HelpMe94
Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function:
f(x)= -5x^3+X^2-X+5
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mathmate
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How many times did the expression changed sign?
HelpMe94
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3
HelpMe94
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i think
mathmate
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There is then a MAXIMUM of 3 positive roots, but it could also be 1.
HelpMe94
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why 1?
mathmate
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Now flip the sign of the coefficients of the odd powers.
How many times does the new expression change sign?
HelpMe94
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Is that 5X^3+X^2+X+5
mathmate
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The number of sign changes is the maximum number of positive roots, including complex.
If there are two complex roots (as is the case), then there is only one (real) positive root.
HelpMe94
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ok
HelpMe94
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so the negative right converted right?
mathmate
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Yes, the number of sign changes of the new expression (i.e. replace x by -x) gives the MAXIMUM number of negative roots.
HelpMe94
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So for this one there is none for the negative and 1 for the positive
HelpMe94
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so how do you know there is one and not 3 for the positive?
mathmate
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Using the Descartes rule of signs, we do know that there is no negative root.
But we do not know if there is ONE or THREE positive roots.
HelpMe94
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ok so the answer for positive is 1 and 3
and negative is 0
mathmate
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You can find that out by factoring the expression.
A cubic has at least one real root, and we know that it is positive in this case.
mathmate
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correct.
HelpMe94
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one other question:
HelpMe94
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Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function:
f(x)=3X^2+2X^2+x+3
HelpMe94
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how does this work then?
HelpMe94
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the positive looks like there is none and same with the negative?
mathmate
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Now, let's see.
(1) The expression as is has no change in sign, so what do you conclude?
(2) flip the sign of the odd powers to give -3X^3+2X^2-x+3, sign changed 3 times, what do you conclude?
HelpMe94
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wait why is 3X^2 flipped?
mathmate
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I supposed you have a typo!
It's 3x^3, ... or not?
HelpMe94
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let me see...
HelpMe94
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oh srry your right
HelpMe94
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Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function:
f(x)=3X^3+2X^2+x+3
HelpMe94
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ok so the positive is none
then the negative is -3X^3+3X^2-X+3, so 2 and 1
mathmate
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No positive is correct.
Can you repeat the conclusion for negative?
HelpMe94
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i don't understand
HelpMe94
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of so negative is 3 and 1
mathmate
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The number of possible roots always differ by an even number because complex roots, if any, come in pairs.
1 and 2 will not be correct.
1 and 3 would be.
HelpMe94
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oh ok
mathmate
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Cool!
HelpMe94
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cool?
mathmate
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Perhaps more exactly, 1 OR 3.
HelpMe94
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ok thanks that makes sense
mathmate
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You're welcome! :)