# How do you solve (2- 3x ) ( x^{3} + 4x ) \leq 0?

How do you solve (2- 3x ) ( x^{3} + 4x ) \leq 0?

Answer 1

$x < - 1.587401 \mathmr{and} x \ge \frac{2}{3}$

#### Explanation:

$\left(2 - 3 x\right) \left({x}^{3} + 4\right) \le 0$
Use the distributive property.
$\left(2\right) \left({x}^{3}\right) + \left(2\right) \left(4\right) + \left(- 3 x\right) \left({x}^{3}\right) + \left(- 3 x\right) \left(4\right) \le 0$
$2 {x}^{3} + 8 - 3 {x}^{4} - 12 x \le 0$
$- 3 {x}^{4} + 2 {x}^{3} - 12 x + 8 \le 0$
Let's find the critical points of the inequality.
$- 3 {x}^{4} + 2 {x}^{3} - 12 x + 8 = 0$
Factorize the left side
$\left(- 3 x + 2\right) \left({x}^{3} + 4\right) = 0$
Set factors equal to 0
$- 3 x + 2 = 0 \mathmr{and} {x}^{3} + 4 = 0$
$x = \frac{2}{3} \mathmr{and} x = - 1.587401$

#### Earn Coin

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions