Bob, I did eventually find a factorization for the expression
x^4 + 2x^3 + 4x^2 + 8x +16
The original problem was probably something like x^5-32=0. I think the poster divided (x-2) into it to get the expression above. This of course is the 5th roots of 32, x=2 being the only real one. Realizing that this was probably the original question, I looked up the 5th roots of unity and got the facorization as:
(x^2 + (1+sqrt(5))x + 4)(x^2 + (1-sqrt(5))x + 4)
I would not expect a high school algebra course to cover the roots of unity, so I doubt the poster should be able to factor this. Just thought I'd mention this.
It seems like you have already found the factorization of the expression x^4 + 2x^3 + 4x^2 + 8x + 16. Based on your explanation, it appears that the expression originally came from solving the equation x^5 - 32 = 0, which implies finding the fifth roots of 32.
To obtain the factorization, you made use of knowledge about the 5th roots of unity. The 5th roots of unity are complex numbers that satisfy the equation z^5 = 1. In this case, you found that x = 2 is the only real root of x^5 - 32 = 0, but there are four complex roots.
A general approach to find the factorization of an expression like x^4 + 2x^3 + 4x^2 + 8x + 16 is to use techniques such as factoring by grouping, synthetic division, or long division to factor out any linear or quadratic terms that might be present. However, in this specific case, the factorization involves complex numbers and the roots of unity.
The factorization you obtained, (x^2 + (1+sqrt(5))x + 4)(x^2 + (1-sqrt(5))x + 4), results from the application of techniques related to the complex roots of unity. It is less likely to be covered in a high school algebra course, as you mentioned.
In summary, your explanation shows a more advanced approach to factorizing the expression based on the roots of unity, which may not be typically taught in high school algebra.