Find the real solutions (if any) of $x^{12}=3 x^9+4 x^6+4 x^3+2 x+2$ 

Find the real solutions(if any) of $$x^{12}=3 x^9+4 x^6+4 x^3+2 x+2$$


Note: I need only HINT:
My effort:
I let $v=x^3$, then we get
$$v^4=3v^3+4v^2+4v+2\sqrt[3]{v}+2 \Rightarrow v^4-3v^3-4v^2-4v=2(1+\sqrt[3]{v})$$
$$\begin{aligned}
& \Rightarrow v^4-3 v^3-4 v^2-3 v+1=(v+1)+2(\sqrt[3]{v}+1) \\
& \Rightarrow v^2\left(\left(v+\frac{1}{v}\right)^2-3\left(v+\frac{1}{v}\right)-6\right)=(v+1)+2(\sqrt[3]{v}+1)
\end{aligned}$$
Any way to proceed with an hint?
 A: Let $$f(x)=x^{12}-3x^9-4x^6-4x^3-2x-2$$
By Intermediate value theorem, we have $f(1)<0, f(2)>0$, so $\exists$ atleast one root on $(1,2)$. But by Descarte's rule of signs, the number of changes in $f(x)$ is one, so only one positive root. Obviously there should be one more real root. So there exists a quadratic factor $x^2+px+q$ with positive discriminant, so we have:
$$f(x)=\left(x^2+p x+q\right)\left(x^{10}-p x^9+b x^8+c x^7+d x^6+e x^5+f x^4+g x^3+h x^2+m x-\frac{2}{q}\right)$$
Comparing the coefficients:
$$b=p^2-q$$
$$c+pb-pq=-3$$
$$d+pc+qb=0$$
$$e+pd+qc=0$$
$$f+pe+qd=-4$$
$$g+pf+qe=0$$
$$h+pg+qf=0$$
$$m+ph+qg=-4$$
$$\frac{-2}{q}+pm+hq=0$$
$$\frac{-2p}{q}+mq=-2$$
Now let us assume or try whether the factorization is possible over $\mathbb{Z}$. So let us start with $q=2$. So the last equation above becomes $$-p+2m=-2 \Rightarrow p=2(m+1)$$ Also the last but one equation will be $$-1+2m(m+1)=-hq$$ This is impossible since LHS is odd, RHS is even. The same reasoning applies when $q=-2$.
\newline
Now let $$q=1$$
Since $x^2+px+q$ should have real roots, we have
$$p^2 \geq 4$$
Now if $q=1$, we get $$-2p+m=-2 \Rightarrow m=2p-2 \Rightarrow h=2-2p^2+2p$$
$$\Rightarrow g=2p^3-2p^2-4p-2 ,f=-2p^4+2p^3+6p^2-2\Rightarrow e=2p^5-2p^4-8p^3+2p^2+6p+2$$
$$\Rightarrow d=-2p^6+2p^5+10p^4-4p^3-12p^2-2p-2-----(*)$$
Also from first two equations, we have
$$b=p^2-1,c=2p-3-p^3$$
From third equation, we get
$$\Rightarrow d=p^4-3p^2+3p+1------(**)$$
From $(*),(**)$, we get
$$2p^6-2p^5-9p^4+4p^3+9p^2+5p+3=0  $$
So that means $p$ can only be odd integer. Since $p^2 \geq 4$, we have $p=3,5,7,...$ or $p=-3,-5,-7,...$. By growth argument, it is easy to say that we cannot find any root of the above sixth degree polynomial with $p \geq 3, p \leq -3$
Thus we have $$\boxed{q \ne \pm 2, 1}$$
Now the only choice left over for $q$ is $q=-1$.For $q=-1$, the discriminant of $x^2+px+q$ is $D=p^2+4>0$.Now
Repeating the above backward substitutions, we get
$$2p^6+2p^5+9p^4+12p^3+9p^2+11p+5=0$$ Obviously the above sixth degree polynomial cannot have non negative roots. So $p<0$. By Inspection $p=-1$ satisfies. Hence $$p=-1,q=-1 \Rightarrow x^2+px+q=x^2-x-1 | f(x)$$ Also if at all there are any integer roots of the above sixth degree polynomial, the root should be an odd integer, that is $p=-1,-3,..$. We have already figured out $p=-1$. Again by growth argument we cannot have any root $p \leq -3$.Thus finally we have successfully found the quadratic factor of $f(x)$ which is $x^2-x-1$.
