How do you factor and solve 3x^2+ 2x = 8?

Apr 29, 2017

See a solution process below:

Explanation:

First, subtract $\textcolor{red}{8}$ from each side of the equation to put in standard quadratic form:

$3 {x}^{2} + 2 x - \textcolor{red}{8} = 8 - \textcolor{red}{8}$

$3 {x}^{2} + 2 x - 8 = 0$

Next, playing with factors of 3 (1 x 3 and 3 x 1) and factors of 8 (1 x 8, 2 x 4, 4 x 2 and 8 x 1) we can factor this as:

$\left(3 x - 4\right) \left(x + 2\right) = 0$

Now, solve each term for $0$:

Solution 1)

$3 x - 4 = 0$

$3 x - 4 + \textcolor{red}{4} = 0 + \textcolor{red}{4}$

$3 x - 0 = 4$

$3 x = 4$

$\frac{3 x}{\textcolor{red}{3}} = \frac{4}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}} = \frac{4}{3}$

$x = \frac{4}{3}$

Solution 2)

$x + 2 = 0$

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

$x + 0 = - 2$

$x = - 2$

The solution is: $x = \frac{4}{3}$ and $x = - 2$