Here is a solution without calculus or Descartes's Rule of Signs. However, some knowledge about continuity of polynomial functions is required.
Let $a,b,c,d,e$ be the roots of this polynomial. Using Vieta's Formulas, we have $$a+b+c+d+e=0$$ and $$ab+ac+ad+ae+bc+bd+be+cd+ce+de=0\,.$$ This means
Since $0$ is not a root of this polynomial, we conclude that not all roots are real (otherwise, it must hold that $a^2+b^2+c^2+d^2+e^2>0$). Thus, the polynomial has either one or three real roots. Since the polynomial has at least one root in each of the three intervals $(-2,-1)$, $(0,1)$, and $(1,2)$, we conclude that there are exactly three real roots.