Question

Find The number of complex zeros that F(x) has in F(x)=x^7+x^9+x^12-2x^2

Answer 1

apparently it has 8, but how you are supposed to do this i have no idea

Answer 2

what method were you supposed to use?

Answer 3

Any

Answer 4

http://www.wolframalpha.com/input/?i=x^7%2Bx^9%2Bx^12-2x^2

Answer 5

How did you find it will have 8??
@satellite73

Answer 6

it will have 8 complex roots

Answer 7

you have a twelfth degree polynomial for heaven sake
we can factor out \(x^2\) but you still have a tenth degree polynomial
how can you do it? i have no idea

Answer 8

but i will be quiet and learn if @k.rajabhishek has a way to know it without graphing

Answer 9

since the highest degree is 12
thus it will have 12 zeros

Answer 10

And out of 12 zeros how many are complex??? @k.rajabhishek

Answer 11

got that
and 0 is a zero with multiplicity 2

Answer 12

by using descarte's rule we will have 2 negative and 1 positive zeros also 0 is one of the zero of equation

Answer 13

thus a total of 4 real zeros and then subtracting 4 by 12
we have 8 complex roots

Answer 14

wow that is impressive

Answer 15

except i think that there is only one negative zero
0 is a zero with multiplicity 2
descartes rule of sign is going to tell us there is one positive root right?

Answer 16

there is some mistake with this method

Answer 17

ok descartes rule of sign will give it, but we have to change it a bit

Answer 18

first we need to write in standard form
then we need to get rid of the \(x^2\) and then descartes rule of sign will tell use there is one positive root and one negative root, leaving 8 complex roots

Answer 19

@Theanitrix you got this? we can walk through it if you like

Answer 20

\[F(x)=x^7+x^9+x^{12}-2x^2\]

\[F(x)=x^2(x^{10}+x^7+x^5-2)\]
now we know it has a zero at \(x=0\) with multiplicity 2 that leaves 10 more to find

Answer 21

start with\(f(x)=x^{10}+x^7+x^5-2\) there is one change in sign, from \(1x^5\) to \(-2\) telling us that there is one positive root

Answer 22

then \(f(-x)=x^{10}-x^7-x^5-2\) and this has one change in sign from \(1x^{10}\) to \(-x^7\) telling us there is one negative root

Answer 23

that leaves 8 complex roots
@k.rajabhishek had method but a mistake in the execution

Answer 24

@satellite73 I am not getting it..
Will you clear my doubt??

How to find the positive and negative roots according to the sign change as you did..??

Answer 25

fantastic discussion on complex roots of polynomials!

Answer 26

there is one change in sign of the coefficients
that means there is one positive root,
if there were two then there could be two or none, you count down by twos
but you can't count down by twos if there is only one
similarly there is one change in sign of the coefficients of \(f(-x\) so again there must be one negative root
ten in total, leaves 8 complex roots
descartes rule of sign explained well here
http://www.purplemath.com/modules/drofsign.htm