Factorise fully (x^(12) - 1), please.?

Question

All I have done is (x^(6)+1)(x^(3)+1)(x^(3)-1)

Why does (x^(3) - 1) factorise as that? How did you know?

Answer 1

Its FULL FACTORIZATION over real numbers:
x^12-1 =
(x-1) (x+1)
(x^2+1) (x^2+x+1) (x^2-x+1)
(x^2-sqrt(3)*x+1) (x^2+sqrt(3)*x+1)

All of the other answers were correct
but THEY MISSED the factorization of
x^4-x^2+1
= x^4-x^2+3x^2+1 -3x^2
= x^4+2x^2+1 - 3x^2
= (x^2+1)^2 - (sqrt(3)*x)^2
= (x^2-sqrt(3)*x+1) (x^2+sqrt(3)*x+1)

More details:
x^12-1 = (x^6-1)(x^6+1)

x^6-1 = (x^3-1) (x^3+1)
x^3-1 = (x-1) (x^2+x+1)
x^3+1 = (x+1) (x^2-x+1)

x^6+1 = (x^2+1) (x^4-x^2+1)
x^4-x^2+1 = (x^2-sqrt(3)*x+1) (x^2+sqrt(3)*x+1)

For understanding the following basic identities
x^3-1 = (x-1) (x^2+x+1)
x^3+1 = (x+1) (x^2-x+1)

You can simply multiply the Right Hand Sides.
or
you can divide x^3-1 over x-1
or x^3+1 over x+1 and watch the remainder.

Answer 2

x^12 - 1

turn the expression in perfect squares
(x^6)^2 - 1^2

a^2 - b^2 = (a + b) (a - b)

use the formula above ou have:
(x^6 - 1) (x^6 + 1)

turn the expression to perfect cubes
[ (x^2)^3 - 1^3 ] [ (x^2)^3 + 1^3 ]

a^3 - b^3 = (a - b) (a^2 + ab + b^2)
and
a^3 + b^3 = (a + b) (a^2 - ab + b^2)

use the formulas above, you have:
(x^2 - 1) (x^4 + x^2 + 1) (x^2 + 1) (x^4 - x^2 + 1)

factor (x^2 - 1)
(x - 1) (x + 1) (x^4 + x^2 + 1) (x^2 + 1) (x^2 - x^2 + 1)

factor (x^4 + x^2 + 1)

(x^2)^2 + x^2 + 1^2

(x^2 + x + 1) (x^2 - x + 1)


the answer is:

(x- 1) (x + 1) (x^2 + 1) (x^2 + x + 1) (x^2 - x + 1) (x^4 - x^2 + 1)

Answer 3

x^12 - 1
(x^6 + 1)(x^6 - 1)
(x^2 + 1)(x^4 - x^2 + 1)(x^3 - 1)(x^3 + 1)
(x^2 + 1)(x^4 - x^2 + 1)(x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)

Answer 4

hi,
well, you have already a part of the answer, but let's do all again.

(x^12 - 1)
turning the expression in perfect squares:

(x^6 + 1)(x^6 - 1)
now turning to perfect cubes, but working by parts:

1.- (x^6 + 1) = ((x^2)^3 + 1), so as you can see this expression allow us to work with perfect cubes.
so, (x^2 + 1)(x^4 - x^2 + 1), now we can't do anymore with this.

2.- (x^6 - 1) = (x^3 -1)(x^3+1) now we turn those expression in perfect cubes:
(x^3-1) = (x-1)(x^2 + x + 1), and (x^3+1) = (x+1)(x^2 - x + 1)

therefore, the whole expression is:

(x^2 + 1)(x^4 - x^2 + 1)(x-1)(x^2 + x + 1)(x+1)(x^2 - x + 1)

I hope it helped

Answer 5

So far so good. (x^3+1) = (x+1)*(x^2-x+1)
(x^3-1) = (x-1)*(x^2 +x +1)
Now for that nasty (x^6+1) term. Let y = x^2. Then

(x^6+1) = (y^3 +1) = (y+1)*(y^2 - y +1) = (x^2+1)*(x^4 - x^2+1)

So: X^12+1 = (x^2+1)*(x^4 - x^2+1)*(x+1)*(x^2-x+1)*(x^2 +x +1)

Answer 6

a + 2b - 1 - 2ab Notice 2b is a factor of two of the terms : 2b - 2ab = 2b(1 - a) We already have a - 1 in the original expression, so to make the above factorisation the same, just multiply 2b by -1 and (1 - a) by -1. Then you get : 2b(1 - a) = -2b(a - 1). Now the whole expression is : a - 1 - 2b(a - 1) and we have a - 1 in two terms, so take it out as a factor. = (a - 1)(1 - 2b) which is the final answer.

Answer 7

Good stuff.
These guys are basically just using properties of binomials
difference of two squares: a^2-b^2 = (a + b) (a - b)
difference of two cubes: a^3 – b^3 = (a – b)(a2 + ab + b2)
Sum of two cubes: a3 + b3 = (a + b)(a2 – ab + b2)

and then manipulating them etc to get them down to a point where they can factor them using foil if needed or the quadratic equation.

Remember that 1 is 1^2, 1^3, 1^10000 if you need it.

Answer 8

well, thats an identity you must know...

if n is odd then
a^n - b^n = [a-b][a^(n-1) + a^(n-2) b + a^(n-3) b^2 +.....+ a b^(n-2) + b^(n-1)]

basically, in the "bigger" bracket, the power of a decreases and that of b increases but their sum remains the same for each term

in fact there is another:

if n is odd then
a^n + b^n = [a+b][a^(n-1) - a^(n-2) b + a^(n-3) b^2)-.....+/- a b^(n-2) -/+ b^(n-1)]
here, the signs alternate for each term in the "bigger" bracket.

you can prove these two easily by simply expanding the RHS and checking that terms cancel each other leaving only the two terms on the LHS.

Answer 9

(x^6 -1)......................................... (x^6+1)
(x^3-1) (x^3+1)............................. ((x^2)^3 +1)
(x-1)(x^2 +x +1)(x+1)(x^2-x+1)....(x^2+1)(x^4 -x^2+1)

Answer 10

That's almost it. x^3-1 factors as:
(x - 1)(x^2 + x + 1)

... which leaves you with:
(x^6+1)(x^3+1)(x - 1)(x^2 + x + 1)

x^6+1 factors as:
(x^2 + 1)(x^4 - x^2 + 1)

... which results in:
(x^2 + 1)(x^4 - x^2 + 1)(x^3+1)(x - 1)(x^2 + x + 1)

and finally x^3+1 factors as:
(x + 1)(x^2 - x + 1)

... which yields:
(x^2 + 1)(x^4 - x^2 + 1)(x + 1)(x^2 - x + 1)(x - 1)(x^2 + x + 1)