Solve by Factoring (x+2)(x+2)=1

$(x + 2) (x + 2) = 1$

Step 1

Subtract $1$ from both sides of the equation.

$(x + 2) (x + 2) - 1 = 0$

Step 2

Simplify $(x + 2) (x + 2) - 1$ .

Simplify each term.

Expand $(x + 2) (x + 2)$ using the FOIL Method.

Apply the distributive property.

$x (x + 2) + 2 (x + 2) - 1 = 0$

Apply the distributive property.

$x \cdot x + x \cdot 2 + 2 (x + 2) - 1 = 0$

Apply the distributive property.

$x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 - 1 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $2$ to the left of $x$ .

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

Multiply $2$ by $2$ .

$x^{2} + 2 x + 2 x + 4 - 1 = 0$

Add $2 x$ and $2 x$ .

$x^{2} + 4 x + 4 - 1 = 0$

Subtract $1$ from $4$ .

$x^{2} + 4 x + 3 = 0$

Step 3

Factor $x^{2} + 4 x + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$(x + 1) (x + 3) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$x + 3 = 0$

Step 5

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

Solve by Factoring (x+2)(x+2)=1

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