Descartes' Rule of Signs does not actually solve the equation
it merely tells you how many real roots you can expect
http://www.purplemath.com/modules/drofsign.htm
try x = ± 1, to quickly see that neither works
So we will have "nasty" solutions.
I ran it through
http://www.wolframalpha.com/input/?i=x%5E3%2Bx%5E2-x%2B1%3D0
and you can see yourself that there are
1 real root and two complex roots
Using Descartes' Rule of Signs, solve:
f(x)=x^3+x^2-x+1.
1 answer