Galois group of $x^6+22x^5-9x^4+12x^3-37x^2-29x-15$ (Lang's exercise)

An exercise in Lang asks us to find the Galois group of $$f=x^6+22x^5-9x^4+12x^3-37x^2-29x-15$$ over the rationals.

I am going to write as far as I can. Then, I will ask how to proceed. I also welcome any other suggestions to solve this problem.

Step 1:

Reducing mod 2, we get $$f=x^6+x^4+x^2+x+1.$$

I am not able to proceed with this polynomial, so I change the variable $x \mapsto x+1$ and get $$g=(x+1)^6+x^4+1+x^2+1+x+1+1=(x^2+1)^3+x^4+x^2+x=x^6+x+1.$$

If I consider $\alpha$ a root of this polynomial and apply the Frobenius map $x\mapsto x^2$, I get roots of $g$: $$\{\alpha,\alpha^2,\alpha^4,\alpha^2+\alpha^3,1+\alpha+\alpha^4,1+\alpha^3\}.$$ To check whether they are distinct or not, I only need to check whether $\alpha$ is equal to $\alpha^2$, $\alpha^4$, and $\alpha^2+\alpha^3$, which is impossible. Hence, $g$ and consequently $f$ are irreducible. Their Galois group over $\mathbb{F}_2$ is $Z/6Z$, so the Galois group over the rationals contains a $6$-cycle.

Step 2:

Reducing mod 3, we get $$f=x^6+x^5-x^2+x=x(x^5+x^4-x+1).$$

If $x^5+x^4-x+1$ is reducible, then it has irreducible degree-$2$ and degree-$3$ terms. The irreducible degree-2 term implies there is a root $\alpha \in \mathbb{F}_9$. Hence, $\alpha^8=1$ which implies $\alpha^4=-1$ which in turn implies $-\alpha-1-\alpha+1=0$; a contradiction. Thus, $x^5+x^4-x+1$ is irreducible over $\mathbb{F}_3$. The Galois group of $f$ over $\mathbb{F}_3$ is $Z/5Z$, and so its Galois group over the rationals contains a $5$-cycle.

So far, how much are we able to detect the Galois group?

Step 3:

I think we need more information, so let's continue with mod 5; then we get $$f=x^6+2x^5+x^4+2x^3-2x^2+x=x(x^5+2x^4+x^3+2x^2-2x+1).$$ Putting numbers into $f$ helps us factor it, which results in $$f=x(x-1)(x+1)(x+2)(x^2+2).$$ Hence, the Galois group of $f$ over $\mathbb{F}_5$ is $Z/2Z$, and so its Galois group over the rationals contains a transposition.

We conclude that the Galois group $G$ over the rationals contains a transposition, $5$-cycle, and $6$-cycle, as a subgroup of $S_6$. I know that a simple transposition with a $6$-cycle generates $S_6$. However, here the transposition may not be simple. We probably need to use the existence of a $5$-cycle, but I am not sure how.

my questions:

1- Do we have enough information to detect the Galois group $G$?

2- If yes, probably the answer is the whole $S_6$, but how do we need to show that?

3- The current approach is fairly long. Since this is a Lang's exercise, it is possible to be this long, but I am wondering whether there is a better way to solve this problem.