Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$ (Heuristic Solution)

Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$.

My solution:

Let $f(x)=x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=(x^2+ax+b)(x^2+px+q)(x^2+mx+n)$, we notice then that

1.

where each of the discriminant for each quadratic factor is less than zero (since we're told $f(x)$ has all 6 complex roots) and

2.

$a,\,b,\,p,\,q,\,m,\,n \in N$ since the the coefficient on the leading term is $1$.

When $x=-1$, we get:

$\color{black}845=\color{yellow}\bbox[5px,purple]{(b-a+1)}\color{black}\bbox[5px,orange]{(q-p+1)}\color{orange}\bbox[5px,blue]{(n-m+1)}[/MATH]

$5\cdot 13 \cdot 13 =(b-a+1)(q-p+1)(n-m+1)$

When $x=1$, we have:

$4641=(b+a+1)(p+q+1)(m+n+1)$

$3\cdot 7 \cdot 13 \cdot 17 =(b+a+1)(q+p+1)(n+m+1)$


If we let $\color{yellow}\bbox[5px,purple]{b-a+1=5}$, $\color{black}\bbox[5px,orange]{q-p+1=13}$ and $\color{orange}\bbox[5px,blue]{n-m+1=13}$, we obtain:

$\begin{align*}3\cdot 7 \cdot 13 \cdot 17&=(b+a+1)(q+p+1)(n+m+1)\\&=(b-a+1+a+a)(q-p+1+p+p)(n-m+1+m+m)\\&=(\color{yellow}\bbox[5px,purple]{(b-a+1)}\color{black}+2a)(\color{black}\bbox[5px,orange]{(q-p+1)}\color{black}+2p)(\color{orange}\bbox[5px,blue]{(n-m+1)}\color{black}+2m)\\&=(5+2a)(13+2p)(13+2m)\end{align*}$

Now, if we consider for one more case that is when $x=-2$, that gives:

$x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=(x^2+ax+b)(x^2+px+q)(x^2+mx+n)$

$672=(4-2a+b)(4-2p+q)(4-2m+n)$

$672=(3-a+1-a+b)(3-p+1-p+q)(3-m+1-m+n)$

$672=(3-a+\color{yellow}\bbox[5px,purple]{(b-a+1)}\color{black})(3-p+\color{black}\bbox[5px,orange]{(q-p+1)}\color{black})(3-m+\color{orange}\bbox[5px,blue]{(n-m+1)}\color{black})$

$672=(8-a)(16-p)(16-m)$

$2^5\cdot 3 \cdot 7=(8-a)(16-p)(16-m)$

Now, if we focus solely on the conditions

[math]\color{yellow}\bbox[5px,green]{3\cdot 7 \cdot 13 \cdot 17 =(5+2a)(13+2p)(13+2m)}[/math] and [math]\color{black}\bbox[5px,yellow]{2^5\cdot 3 \cdot 7=(8-a)(16-p)(16-m)}[/math],

It's easy to check that $a=4,\,p=2,\,m=4$ satisfy the condition and that yields $b=8,\,p=14,\,m=16$ and hence

$x^2+ax+b=x^2+4x+8=0$ gives the complex roots of $-2 \pm 2i$.

$x^2+px+q=x^2+2x+14=0$ gives the complex roots of $-1 \pm \sqrt{13}i$.

$x^2+mx+n=x^2+4x+16=0$ gives the complex roots of $-2\pm2\sqrt{3}i$.