x^32-1=(x^16+1)(x^8+1)(x^4+1)(x^2+1)(X+1)(?) Find ? .. How do I do this?
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)
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.
To find the missing term in the expression \(x^{32} - 1 = (x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(?)\), let's consider the pattern of the powers of \(x\) in the given expression.
The expression can be factored into several terms, where each term has a power of \(x\). Notice that each term has a power equal to a power of 2. Let's expand the expression:
\[x^{32} - 1 = (x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(?)\]
First, we can factor out the difference of squares using the formula \(a^2 - b^2 = (a + b)(a - b)\):
\[(x^{16} + 1)(x^{16} - 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(?)\]
Now, we can see that \((x^{16} - 1)\) can be factored further as \((x^{8} + 1)(x^{8} - 1)\). Continuing this process, we get:
\[(x^{16} + 1)(x^{8} + 1)(x^{8} - 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(?)\]
Further factoring \((x^{8} - 1)\) gives us \((x^{4} + 1)(x^{4} - 1)\):
\[(x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{4} - 1)(x^{2} + 1)(x + 1)(?)\]
We can then factor \((x^{4} - 1)\) as \((x^{2} + 1)(x^{2} - 1)\):
\[(x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x^{2} - 1)(x + 1)(?)\]
Further factoring \((x^{2} - 1)\) gives us \((x + 1)(x - 1)\):
\[(x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(x - 1)(?)\]
Now, we see that \(x^{32} - 1\) can be factored completely as \((x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(x - 1)\). Thus, the missing term is \((x - 1)\).
Therefore, the final factorization of \(x^{32} - 1\) is:
\[(x^{16} + 1)(x^{8} + 1)(x^{4} + 1)(x^{2} + 1)(x + 1)(x - 1)\]