I have a polynomial (degree 6) that is form by multiplication of two smaller polynomials (degree 4 and 2)
$H(x) = 1+ax+bx^2+cx^3+dx^4)(u+vx+x^2) = \\u+ (a u+v)x + (a v+b u+1) x^2 + (a+b v+c u)x^3 +(b+c v+d u)x^4+(c+d v)x^5+ d x^6$
My goal so far has been to try and find out if an analytic expression for the coefficients of the two smaller polynomials $(a,b,c,d,u,v)$ can be found given that I know the coefficients in the normalized polynomial
$H(x)/u = 1+ (a+v/u)x + (a v/u+b+1/u) x^2 + (a/u+bv/u+c)x^3 +(b/u+cv/u+d)x^4+(c/u+dv/u)x^5+ d/u x^6$
The coefficients are given by the following expressions
$ d/u = \frac{(W+1)^3-B^3}{(W+1)^3 + B^3} \\ c/u+dv/u = \frac{6(W-1)(W+1)^2}{(W+1)^3+B^3} \\ b/u+cv/u+d = \frac{3 (5 W^3-W^2-W+5)+3B^3}{(W+1)^3+B^3} \\ a/u+bv/u+c = \frac{4 (5 W^3-3 W^2+3 W-5)}{(W+1)^3+B^3} \\ a v/u+b+1/u = \frac{3 (5 W^3-W^2-W+5)-3B^3}{(W+1)^3+B^3} \\ a+v/u = \frac{6(W-1)(W+1)^2}{(W+1)^3+B^3} $
where $W$ and $B$ are constants. Note that the expressions for the coefficients of the $x$ and $x^5$ terms are equal. Don't know if this is useful but if the symmetric polynomial, formed by subtracting a 'left-right flipped' normalized polynomial from itself, can be put on the form $-2B^3(x^2-1)^3$.
I will accept any answer that provides the slightest clue to how I can make some progress on this problem. I have tried using the Wolfram-alpha website but it complained about input being to long and I never found out how to work around that.
Any polynomial with real coefficients factors as a single constant times a product of some number of quadratic terms $x^2 + v x + u,$ times a number of linear terms $x - w.$ Furthermore, since each quadratic term is a pairing of complex roots $ (x -\lambda) ( x - \bar{\lambda})$ there is a sort of positivity built in to those, with your letters $u > 0$ and $u + v + 1 \geq 0$ may do it.
The problem I see is that there is no guarantee, having arranged each term to be monic, that you can fix the constant terms. That is, with your letters, I do not see how you can take $$ e + a x + b x^2 + c x^3 + d x^4 $$ and guarantee that $e=1.$ So, any attempt to solve with $e=1$ is going to lead to gibberish or a dead end, as long as you are using letters.