Solve for x (x+2)(x+5)=-2

$(x + 2) (x + 5) = - 2$

Step 1

Simplify $(x + 2) (x + 5)$ .

Expand $(x + 2) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (x + 5) + 2 (x + 5) = - 2$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 2 (x + 5) = - 2$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 2 x + 2 \cdot 5 = - 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 5 + 2 x + 2 \cdot 5 = - 2$

Move $5$ to the left of $x$ .

$x^{2} + 5 \cdot x + 2 x + 2 \cdot 5 = - 2$

Multiply $2$ by $5$ .

$x^{2} + 5 x + 2 x + 10 = - 2$

Add $5 x$ and $2 x$ .

$x^{2} + 7 x + 10 = - 2$

Step 2

Add $2$ to both sides of the equation.

$x^{2} + 7 x + 10 + 2 = 0$

Step 3

Add $10$ and $2$ .

$x^{2} + 7 x + 12 = 0$

Step 4

Factor $x^{2} + 7 x + 12$ 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 $12$ and whose sum is $7$ .

$3, 4$

Write the factored form using these integers.

$(x + 3) (x + 4) = 0$

Step 5

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

$x + 3 = 0$

$x + 4 = 0$

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

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

The final solution is all the values that make $(x + 3) (x + 4) = 0$ true.

$x = - 3, - 4$

Solve for x (x+2)(x+5)=-2

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