How many real roots does $x^4 - 4x^3 + 4x^2 - 10$ have? 
How many real roots does this polynomial have?
  $$x^4 - 4x^3 + 4x^2 - 10$$

Because non-real roots come in pairs, it must have 4, 2 or 0 real roots.
Following Descartes' rules of signs, it either has one negative (real) number and one or three positive numbers.
How can I tell if it has 2 or 4 real roots?
Thank you very much in advance.

As $x\to -\infty$ $f(x)\to +\infty>0$.  And $f(-1)= -1$ so $f(x)$ "crosses" the $x$-axis at some $x < -1$.  That's our one and only negative root.
Now $f(x) = x^2(x-2)^2 - 10$.  If $x > 0$ then $x^2$ is strictly increasing and positive and $(x-2)^2$ is strictly increasing and non-negative so $f(x)$ is strictly increasing.  As $x\to \infty$ we have $f(x)\to +\infty$ and $f(2)=-10 < 0$.  For some $x > 2$, $f(x)$ "crosses" the axis, but as $f(x)$ is strictly increasing at that point it will only "cross" once and that will be our very last root.
So all that remains is to figure out if $f(x)$ has any roots between $0$ and $2$.
but for $0 \le 1$ we have $0 \le x^2 \le 1$ and $-3 \le x-2 \le -1$ and so $1\le (x-2)^2 \le 9$ and so $f(x) < 9-10 =-1 < 0$.
And if $1\le 2$ we have $1\le x^2 \le 4$ and $-1\le x-2 \le 0$ so $0 \le (x-2)^2 \le 1$ and so $f(x) < 4-10 =-6$.
So for $0\le x \le 2$ we have $f(x) < 0$ so there are no roots there.
So there is one negative root where $x<-1$ and one positive root where $x> 2$.
