Question

Factor the polynomial completely. (If the expression is not factorable using integers, enter PRIME.)

6x^6 − 6

my answer (2x^3-2)(3x^3+4x^3+3) what did I do wrong?

Answer 1

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

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

Answer 2

\[6x^6 - 6 = 6(-1+x) (1+x) (1-x+x^2) (1+x+x^2)\]

Answer 3

\[6x ^{6}-6 =6(x ^{6}-1)=6(x ^{2})^{3}-1=6(x-1)(x ^{2}+x+1)\]

Answer 4

sorry

Answer 5

\[6(x ^{2}-1)(x ^{4}+x ^{2}+1)\]

Answer 6

this will be right , correct

Answer 7

the last answer

Answer 8

pa

Answer 9

pard

Answer 10

pardon

Answer 11

hope so much that is understandably

Answer 12

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

Answer 13

It's the same as my answer, except my answer is more simplified

Answer 14

no moneybird us formula a3 -b3 =(a-b)(a2+ab+b2)

Answer 15

but you can using again the same 6(x6-1)=6((x3)2 -1 =6(x3-1)(x3+1) what you can continue

Answer 16

and finaly will get what i have wrote firstly

Answer 17

No there's a formula for all n numbers
(a^n - b^n = (a-b)(a^(n-1) + a^(n-2)b...b^(n-1)

Answer 18

most regular schools won't teach you that. It doesn't mean this property does not exist

Answer 19

the right answer is 6(x-1)(x+1)(x4+x2+1)

Answer 20

no. You can still simplify x^4+x^2+1

Answer 21

how ?

Answer 22

ok let x^2 = a
x^4+x^2+1 = a^2 + a + 1
factor this and plug x^2 back after

Answer 23

h

Answer 24

how ?

Answer 25

can you make it ?

Answer 26

camon please i like seeing how

Answer 27

how you can factorizing a2+a+1

Answer 28

(1-x+x^2) (1+x+x^2)
if you expand this, you will get x^4+x^2+1)

Answer 29

ok bye

Answer 30

thank you