SOLUTION: How many positive zeros does f(x)= x^5-2x^4-3x^3+x^2+x-1 have?

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Question 1158937: How many positive zeros does f(x)= x^5-2x^4-3x^3+x^2+x-1 have?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=+x%5E5-2x%5E4-3x%5E3%2Bx%5E2%2Bx-1
use Descartes' Rule of Signs
f+%28x%29+=+x%5E5-2x%5E4-3x%5E3%2Bx%5E2%2Bx-1
first sign: +
second sign: -
third sign: -
fourth sign:+
fifth sign:-
Then I count the number of changes:
from first sign to second sign:+change
from second sign to third sign: no+change
from third to fourth sign:change
from fourth to fifth sign: change
There are three sign changes in the positive-root case. This number "three" is the maximum possible number of positive zeroes
so now I look at f%28-x%29. That is, having changed the sign on+x, I'm now doing the negative-root case:
recall:%28-x%29%5E5=-x%5E5,%28-x%29%5E4=x%5E4
f%28-x%29+=+-x%5E5-2x%5E4%2B3x%5E3%2Bx%5E2-x-1
from first sign to second sign:no +change
from second sign to third sign: +change
from third to fourth sign:no change
from fourth to fifth sign: change
There are two sign changes in the negative-root case
so, since we have three sign changes in the positive-root case, there will be+one positive root for sure