Question

It's been a while so can somebody show me how to figure this out?

How do I find how many zeroes a function has over a set of complex numbers? (I'll post the specific function I need to work on in the comments)

Answer 1

Okay well the equation thing isnt working for me so this is going to look really sloppy:

f(x) = x^5 - 2x^4 + 8x^2 - 13x + 6

Answer 2

I also need help on listing the possible rational zeroes, and factoring the polynomial completely over the set of complex numbers... but first things first.

Answer 3

I worked this problem with someone else a few days ago: http://openstudy.com/study#/updates/52d71f53e4b0e63d49ac287c

Unfortunately it will be a bit harder to read with OpenStudy not formatting the equations :-(

Answer 4

Thank you! It's loading the equations for me just fine, it's just not letting me type them like I usually do.

Answer 5

At the end, I need help completeing the square for x^2 - 2x + 3... I've never really been good at completeing squares

Answer 6

Okay, to complete the square, we need to get the x^2 and x terms on one side and any solely numerical term on the other.

x^2-2x = -3

Answer 7

okay i understand that much c:

Answer 8

Now we take half the value of the coefficient of the x term, square it, and add it to both sides.

Answer 9

so, 1?

Answer 10

Remember that (x+a)(x+a) = x^2 + ax + ax + a^2 = x^2 + 2ax + a^2

So we are just taking that middle term and making it equal to 2ax

Dividing by 2 and squaring it gives us a^2

Answer 11

So would it be x^2 - 2x + 1 and the factors be (x - 1)(x - 1)?

Answer 12

and if we added 1 to the other side then that would be -2

Answer 13

As you said, we get 1 after dividing by 2 and squaring. Now we add that to each side of the equation:

x^2 -2x +1 = -3 + 1

We can rewrite the left side as a square:
(x-1)(x-1) = 2
Take the square root of both sides and we get
(X-1) = sqrt(2)
But we need to use both the positive and negative square roots to get all the solutions

Answer 14

Sorry, I'm being pestered for attention by a demanding little kid which s slowing down my typing :-)

Answer 15

x-1 = sqrt(2)
x = 1+sqrt(2) (Solution 1)
x-1 = -sqrt(2)
x = 1-sqrt(2) (Solution 2)

Answer 16

That's fine, lol take your time. Oh so then two of the complex solutions are 1+sqrt(2) and 1-sqrt(2)?

Answer 17

but wait that 2 was negative, right? we can't find the square root of negatives?

Answer 18

Whoops, we dropped the minus sign on the 2! Those should be square root of -2, or we could keep it as I wrote it but multiply each radical by i

Answer 19

1+i*sqrt(2) and 1-i*sqrt(2)

Answer 20

ohhh okay I see

Answer 21

I wish they would fix the %}%%## equation formatting

Answer 22

Anyhow, do you understand completing the square now?

Answer 23

yeah me too /: it's alright though, i still understand everything you said (: thank you very much! and yes, it makes a lot more sense to me now

Answer 24

So the roots of this equation are (x - 1)(x - 1)(x + 2)(1 + i*sqrt(2))(1 - i*sqrt(2))?

Answer 25

Is that how I would write it out factored?

Answer 26

Close: (x-1)(x-1)(x+2)(x-1-i*sqrt(2))(x-1-i*sqrt(2)) assuming I didn't make any mistakes typing

Answer 27

Uh, I did, the second one should have the other sign for the i*sqrt(2) part

Answer 28

okay, should the last two be (x - 1 - i*sqrt(2)) or should one of them be (x - 1 + i*sqrt(2))?

Answer 29

Oh okay (:

Answer 30

Thank you very much!

Answer 31

Yeah, that second one you wrote is correct