The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is   
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.

If $x \ge 0$ the $x^5 + 2x^3 + x^2 + 2 \ge 2$ so there are no solutions for $x \ge 0$.
If $x = -humongouseffinglargenumber$ then $x^5 + 2x^3 + x^2+2 < 0$ so somewhere between $-humongouseffinglargenumber$ and $0$ there must be an $x$ where the term is $0$.  Let's try to narrow that range down a little.
As we are only looking at negative values of $x$ we want to find values where $|x|^5 + 2|x|^3 = x^2 + 2$ and one such obvious place is when $|x| = 1$ i.e. when $x = -1$ thus $x^5 + 2x^3 + x^2 + 2 = -1 + -2 + 1 + 2 = 0$.
So that's one root.  We can factor out $(x - (-1)) = x + 1$ to get $x^5 + 2x^3 + x^2 + 2 = (x + 1)(x^4 - x^3 + 3x^2 -2x + 2)$.  So if there are any more real roots they are real roots to $x^4 - x^3 + 3x^2 - 2x + 2 = 0$.
As we are only considering negative $x$ these are the same as solving for $x^4 + |x|^3 + 3x^2 + 2|x| + 2 = 0$.  As $|x| \ge 0$, $x^4 + |x|^3 + 3x^2 + 2|x| + 2 \ge 2$ so there are no solutions.
So $x = -1$ is the only solution.
