How do you factor `x^7-8x^4-16x^3+128`?

We will employ the method of "factor by grouping", in which we take a common term from the first two and last two terms of the expression.

Group the expression into two sets of two:

`=(x^7-8x^4)+(-16x^3+128)`

Factor a common term from both sets:

`=x^4(x^3-8)-16(x^3-8)`

Factor an `(x^3-8)` term from the `x^4` and `-16` terms.

`=(x^3-8)(x^4-16)`

Here, we have two factoring identities. The first we'll tackle is `(x^3-8)`, which is a difference of cubes, as both of its terms are cubed: `x^3=(x)^3` and `8=(2)^3`.

Differences of cubes factor as follows:

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

Here, we have `a=x` and `b=2`, so

`x^3-8=(x+2)(x^2+x(2)+2^2)=(x+2)(x^2+2x+4)`

We can substitute this back into the expression we had earlier:

`=(x-2)(x^2+2x+4)(x^4-16)`

Now, note that `(x^4-16)` is a difference of squares. Differences of squares factor as follows:

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

In `x^4-16`, we see that `x^4=(x^2)^2` and `16=(4)^2`, so `a=x^2` and `b=4`. This gives us a factorization of

`x^4-16=(x^2+4)(x^2-4)`

This gives us the following factorization of the original expression:

`=(x-2)(x^2+2x+4)(x^2+4)(x^2-4)`

However, notice that `(x^2-4)` is also a difference of squares. It factors into

`x^2-4=(x+2)(x-2)`

Which allows us to obtain the factorization of

`=(x-2)(x^2+2x+4)(x^2+4)(x+2)(x-2)`

Note that there are two `(x-2)` terms, so this can be rewritten as

`=(x-2)^2(x+2)(x^2+4)(x^2+2x+4)`