Question

Solve the inequality algebraically

$2x^3>-8x^2$

Short answer

Required solution set is $(-4,0)\cup (0,\infty )$.

Step-by-step solution

Step 1. Given information  

We are given an inequality,

$2x^3>-8x^2$

Step 2.Finding the zeros   

The zeros is given by,

$$\begin{gathered} 2x^3=-8x^2 \\ 2x^3+8x^2=0 \\ 2x^2(x+4)=0 \\ 2x^2=0 \\ x=0 \\ x+4=0 \\ x=-4 \end{gathered}$$

So, the intervals will be,

$(-\infty ,-4),(-4,0),\text{and}(0,\infty )$

Step 4. Selecting a test number in each interval  

The table is as follows,

Interval	$(-\infty ,-4)$	$(-4,0)$	$(0,\infty )$
Chosen Number	$-5$	$-1$	$1$
Value of Function	$f(-5)=-50$	$f(-1)=6$	$f\left(1\right)=10$
Check	Negative	Positive	Positive

Hence the solution set is localid="1646304433911" $(-4,0)\cup (0,\infty )$.