Question

I need to find the roots to the function x^6-3x^4+3x^2-1 and then write it as a product of linear factors and I can't see where to begin.

Answer 1

have you tried useing x=1 ?

Answer 2

since they are all evens, x=-1 should work as well

Answer 3

yeah, x = 1 is one "obvious" root, however there are more roots

Answer 4

do all the roots need to be real? or can they be complex values too?

Answer 5

no they should be only real

Answer 6

well (x-1)(x+1) is a start .... and it might be that you can use multiples of them as well

Answer 7

x^6 +0x^5 -3x^4 +0x^2 +3x^2 +0x -1
0 1 1 -2 -2 1 1
____________________________________________
1 ) x^5 + x^4 -2x^3 -2x^2 + x + 1 R0


x^5 + x^4 -2x^3 -2x^2 + x + 1
0 -1 0 2 0 -1
-------------------------------
-1 ) x^4+ 0x^3 -2x^2 +0x + 1 R0


x^4 -2x^2 +1 ; this gives us a perfect square "quadratic" to reslove

Answer 8

(x^2-1) (x^2-1); x=+- 1 again

Answer 9

(x+1)(x+1)(x+1)(x-1)(x-1)(x-1)
or
(x+1)^3 (x-1)^3

do you agree?

Answer 10

sort of, kind of lost at your post where you got out a perfect square quadric

Answer 11

i used synthetic division instead of longhand division to come to the last setup

Answer 12

synthD is simpler to display in text than longhand

Answer 13

yeah, now I see what you did, thanks!

Answer 14

good luck