SOLUTION: Find the exact number of real roots of the equation x^6-x^3+2x^2-3x-1=0 .

Question 631295: Find the exact number of real roots of the equation x^6-x^3+2x^2-3x-1=0 .
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the exact number of real roots of the equation
x^6-x^3+2x^2-3x-1=0 .
**
Use rational roots theorem to solve:

...0...|.....1.......0.......0......-1......2......-3.......-1.
...1..|......1.......1.......1........0......2......-1.......-2...
...2...|.....1.......2.......4........7.....16......29......57 (2 is upper limit, all numbers>0)
Also, real root exists between 1 and 2 (sign change of last number in 2nd and 3rd rows)
============================
...0...|.....1.......0.......0......-1......2......-3.......-1
.-1...|.....1.....-1......1.......-2......0.......-3.......3 (-1 is lower limit, numbers alternate in sign)
Also, real root exists between 0 and -1 (sign change of last number in 1st and 2nd rows)
..
exact number of real roots=2
which means the other 4 roots are 2 pairs of non-real or imaginary roots

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