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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
Found 3 solutions by Edwin McCravy, stanbon, venugopalramana:
Answer by Edwin McCravy(20086)   (Show Source):
You can put this solution on YOUR website! Need help....please.
Use Descartes Rule to determine how many positive and negative zeros.
Do not need to find the zeros.
f(x)= -6x5 + x4 + 5x3 + x + 1
Rule:
Write the terms in descending order, then list the signs of the terms:
f(x)= -6x5 + x4 + 5x3 + x + 1
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- + + + +
In going from the 1st sign - to the 2nd sign +, that is 1 sign
change.
In going from the 2nd sign + to the 3rd sign +, there is no
sign change. So there is still only one sign change.
In going from the 3rd sign + to the 4th sign +, there is no
sign change. So there is still only one sign change.
In going from the 4th sign + to the 5th sign +, there is no
sign change. So there is still only one sign change.
So there is one sign change, so there is just one positive
zero.
*If there had been more than 1 sign change, the number of
positive zeros might have been reduced by a multiple of 2,
but we can't reduce 1 by a multiple of 2, so there must be
exactly one positive zero.
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Next to find the number of negative zeros we must find f(-x).
The positive zeros of f(-x) will be the zeros of f(x) with
their sign changed, so we just need to do the same thing with
f(-x) as we did above.
f(-x)= -6(-x)5 + (-x)4 + 5(-x)3 + (-x) + 1
f(-x) = +6x5 + x4 - 5x3 - x + 1
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+ + - - +
In going from the 1st sign + to the 2nd sign +, there is no
sign change.
In going from the 2nd sign + to the 3rd sign +, there is one
sign change.
In going from the 3rd sign - to the 4th sign -, there is no
sign change. So there is still only one sign change.
In going from the 4th sign - to the 5th sign +, there is one
sign change. So we end up with two sign changes.
Since there are two sign change, there will either two of
0 negative zeros. However the rule can't tell us whether it
is 2 negative zeros or 0 negative zeros.
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Note that if the signs of some polynomial were:
- + - + - + - -
There would be 6 sign changes, so there would be
either 6 or 4 or 2 or 0 positive zeros.
If the signs of some polynomial were
+ - + - + - + -
There would be 7 sign changes, so there would be
either 7 or 5 or 3 or 1 positive zeros.
Note that if there are an odd number of sign changes,
there must be at least one positive zero. However if
there are an even number of sign changes there may
not be any positive zeros at all.
Edwin
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! f(x)= -6x^5+x^4+5x^3+x+1
# of changes in signs = 1
Therefore one positive real zero.
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f(-x)=6x^+x^4-5x^3-x+1
# of changes in signs = 2
Therefore 2 or 0 negative real zeroes.
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Cheers,
Stan H.
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! 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.
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