out of $8 x$ .

$8 (x) + 64$

Step 2.2

Factor $8$ out of $64$ .

$8 x + 8 \cdot 8$

Step 2.3

Factor $8$ out of $8 x + 8 \cdot 8$ .

$8 (x + 8)$

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

The prime factors for $8$ are $2 \cdot 2 \cdot 2$ .

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Step 5.1

$8$ has factors of $2$ and $4$ .

$2 \cdot 4$

Step 5.2

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

$2 \cdot 2 \cdot 2$

Step 6

Multiply $2 \cdot 2 \cdot 2$ .

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Step 6.1

Multiply $2$ by $2$ .

$4 \cdot 2$

Step 6.2

Multiply $4$ by $2$ .

$8$

Step 7

The factor for $x + 8$ is $x + 8$ itself.

$(x + 8) = x + 8$

$(x + 8)$ occurs $1$ time.

Step 8

The factor for $x - 8$ is $x - 8$ itself.

$(x - 8) = x - 8$

$(x - 8)$ occurs $1$ time.

Step 9

The factor for $x + 8$ is $x + 8$ itself.

$(x + 8) = x + 8$

$(x + 8)$ occurs $1$ time.

Step 10

The LCM of $x + 8, x - 8, x + 8$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 8) (x - 8)$

Step 11

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

$8 (x + 8) (x - 8)$