Find the Roots (Zeros) (-5x-1)(2x+8)

$(- 5 x - 1) (2 x + 8)$

Set $(- 5 x - 1) (2 x + 8)$ equal to $0$ .

$(- 5 x - 1) (2 x + 8) = 0$

Solve for $x$ .

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

$- 5 x - 1 = 0$

$2 x + 8 = 0$

Set $- 5 x - 1$ equal to $0$ and solve for $x$ .

Set $- 5 x - 1$ equal to $0$ .

$- 5 x - 1 = 0$

Solve $- 5 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$- 5 x = 1$

Divide each term in $- 5 x = 1$ by $- 5$ and simplify.

Divide each term in $- 5 x = 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{1}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{1}{- 5}$

Divide $x$ by $1$ .

$x = \frac{1}{- 5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

Set $2 x + 8$ equal to $0$ and solve for $x$ .

Set $2 x + 8$ equal to $0$ .

$2 x + 8 = 0$

Solve $2 x + 8 = 0$ for $x$ .

Subtract $8$ from both sides of the equation.

$2 x = - 8$

Divide each term in $2 x = - 8$ by $2$ and simplify.

Divide each term in $2 x = - 8$ by $2$ .

$\frac{2 x}{2} = \frac{- 8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 8}{2}$

Divide $x$ by $1$ .

$x = \frac{- 8}{2}$

Simplify the right side.

Divide $- 8$ by $2$ .

$x = - 4$

The final solution is all the values that make $(- 5 x - 1) (2 x + 8) = 0$ true.

$x = - \frac{1}{5}, - 4$