x^32-1=(x^16+1)(x^8+1)(x^4+1)(x^2+1)(X+1)(?) Find ? .. How do I do this?

Answer 1

The x^32 can only come from multiplying the first tems of all those binomials
The -1 can only come from multiplying the constants

notice that if we multiply the first terms , the x's, in the right side we get
(x^16)(x^8)(x^4)(x^2)(x) = x^31
so the missing bracket must be (x + a)

In the expansion in the right side, if we multiply only the constants in all those binomials, we should get -1 (in the x^32 - 1 answer)

and (1)(1)(1)(1)(1)(a) = -1
a = -1

so the missing factor is (x-1)

Answer 2

or, note that

x^32-1
= (x^16+1)(x^16-1)
= (x^16+1)(x^8+1)(x^8-1)
...
and if you keep expanding the differences of squares, you wind up with (x-1) at the end.