# How do you solve (8z+4)(5z+10)=0?

Dec 21, 2016

To solve this, you can set each set of parenthesis equal to zero.

$8 z + 4 = 0$ and $5 z + 10 = 0$

Solve for $z$

$8 z + 4 = 0$

Subtract $4$ on both sides of the equation

$8 z = - 4$

Divide $8$ on both sides of the equation

$z = - \frac{1}{2}$

Solve for $z$

$5 z + 10 = 0$

Subtract $10$ on both sides of the equation

$5 z = - 10$

Divide both sides of the equation by $5$

$z = - 2$

So, the solutions are:

$z = - \frac{1}{2} \mathmr{and} z = - 2$

Dec 21, 2016

$z = - \frac{1}{2} , z = - 2$

#### Explanation:

$\left(8 z + 4\right) \left(5 z + 10\right) = 0$
Zero Product Principle: If $a b = 0$ then $a = 0$, $b = 0$ OR both equal zero.

Therefore, in terms of a factored polynomial, you would have to assume both have the chance of being equal to 0.

$\left(8 z + 4\right) = 0$ and $\left(5 z + 10\right) = 0$...

$8 z + 4 = 0$
$8 z \setminus \stackrel{\setminus \cancel{- 4}}{\setminus \cancel{+ 4}} = 0 - 4$
$8 z = - 4$
$z = - \frac{4}{8} = - \frac{1}{2}$ or $- 0.5$

$5 z + 10 = 0$
$5 z \setminus \stackrel{\setminus \cancel{- 10}}{\setminus \cancel{10}} = 0 - 10$
$5 z = - 10$
$z = - \frac{10}{5} = - 2$

$z = - \frac{1}{2} , z = - 2$