equation #1

gourish · 2014-03-15 16:48:37

hi i am back with another question on equations Let p(x) = x8 − x6 + x5 + x4 + x2 − x + 1=0 then p(x) has roots:
A)atmost 4 positive real roots
B)atmost 2 negative real roots
C)maximum of 6 real roots
D)minimum of 2 complex roots
answer is all of the above as explained below but i have no idea how it is justified can anybody please explain me what is going on below
or just tell me how can the answer be found

"Let p(x) = x8 − x6 + x5 + x4 + x2 − x + 1 = 0
Here, the coefficients in p(x) have 4 changes in sign so p(x) = 0 have atmost 4 positive real roots.
Now, p(−x) = x8 − x6 − x5 + x4 + x2 − x + 1 = 0
Since the coefficients in p (−x) has 2 changes in sign, so p(−x) = 0 have atmost 2 negative real roots.
Thus, p(x) = 0 can have a maximum of (2 + 4) = 6 real roots.
The coefficients in p(x) = 0 are real, therefore, imaginary roots occurs in conjugate pair.
Since p(x) = 0 can have a maximum of 6 real roots, so the minimum number of complex roots of p(x) = 0 is 2."

bobbym · 2014-03-15 19:30:49

Hi;

That is what I am getting. To do more you would need a Cauchy bound and Sturm sequence.

Bob · 2014-03-15 20:27:30

Try this:

http://www.mathsisfun.com/algebra/polyn … signs.html

Bob

gourish · 2014-03-16 02:56:32

hi bobbym,
i did not understand what how that answer was found i did not understand why is it that "there are 4 change of signs" in first equation and what is "changes in sign" in this context mean?

gourish · 2014-03-16 02:59:00

hi bob and bobbym i used the link to mathisfun.com now i am clear about the answer thanks guys

bobbym · 2014-03-16 03:08:19

To let it sink in:

x^8 − x^6 + x^5 + x^4 + x^2 − x + 1=0

The x^8 has a + sign, we just leave it out for convention. The x^6 has a - sign. So that is your first change of sign.

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

So that polynomial can have 4, 2 or 0 positive roots.

gourish · 2014-03-16 03:14:00

hi bobbym, the p(x) has only two change of signs for x=x i.e. x^6 and x but you are saying that there are 4?? well is my count wrong?

Last edited by gourish (2014-03-16 03:15:14)

bobbym · 2014-03-16 03:15:35

There is another one between x^6 and x^5 so that is already 2. There are more.

gourish · 2014-03-16 03:17:33

sorry to bother but can you just list them out. i think i am still struggling with the change of signs deal... sorry i will work on it

bobbym · 2014-03-16 03:22:26

Hi;

x^8 − x^6 + x^5 + x^4 + x^2 − x + 1=0

x^8 and x^6

x^6 and x^5

x^2 and x

x and 1

gourish · 2014-03-16 03:25:11

oh.. silly me.. well i will look into more of those questions thanks for helping me out... you are an awesome admin bobbym... yoda is probably proud of you...

bobbym · 2014-03-16 03:34:44

He is too uptight. Too many rules. I only have that avatar because some of the kids insisted on it.

Let me know how you do with the some more examples.

Math Is Fun Forum

#1 2014-03-15 16:48:37

equation #1

#2 2014-03-15 19:30:49

Re: equation #1

#3 2014-03-15 20:27:30

Re: equation #1

#4 2014-03-16 02:56:32

Re: equation #1

#5 2014-03-16 02:59:00

Re: equation #1

#6 2014-03-16 03:08:19

Re: equation #1

#7 2014-03-16 03:14:00

Re: equation #1

#8 2014-03-16 03:15:35

Re: equation #1

#9 2014-03-16 03:17:33

Re: equation #1

#10 2014-03-16 03:22:26

Re: equation #1

#11 2014-03-16 03:25:11

Re: equation #1

#12 2014-03-16 03:34:44

Re: equation #1

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