Q. 31 - Precalculus Enhanced with Graphing Utilities

Question

In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$x^3+x^2-4x-4=0$

Step-by-Step Solution

Answer

The required solutions are $x_1=-1,\hspace{0.17em}{x}_2=-2,\hspace{0.17em}{x}_3=2$ and the graph of the solution is given below:

Step 1. Given information.

We have:

$x^3+x^2-4x-4=0$

Step 2. Solve the equation algebraically.

The equation is $x^3+x^2-4x-4=0$.

Take the common out from the groups:

$$\begin{gathered} x^2\left(x-1\right)-4\left(x-1\right)=0 \\ \left(x^2-4\right)(x-1)=0 \\ \left(x^2-2^2\right)\left(x-1\right)=0 \end{gathered}$$

Apply the difference of two squares formula: $x^2-y^2=\left(x+y\right)\left(x-y\right)$

$$\begin{gathered} \left(x+2\right)\left(x-2\right)(x-1)=0 \\ x=-2,1,2 \end{gathered}$$

Step 3. Verify the solution using a graphing utility.

The solution of the equation are $x=-1,\hspace{0.17em}x=-2,\hspace{0.17em}x=2$.
Draw the graph of the equation $x^3+x^2-4x-4=0$ using the graphing utility: