Solve for x x^2-1=(x+1)(x-1)

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

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

Simplify $(x + 1) (x - 1)$ .

Rewrite.

$0 + 0 + (x + 1) (x - 1) = x^{2} - 1$

Simplify by adding zeros.

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

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 (x - 1) = x^{2} - 1$

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Add $- x$ and $x$ .

$x^{2} + 0 - 1 = x^{2} - 1$

Add $x^{2}$ and $0$ .

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

Step 3

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

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

Combine the opposite terms in $x^{2} - 1 - x^{2}$ .

Subtract $x^{2}$ from $x^{2}$ .

$0 - 1 = - 1$

Subtract $1$ from $0$ .

$- 1 = - 1$

Step 4

Since $- 1 = - 1$ , the equation will always be true for any value of $x$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x x^2-1=(x+1)(x-1)

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