Question

Solve each inequality. Graph the solution set.

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

Short answer

The real zeroes are $-1,2,-2$and the solution set can be graphed as

Step-by-step solution

Step 1. Given Information

The given inequality is$x^3+x^2<4x+4$

Step 2. Calculation

Solve the given inequality as follows,

$$\begin{gathered} x^3+x^2-4x-4<4x+4-4x-4 \\ {x}^3+x^2-4x-4<0 \\ {x}^3+x^2-4x-4=0 \\ {x}^2(x+1)-4(x+1)=0 \\ (x+1)(x^2-4)=0 \\ x+1=0\Rightarrow x=-1 \\ {x}^2-4=0\Rightarrow x=\pm 2 \end{gathered}$$

Thus, the real zeroes of function are$-1,2and-2$

Step 3. Graph

Construct the table by using a number from each interval to check whether the function is positive or negative.

Thus, the solution set is $(-\infty ,-2)\cup (-1,2)$

And the solution set can be graphed as