Calculation of roots of a polynomial

Question

Let $a,b>0$. How can I calculate the roots of the following polynomial?

$$2bx^6 + 4bx^5+(4b-a)x^4 + 4(b+a)x^3 + (2b-6a)x^2+4ax-a=0$$

Answer 1

There is really only one parameter here rather than two: if $s = a/b$, you can divide by $b$ and write the equation as $$ 2\,{x}^{6}+4\,{x}^{5}- \left( s-4 \right) {x}^{4}+4\, \left( s+1 \right) {x}^{3}-2\, \left( 3\,s-1 \right) {x}^{2}+4\,sx-s=0$$

But in general you're not going to get a nice closed-form solution. E.g. for $s=1$ this is an irreducible sextic over the rationals with Galois group $S_6$. There is no solution in radicals. Of course you can use numerical methods for specific values of $s$. Or you might use a series expansion: for small $s$, one root is $$ {\frac {\sqrt {2}\sqrt {s}}{2}}-{\frac {3s}{2}}+{ \frac {25\,\sqrt {2}{s}^{3/2}}{8}}-{\frac {61\,{s}^{2}}{4 }}+{\frac {2615\,\sqrt {2}{s}^{5/2}}{64}}-{\frac { 1863\,{s}^{3}}{8}}+{\frac {177433\,\sqrt {2}{s}^{7/2}}{256}}-{\frac {68165\,{s}^{4}}{16}}+{\frac {54967659\, \sqrt {2}{s}^{9/2}}{4096}}-{\frac {2758467\,{s}^{5}}{32}}+{\frac {4607808079\,\sqrt {2}{s}^{11/2}}{16384}}+\ldots $$ and another is obtained by replacing $s^{k/2}$ by $-s^{k/2}$ for odd $k$.

Answer 2

Note that the given equation is equivalent to $$s(x-1)^4=2x^2(x+1)^2(x^2+1).$$ where $s=a/b>0$. I guess there is no "easy" closed formula for the roots, but, for example, we always have at least two real roots: one in $(0,1)$ because $$s=\text{LHS}(0)>\text{RHS}(0)=0 \quad\text{ and }\quad 0=\text{LHS}(1)<\text{RHS}(1)=16$$ and another one in $(-\infty,-1)$ because for sufficiently large $t>1$, $$\text{LHS}(-t)<\text{RHS}(-t) \quad\text{ and }\quad 16s=\text{LHS}(-1)>\text{RHS}(-1)=0.$$ Numerical methods can be used to find approximations of such roots.