Answer
See the explanation below.
Work Step by Step
Interpret the equation as $f(x)=0$.
Create a table of function values for some chosen $x\in[-4,4]$
$\left[\begin{array}{c|c}
x & f(x)\\
\hline -4 & -3\\
-2 & 23\\
2 & -21\\
4 & 5
\end{array}\right]$
For each of the intervals $[-4,-2]$ , $[-2,2]$ , $[2,4]$ ,
$f$(left border) has a different sign to $f$(right border).
Apply the IVT on each interval.
The given function (a polynomial) is continuous everywhere.
In each interval, there exists a $c$ such that $f(c)=0$,
because $0$ is a value between $f$(left border) and $f$(right border).
In other words, each interval contains a solution of $f(x)=0$.
Since the intervals have no common points (except the borders, which are not solutions anyway), there are three solutions to the equation in the interval $[-4,4]$.
