Question

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

Answer 1

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)

Answer 2

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

Answer 3

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

Answer 4

\[f(x)=x^3+x^2-x+1 \]
\[f(-x)=-x^3+x^2+x+1 \] 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)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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"

Answer 9

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