can anyone factor this?!?

Question

can anyone factor this?

x^6-2x^3 +1

anyone who can EXPLAIN to me how to factor that gets 10 points =]

Answer 1

Dear cate,
Its very simple
You can use identity (a – b)/\2 = a/\2 – 2ab + b/\2

you can treat x/\3 as 'a' and 1 as 'b'.
x/\6 - 2x/\3 + 1
= (x/\3)/\2 - 2(x/\3)(1) + (1)/\2
= (x/\3 - 1)/\2

=[(x - 1)(x/\2 + x + 1)]/\2 as x/\3 - 1 = (x - 1)(x/\2 + x + 1)

Therefore the factors are
[(x - 1)/\2] [(x/\2 + x + 1)/\2]

Answer 2

When you factor that you get (x^3-1)*(x^3-1), or (x^3-1)^2. Well I'm not sure how to explain it...but it's just like factoring x^2-2x+1 (which is (x-1)*(x-1)), there's a pattern. The coefficients are 1, -2, 1 and so you always multiply two of the same terms together. don't know if this helps

Answer 3

(x^3-1)(x^3-1)=
x^6-1x^3-1x^3-(-1)= product of first two, outer, inner, last factors

x^6-2x^3-(-1)= combine like values
x^6-2x^3-(-1) inverse of negative is positive

when the third value in a trinomial is positive its factors can be both negative or both positive.

Answer 4

First, simplify using U = x^3

U^2 - 2U +1

This factors as (U -1)(U - 1)

Plug back in U = x^3

(x^3 - 1)^2