Question 63890
Need help....please. 
Use Descartes Rule to determine how many positive and negative zeros.
Do not need to find the zeros. 
f(x)= -6x^5+x^4+5x^3+x+1

I.POSITIVE ROOTS

TAKE F(X)
ARRANGE IT IN DESCENDING ORDER...IT IS ALREADY SO.
NOTE HOW MANY CHANGES ARE THERE IN SIGNS OF COSECUTIVE TERMS
THERE IS ONLY 1 CHANGE FROM -6(COEFFICIENT OF X^5) TO 
+5(COEFFICIENT OF X^4).
IF THE NUMBER OF CHANGES IS N THEN THE MAXIMUM NUMBER
 OF POSITIVE ZEROS ARE N.THE ACTUAL NUMBER COULD BE 
= N OR N-2 OR N-4 ETC>=0
HERE N=1...HENCE THERE IS ONE POSITIVE ZERO.

II.NEGATIVE ROOTS

TAKE F(-X)....= 6X^5+X^4-5X^3-X+1
ARRANGE IT IN DESCENDING ORDER...IT IS ALREADY SO.
NOTE HOW MANY CHANGES ARE THERE IN SIGNS OF COSECUTIVE TERMS
THERE ARE 2 CHANGES FROM 1(COEFFICIENT OF X^4) TO 
-5(COEFFICIENT OF X^3)AND FROM -1(COEFFICIENT OF X)TO 
+1 (CONSTANT TERM).
IF THE NUMBER OF CHANGES IS N THEN THE MAXIMUM NUMBER OF 
NEGATIVE ZEROS ARE N.THE ACTUAL NUMBER COULD BE
= N OR N-2 OR N-4 ETC>=0
HERE N=2...HENCE THERE  COULD BE 2 OR NIL NUMBER OF 
NEGATIVE ZEROS.