Question

Factorise f(x)= x^6+x^4+x^2+1 by grouping in pairs. Hence find all the roots of f(x) = 0, using complex numbers.Please do not use pie as i havent learnt that yet :)

Answer 1

Write it as a geometric series under the substitution u:x^2, then factor it into roots of unity.

Answer 2

Lol, wtf? Bit much for this.

Answer 3

how is that a bit much, I don't see any other way of doing it

Answer 4

i get how to factorise by grouping in pairs i get (x^2+1)(x^4+1), then what do i do?

Answer 5

Factor each of those expressions

Answer 6

Set each factor equal to 0.
x^2 + 1 = 0
x^4 + 1 = 0

Answer 7

There going to be roots of unity

Answer 8

Of the form [\e^{2\pi i t}\]

Answer 9

The first one has roots $$-i,i$$

Answer 10

yeah for x^2+1=0, i got x= -i and x=i, and i havent learn that stuff with pie yett and for x^4 + 1 = 0 i get square root of i, and sqaure root of -i

Answer 11

You try the second one

Answer 12

That's all ya gotta do. Your zeros are +/- i.

Answer 13

but the answer says \[\pm(1/\sqrt{2} + 1/\sqrt{2} i ) , \pm i, \pm(1/\sqrt{2} - 1/\sqrt{2} i )\]

Answer 14

Hmm.....maybe this is something I haven't learned yet. Is just regular algebra or higher level?

Answer 15

@rebeccapecovski
I looked it up, guess it isn't something I would have thought of. I have no idea about the pi thing either you mentioned, but I do have a way to go about showing the answer that has nothing to do with pi.

Answer 16

I think I seeyour mistake, you factored wrong

Answer 17

Nah, she factored correctly, it's just a bit.....more complicated than that, lol.

Answer 18

You sure about that?
I may be wrong but just making sure.. it's 1 AM right now so

Answer 19

Well, what do you think then?

Answer 20

I can't think at the moment, my brain is on an extended vacation -_-

Answer 21

Haha. Well, I would say that you have
\[\pm i, \pm \sqrt{i}\] and then getting +/- sqrt i is it's own adventure.

Answer 22

x^6 + x^4 + x^2 + 1 = 0
x^4(x^2 + 1) + 1(x^2 + 1) = 0
(x^2 + 1)(x^4 + 1) = 0

x^2 + 1 = 0
x^4 + 1 = 0

x^2 = -1
x^4 = -1

x = +/- sqrt(-1)
x + +- qdrt(-1)