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

- anonymous

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- mathmate

How many times did the expression changed sign?

- anonymous

3

- anonymous

i think

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- mathmate

There is then a MAXIMUM of 3 positive roots, but it could also be 1.

- anonymous

why 1?

- mathmate

Now flip the sign of the coefficients of the odd powers.
How many times does the new expression change sign?

- anonymous

Is that 5X^3+X^2+X+5

- 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

ok

- anonymous

so the negative right converted right?

- 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

So for this one there is none for the negative and 1 for the positive

- anonymous

so how do you know there is one and not 3 for the positive?

- 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

ok so the answer for positive is 1 and 3
and negative is 0

- 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

correct.

- anonymous

one other question:

- 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

how does this work then?

- anonymous

the positive looks like there is none and same with the negative?

- 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

wait why is 3X^2 flipped?

- mathmate

I supposed you have a typo!
It's 3x^3, ... or not?

- anonymous

let me see...

- anonymous

oh srry your right

- 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

ok so the positive is none
then the negative is -3X^3+3X^2-X+3, so 2 and 1

- mathmate

No positive is correct.
Can you repeat the conclusion for negative?

- anonymous

i don't understand

- anonymous

of so negative is 3 and 1

- 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

oh ok

- mathmate

Cool!

- anonymous

cool?

- mathmate

Perhaps more exactly, 1 OR 3.

- anonymous

ok thanks that makes sense

- mathmate

You're welcome! :)

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