# How do you solve 3x^3 -2x^2 -12x + 8 = 0?

Jul 6, 2017

With a large polynomial like this, you'll need to factor in order to isolate the x and solve for it.

#### Explanation:

This is the equation you're given: $3 {x}^{3} - 2 {x}^{2} - 12 x + 8 = 0$

Separate the polynomial into two parts, and factor that way.
$\left(3 {x}^{3} - 2 {x}^{2}\right) + \left(- 12 x + 8\right) = 0$
${x}^{2} \left(3 x - 2\right) - 4 \left(3 x - 2\right)$
$\left(3 x - 2\right) \left({x}^{2} - 4\right) = 0$

You can factor out $\left({x}^{2} - 4\right)$ to $\left(x + 2\right) \left(x - 2\right)$.

The fully factored polynomial is $\left(3 x - 2\right) \left(x + 2\right) \left(x - 2\right)$.

But remember... the question asks you to solve the equation.
Set the factored polynomial to zero and solve for x (you will end up with three values of x):
$\left(3 x - 2\right) \left(x + 2\right) \left(x - 2\right) = 0$

$3 x - 2 = 0$ ---> $x = \frac{2}{3}$
$x + 2 = 0$ ---> $x = - 2$
$x - 2 = 0$ ---> $x = 2$

Your final answers are $x = \frac{2}{3} , - 2 , 2$.

Jul 6, 2017

$x = - 2$

$x = 2$

$x = \frac{2}{3}$

#### Explanation:

$3 {x}^{3} - 2 {x}^{2} - 12 x + 8 = 0$

Common factors

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

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

Three solutions (B.S$=$both sides)

(1) $\text{ "x+2=0" }$ ($- 2$ B.S)

$x = - 2$

(2) $\text{ "x-2=0" }$ ($+ 2$ B.S)

$x = 2$

(3) $\text{ "3x-2=0" }$ ($+ 2$ B.S)

$3 x = 2 \text{ }$ ($\div 3$ B.S)

$x = \frac{2}{3}$

So, the problem has 3 solutions, $x = - 2$, $x = 2$, and $x = \frac{2}{3}$.

-Fish