Question 88624: Solve x^3 - 4x^2 - 2x + 20 = 0 over the set of complex numbers.
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! x^3 - 4x^2 - 2x + 20 = 0
Descartes Rule of Signs:
Pos: 2 . 0
Neg: 1 . 1
Img: 0 . 2
We know that there is a negative root for sure...
Possible roots:
a = 1,2,4,5,10,20
b = 1
+- a/b
+-1, +-2, +-4, +-5, +-10, +-20 are the "real" choices
Synthetic Division (or any division you know at this point):
..-2..|........1.......-4.......-2.......20
....................1.......-6......10........0
(x + 2)(x^2 - 6x + 10) = 0
x^2 - 6x + 10 = 0 and x + 2 = 0
x^2 - 6x = -10
(x - 3)^2 = -1
x = 3 +- i and x = -2
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