anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
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mathmate
  • mathmate
How many times did the expression changed sign?
anonymous
  • anonymous
3
anonymous
  • anonymous
i think

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mathmate
  • mathmate
There is then a MAXIMUM of 3 positive roots, but it could also be 1.
anonymous
  • anonymous
why 1?
mathmate
  • mathmate
Now flip the sign of the coefficients of the odd powers. How many times does the new expression change sign?
anonymous
  • anonymous
Is that 5X^3+X^2+X+5
mathmate
  • mathmate
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.
anonymous
  • anonymous
ok
anonymous
  • anonymous
so the negative right converted right?
mathmate
  • mathmate
Yes, the number of sign changes of the new expression (i.e. replace x by -x) gives the MAXIMUM number of negative roots.
anonymous
  • anonymous
So for this one there is none for the negative and 1 for the positive
anonymous
  • anonymous
so how do you know there is one and not 3 for the positive?
mathmate
  • mathmate
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.
anonymous
  • anonymous
ok so the answer for positive is 1 and 3 and negative is 0
mathmate
  • mathmate
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
  • mathmate
correct.
anonymous
  • anonymous
one other question:
anonymous
  • anonymous
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
anonymous
  • anonymous
how does this work then?
anonymous
  • anonymous
the positive looks like there is none and same with the negative?
mathmate
  • mathmate
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?
anonymous
  • anonymous
wait why is 3X^2 flipped?
mathmate
  • mathmate
I supposed you have a typo! It's 3x^3, ... or not?
anonymous
  • anonymous
let me see...
anonymous
  • anonymous
oh srry your right
anonymous
  • anonymous
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
anonymous
  • anonymous
ok so the positive is none then the negative is -3X^3+3X^2-X+3, so 2 and 1
mathmate
  • mathmate
No positive is correct. Can you repeat the conclusion for negative?
anonymous
  • anonymous
i don't understand
anonymous
  • anonymous
of so negative is 3 and 1
mathmate
  • mathmate
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.
anonymous
  • anonymous
oh ok
mathmate
  • mathmate
Cool!
anonymous
  • anonymous
cool?
mathmate
  • mathmate
Perhaps more exactly, 1 OR 3.
anonymous
  • anonymous
ok thanks that makes sense
mathmate
  • mathmate
You're welcome! :)

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