I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $\phi$), $\phi^3+\phi^{-3}=4\phi-2$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which other numbers does this equation hold for?
The six possible answers are the roots of $x^6-4x^4+2x^3+1=0$. Note that I am not interested in solving for $x$ itself as much as I am interested in a method which would allow me to completely factor out this polynomial into lowest degree factors which still have real coefficients. Note that I am treating this equation as if I had no clue that the golden ratio is one of the solutions. In other words, I am trying to factor this equation as if I never saw it before, so I can't just immediately factor out $(x^2-x-1)$ without a justifiable process, even though it is indeed one of the factors.
I first observed that the equation holds for $x=1$, so I was able to divide out $(x-1)$ to get the factorization of:
$$(x-1)(x^5+x^4-3x^3-x^2-x-1)$$
I tried making an assumption that the quintic reduces to a product of $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$, multiplying out, and equalling coefficients, but I ended up with a system of two extremely convoluted equations which I had no idea how to solve. I also tried to turn the first five terms of the quintic into a palindromic polynomial and then perform the standard method of factoring palindromic polynomials, to no avail.
I am either missing something, or I don't know of a nice method that would let this expression be factored. I'm looking forward to being enlightened, thanks for any help.
Best Answer
Here's a possible way to do it:
$x^6-4x^4+2x^3+1 = (x^6+2x^3+1)-4x^4 = (x^3+1)^2 - 4x^4$
$(x^3+1)^2-4x^4 = [x^3+1-2x^2][x^3+1+2x^2]$
$x^6-4x^4+2x^3+1= [(x^3-x^2)+(1-x^2)][x^3+2x^2+1]$
Then, we have:
$x^6-4x^4+2x^3+1 =[x^3+2x^2+1][x^2(x-1)+(1-x)(1+x)]$
$x^6-4x^4+2x^3+1 = (x-1)(x^2-x-1)[x^3+2x^2+1]$
So that gives you a decently nice factored form.