# How do you solve 4(3n - 2) (4n + 1) = 0?

$n = \frac{2}{3} \mathmr{and} - \frac{1}{4}$

#### Explanation:

You can interpret the equation this way:
The product is 0 when you multiply the three factors 4, (3n-2), and (4n+1). Therefore, either of the three factors must be zero to make the statement true.

you can see it this way
$\left(4\right) \setminus \cdot \left(0\right) \setminus \cdot \left(4 n + 1\right) = 0$
or
$\left(4\right) \setminus \cdot \left(3 n - 2\right) \setminus \cdot \left(0\right) = 0$
or
$\left(4\right) \setminus \cdot \left(0\right) \setminus \cdot \left(0\right) = 0$

Which means to say, that the factor $\left(3 n - 2\right)$ and/or $\left(4 n + 1\right)$ should be zero.

So,
$3 n - 2 = 0$
$\implies 3 n = 2$
$\implies n = \frac{2}{3}$

Also,
$4 n + 1 = 0$
$\implies 4 n = - 1$
$\implies n = - \frac{1}{4}$

Therefore, $n = \frac{2}{3} \mathmr{and} - \frac{1}{4}$

Another method is to expand the whole expression and perform long division. you should arrive with the same answer.