SOLUTION: Please help with:
For f(x)=-2x^5+3x^4-6x^2+1, what does Descartes rule of signs tell you about:
(a) the number of positive real zeros?
(b) the number of negative real ze
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Question 166613: Please help with:
For f(x)=-2x^5+3x^4-6x^2+1, what does Descartes rule of signs tell you about:
(a) the number of positive real zeros?
(b) the number of negative real zeros?
Thank you.
Found 2 solutions by nerdybill, gonzo:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Refer to this site for a good explanation:
http://www.purplemath.com/modules/drofsign.htm
.
f(x)=-2x^5+3x^4-6x^2+1
There are three sign changes.
So, 3 or 1 positive roots
.
f(-x)=-2(-x)^5+3(-x)^4-6(-x)^2+1
f(-x)=2(x)^5+3(x)^4-6(x)^2+1
There are two sign changes.
So, 2 or 0 negative roots
.
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
if you have the time, check this out.
it's worth the visit.
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http://www.purplemath.com/modules/drofsign.htm
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the answer to your question is as follows:
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your equation is:
f(x)=-2x^5+3x^4-6x^2+1
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what you need to do is check the sign changes for f(x).
that gives you the maximum number of positive roots.
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then you need to check the sign changes for f(-x).
that gives you the maximum number of negative roots.
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since all roots are not necessarily real, you can have less than the maximum number of roots.
you will, however, go down in pairs.
example:
maximum number of positive roots is 6.
the actual number of positive roots can be either 6,4,2, or 0.
same for maximum number of negative roots is 6.
the actual number of negative roots can be either 6,4,2, or 0.
if the number of positive roots (or negative roots) is 5 (for example), then the actual number of positive roots (or negative roots) is 5,3, or 1.
they go down by pairs, and only by pairs.
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back to your equation.
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your equation is:
f(x)= - 2(x)^5 + 3(x)^4 - 6(x)^2 + 1
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your signs are in order from left to right: - + - +
you have 3 sign changes.
(- +) (+ -) (- +)
3 is the maximum number of positive roots.
the actual number of positive roots can be 3 or 1.
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f(-x) = -2(-x)^5 + 3(-x)^4 -6(-x)^2 + 1
here you have to watch the order of the exponent.
(-x)^5 give you a negative answer.
(-x)^4 give you a positive answer.
if the exponent is odd, then you keep the sign of (-x)
if the exponent is even, then you make the minus sign a plus.
in your equation here is what happens:
-2(-x)^5 becomes a + since (-x)^5 remains a minus, and a minus times a minus makes a plus.
+3(-x)^4 becomes a + since (-x)^4 becomes a plus, and a plus times a plus makes a plus.
-6(-x)^2 becomes a - since (-x)^2 becomes a plus, and a minus times a plus makes a minus.
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the resulting signs in order from left to right are: + + - +
you have 2 maximum sign changes for negative roots.
the actual number of sign changes would be 2 or 0.
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examining what you have:
you have 3 or 1 positive roots.
you have 2 or 0 negative roots.
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the maximum number of total roots (positive or negative) is 3 + 2 = 5.
this tells you that you are probably right in your calculations since the number of maximum possible roots is equal to the degree of the equation (5).
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descartes rule of signs helps to narrow down the list of possible solutions.
if your list of possible roots includes positive and negative numbers, and descartes rule of signs says there are no negative roots, that can narrow your search down to only the possible positive roots.
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