Question

Solve the inequality algebraically

$x^3+8x^2<0$

Short answer

Required solution set is $(-\infty ,-8)$.

Step-by-step solution

Step 1. Given information  

We are given an inequality,

$x^3+8x^2<0$

Step 2.Finding the zeros   

The zeros are given by,

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

So, the intervals will be,

$(-\infty ,-8),(-8,0)\text{and}(0,\infty )$

Step 4. Selecting a test number in each interval  

The table is as follows,

Interval	$(-\infty ,-8)$	$(-8,0)$	$(0,\infty )$
Chosen Number	$-10$	$-1$	$1$
Value of Function	localid="1646303382470" $f(-10)=-200$	$f(-1)=7$	$f\left(1\right)=9$
Check	Negative	Positive	Positive

Hence the solution set is $(-\infty ,-8)$.