Question

What are the possible number of negative zeros of f(x)=-2x^7-2x^6+7x^5-7x^4+4x^3+4x^2?

Answer 1

Have you studied Descartes' Rule of Signs?

Answer 2

Yes I know that there are 2 sign changes when it's negative?

Answer 3

butt how do I know if 0 is also a negative 0?

Answer 4

@Directrix is that right?!

Answer 5

I'm thinking.

Answer 6

I know to find the negative zero's, you multiply -1 by the numbers with odd exponents.

Answer 7

I would write this: f(x)=-2x^7-2x^6+7x^5-7x^4+4x^3+4x^2

as f(x)=-2x^7-2x^6+7x^5-7x^4+4x^3+4x^2 + 0x + 0 and then replace x by -x to look for sign changes. I do not think +) or -0 counts as a sign change because zero has no sign.

Answer 8

I thought it was f(x)=-2x^7-2x^6+7x^5-7x^4+4x^3+4x^2 Multiply -1 by numbers w/ odd exponents.
f(x)=2x^7-2x^6-7x^5-7x^4-4x^3+4x^2?
Then there are 2?

Answer 9

I have 2 sign changes or less that by 2.
So, I'm thinking that the polynomial has 2 negative real roots or zero real roots.

Answer 10

The options are:
A 7, 5, 3, 0r 1
B 3, 0r 1
C 2 or 0
D 2

Answer 11

I'm following the theorem. Your technique is, I think, equivalent to the theorem. But, you did not reduce the number of negative roots by multiples of 2 possibilities.

Answer 12

so it's c?

Answer 13

Cold turkey, I would go with C 2 or 0.
But because you did not mention studying the multiple of 2 portion of the theorem, I am going to read the theorem from a reference source.

Answer 14

Thanks! That's what I thought it was :)

Answer 15

^ @Directrix you're correct. It's always either the maximum number, or an even number less than that.

Answer 16

I see what @abbie1 means by that zero root. On whose side does it count?

Answer 17

I'm sticking with 2 or zero negative real roots.

A root of 0 is neither positive or negative.

Final Answer: C

Answer 18

Could you help me on the question I just asked? It's at the top! I am so confused!! Thanks for the help :)

Answer 19

Looks like zero would count as neither - the actual function has 2 neg and 1 pos - and it has three sign changes for positive zeroes, so either 3 or 1 positive zeroes. Appears the zero doesn't count as either.

Answer 20

"The number of negative roots of P(x)=0 is either equal to the number of variations in sign of P(-x) or is less than this number by a positive even integer."

Answer 21

On most of the questions about negative real roots, zero is always included.

Answer 22

That I have had recently.

Answer 23

@abbie1 Which question is this --> Could you help me on the question I just asked? It's at the top!

Answer 24

Find the polynomial f(x) that has the roots of -2, 3 of multiplicity 2. Explain how you would verify the zeros of f(x)? It's my open question. On the left side

Answer 25

>On most of the questions about negative real roots, zero is always included.
Why would it not be included as a postive root then? Zero has no sign. Zero separates the positive real numbers from the negative real numbers.

Answer 26

http://mathforum.org/library/drmath/view/69519.html

"When the constant term is zero, at least one of the roots is ZERO,
since you can factor out an x. Descartes' Rule of Signs mentions only
POSITIVE and NEGATIVE roots! So it tells us that your equation has 0
positive roots and 0 negative roots, but that does not mean there are
no REAL roots; 0 is a real number, and is a root. "

Answer 27

So zero does not count as either positive or negative, as the evidence above showed...

the actual function has 2 neg and 1 pos roots - and it has three sign changes for positive zeroes and 2 sign changes for neg zeroes, so either 3 or 1 positive zeros and 2 or 0 negative zeroes.

Answer 28

We should have factored out x^2 at the outset. Then, we could run Descartes' rule.

In real life, there are 7 total roots: Pair of Complex Conjugates, 0 is a double root, two negative real roots, and 1 real root.