# How do you solve the polynomial inequality and state the answer in interval notation given (x^3+2x^2)/2<x+2?

Feb 16, 2017

$\left(- \infty , - 2\right) \bigcup \left(- \sqrt{2} , \sqrt{2}\right)$

#### Explanation:

Multiply both sides by $2$.

${x}^{3} + 2 {x}^{2} < 2 x + 4$

${x}^{3} + 2 {x}^{2} - 2 x - 4 < 0$

Factor by extracting a common factor twice.

${x}^{2} \left(x + 2\right) - 2 \left(x + 2\right) < 0$

$\left({x}^{2} - 2\right) \left(x + 2\right) < 0$

If you rewrite as an equation, you get zeroes of $x = \pm \sqrt{2} \mathmr{and} - 2$.

Now, select test points.

Test point 1: x = -3

((-3)^3 + 2(-3)^2)/2 <^? -3 + 2

$\frac{- 27 + 18}{2} < - 1$

This is absolutely true.

Repeat this process in the following intervals:

(-2, -sqrt(2)); (-sqrt(2), sqrt(2)); (sqrt(2), oo). You should get as a final result that

$x < - 2$ and $- \sqrt{2} < x < \sqrt{2}$ both satisfy the inequality.

Hopefully this helps!