Finding the number of solutions of the equation $2x^5- 6x^3 + 2x = 4x^4 - 6x^2 +1$ in the interval $I = [-2, 2]$ I have to find the number of solutions for the following equation on the interval $I=[-2,2]$ $$2x^5- 6x^3 + 2x = 4x^4 - 6x^2 +1$$
Now I know I have to put them all on one side and then use the intermediate value theorem to show a solution exists wherever the sign changes. But how do I know I plugged in enough values to find all the solutions?
 A: Okay,  resulting polynomial is always negative on ($-\infty, -2$], has at least one root in (-2,-1). the polynomial is positive at -1, 0 and 1.  At 2, the polynomial is negative, so there is at least one root in (1,2).  The polynomial is always positive on [$3, \infty$).  So three roots are accounted for and two are not.  The nature of these roots is analyzable  in (-2,3).
Take the derivative and look for max/mins.  The derivative is positive on $x \le -2$ but at -1 it is negative, indicating a max in the interval.  (So we know the polynomial was negative had a root, reached  a maximum and began to drop.)  
At 0 the derivative is positive again indicating that there was a min in (-1,0).  This means we can analyze the two unaccounted for roots here.  If the polynomial is positive at the min, then the two roots are not real.  If the polynomial is 0 at the min, then the polynomial has a double root.  If the polynomial is negative at the min then there are real roots on either side.
At x = -1/2 the derivative is negative so the min must be in (-1/2,0).  (The polynomial is positive but that doesn't help us. )
At x= -1/4, the derivative is still negative so the min is in (-1/4,0).  Now... we can find an extreme range for the polynomial and evaluating each individual summands at -1/4 and at 0 and adding all the lessor values, and adding all the greater values, we can determine the polynomial must always be in that range.
It turns at that range is always positive.  So the polynomial is positive at the min and the polynomial only has 3 real roots, 2 of which are in (-2,2).
In summary: the polynomial is negative at -2.  It has a root, and then reaches a max and dips and is descending, but positive, at x= -1.  It continues to dip and reaches a minimum (but still positive) somewhere before 0.  We didn't bother to find the next two extrema bu that doesn't matter.  The polynomial stays positive from 0 to 1, and somewhere between 1 and 2 it has a second root and becomes negative again.  Then somewhere between 2 and 3 it has its third and last root and becomes positive again.  Then it is strictly increasing.
