Prove a function has exactly 3 real roots?

Please delete, wrong thread

there is Descartes' rule of signs
which states that the number of sign changes from - to + and + to - in f(x) denotes the maximum number of positive roots
and the number of changes of sign in f(-x) is the the maximum number of negative roots

lets proceed
f(x) has one change of sign from 8x^2 to 93x and then again from 93x to the next term
more specifically
from + to -
and then - to +
so max 2 +ve roots

for f(-x) the function
is -8x^5+77x^4-2x^3+8x^2+93x+42=0
i.e. maximum 3 -ve roots
now see the pic attached and if you have nay problem understanding any of it, just let me know
if pic is not clear, go to this link
https://dl.dropbox.com/u/35279697/rule%20of%20signs.png