Solve for x (x+2)(x-2)-3x=0

$(x + 2) (x - 2) - 3 x = 0$

Step 1

Simplify each term.

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

Apply the distributive property.

$x (x - 2) + 2 (x - 2) - 3 x = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 (x - 2) - 3 x = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2 - 3 x = 0$

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$x \cdot x - 2 x + 2 x + 2 \cdot - 2 - 3 x = 0$

Add $- 2 x$ and $2 x$ .

$x \cdot x + 0 + 2 \cdot - 2 - 3 x = 0$

Add $x \cdot x$ and $0$ .

$x \cdot x + 2 \cdot - 2 - 3 x = 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 2 \cdot - 2 - 3 x = 0$

Multiply $2$ by $- 2$ .

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

Step 2

Factor $x^{2} - 4 - 3 x$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

Step 3

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

$x - 4 = 0$

$x + 1 = 0$

Step 4

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Solve for x (x+2)(x-2)-3x=0

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