Find the LCM x^2-1 , x^2+2x+1

$x^{2} - 1$ , $x^{2} + 2 x + 1$

Step 1

Factor $x^{2} - 1$ .

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

$x^{2} - 1^{2}$

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 $b = 1$ .

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

Step 2

Factor using the perfect square rule.

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

$x^{2} + 2 x + 1^{2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2}$

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

$1$

Step 6

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

$(x + 1) = x + 1$

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

Step 7

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

$(x - 1) = x - 1$

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

Step 8

The factors for $x + 1$ are $(x + 1) \cdot (x + 1)$ , which is $x + 1$ multiplied by itself $2$ times.

$(x + 1) = (x + 1) \cdot (x + 1)$

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

Step 9

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

${(x + 1)}^{2} (x - 1)$