## puppy33 2 years ago help i been try to do this for 2 days i dont understand how do you use Descartes' rule and the Fundamental Theorem of Algebra to predict the number of complex roots of f(x)=x3+x2-x+1 . please explain the process

1. satellite73

fundamental theorem of algebra tells you that there are exactly 3 real zeros of this function, and since it is a cube, either one or all three are complex (a cubic must have on real zero)

2. satellite73

descartes rule of sign: there are two changes in sign for of the coefficients \[1,1,-1,1\] from 1 to -1 and from -1 to 1, so there are either two positive real zeros or no positive real zeros

3. puppy33

what would be another good example do show better because im trying to get a good understanding

4. satellite73

\[f(x)=x^3+x^2-x+1 \] \[f(-x)=-x^3+x^2+x+1 \] and here there is only one change in sign, so there there must be one negative real zero (because you count down by twos, so there must be one)

5. satellite73

so there could be a) one negative, two positive or b) one negative, two complex

6. satellite73

here is a really good explanation with worked out examples http://www.purplemath.com/modules/drofsign.htm

7. satellite73

of course in the modern world we simply check what is right, so this is more of a theoretical (historical) exercise http://www.wolframalpha.com/input/?i=x3%2Bx2-x%2B1 as you can see there is one negative zero and two complex zeros

8. satellite73

descartes rule of sign does not give you the exact answer. in this example it only tells you "two positive, one negative no complex" or "one negative, two complex"

9. puppy33

ok i get that part i just dont understand what i have to do with the fundamental theorem of algebra