FACTOR (X-1)^4 - 1 COMPLETELY. HOW TO DO IT? HELP MATH GENIUS!?

Question

answer is supposed to be x(x-2)(x^2-2x+2). How do I get there?

I expanded (x-1)^4 -1 and i got:

x^4 - 4x^3 - 6x^2 - 4x

I tried factoring by grouping but I got:

x[x^3 - 4x^2 - 6x -4)]

x[x^2 (x-4) + 2 (3x-2)]

and the round bracket don't match up so it doesn't work?

Answer 1

Think simpler.

Difference of two squares where a² - b² = (a - b)(a + b)

(x-1)^4 - 1 is the same as

((x-1)²)² - 1²

[(x-1)² - 1][(x-1)² + 1] ← Another diff of 2 squares in there

[(x-1) - 1][(x-1)+1][(x-1)² + 1]

(x-2)(x)[(x-1)² + 1]

x(x-2)[(x-1)(x-1) + 1]

x(x-2)[x²-2x+1 + 1]

x(x-2)(x²-2x+2) ← Answer

Answer 2

It's a difference of squares.  The "x-1" in parentheses can be treated just like a different variable.

(x-1)^4 - 1 =

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

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

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

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

Answer 3

(x-1)^4-1=((x-1)²-1)((x-1)²+1)

=((x-1)-1)((x-1)+1)((x-1)²+1)

=(x-2)(x)((x-1)²+1)

=(x-2)x(x²-2x+2)

Let's search if there are roots to x²-2x+2

The equation is : x²-2x+2=0

The reduced discriminant is : (-1)²-1.2=-1<0

There are no real solutions.

x²-2x+2 can't be factorized for x as a real value.

(x-1)^4-1=x(x-2)(x²-2x+2)