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

May 15, 2017

See a solution process below:

#### Explanation:

First, divide each side of the equation by $\textcolor{red}{3}$ to eliminate the constant while keeping the equation balanced"

$\frac{3 \left(4 - x\right) \left(2 x + 1\right)}{\textcolor{red}{3}} = \frac{0}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \left(4 - x\right) \left(2 x + 1\right)}{\cancel{\textcolor{red}{3}}} = 0$

$\left(4 - x\right) \left(2 x + 1\right) = 0$

Solve each term on the left side of the equation for $0$:

Solution 1)

$4 - x = 0$

$- \textcolor{red}{4} + 4 - x = - \textcolor{red}{4} + 0$

$0 - x = - 4$

$- x = - 4$

$\textcolor{red}{- 1} \times - x = \textcolor{red}{- 1} \times - 4$

$x = 4$

Solution 2)

$2 x + 1 = 0$

$2 x + 1 - \textcolor{red}{1} = 0 - \textcolor{red}{1}$

$2 x + 0 = - 1$

$2 x = - 1$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{1}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = - \frac{1}{2}$

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

The solutions are: $x = 4$ and $x = - \frac{1}{2}$