Quartic equation with integer solutions I am trying to find the solutions of the following quartic equation given that all the solutions are integers.
$$x^4+22x^3+172x^2+552x+576=0$$
Below is the original phrasing of the problem with hints building up to this equation. I can prove all of the results it asks for before the final part of the question, but I am struggling to actually find the solutions to the equation and require help with explaining the process as well.



I then know that:
(1) $k_1k_2k_3k_4 = 576 = (1)(2^6)(3^2)$
(2) $(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323=(1)(3^3)(7^2)$
(3) $(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = (1)(5^2)(7)$
However, I do not know how to proceed from here and the solution to this problem didn't explain in enough detail for me to either understand the solution, or the approach.

For $$f(x) = x^4+22x^3+172x^2+552x+576,$$
notice that if $x\geq0$ then $f(x) \geq 576 > 0,$
so all roots of the equation $f(x) = 0$ are negative.
Since the roots are $-k_1, -k_2, -k_3, -k_4,$
this tells us that $k_1, k_2, k_3, k_4$ are all positive.
That leaves only a few possible values of $k_i$ that could occur
in the equation
$$(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = 5^2 \cdot 7.$$
The factors of $175$ are $1, 5, 5^2 = 35, 7, 5\cdot7 = 35,$ 
and $5^2 \cdot 7 = 175.$
(I chose this equation to start with because $175$ has fewer factors to consider than either $576$ or $1323$.)
Since each factor $k_i - 1$ must be a factor of $175,$
the possible values of any $k_i$ can only be among the numbers
$2, 6, 26, 8, 36,$ and $176.$
From 
$$k_1k_2k_3k_4 = 576 = 2^6 \cdot 3^2, \tag1$$
we know any of the $k_i$ can have only $2$ or $3$ as prime factors.
So $k_i \neq 26 = 2 \cdot 13$ and $k_i \neq 176 = 16 \cdot 11.$
From
$$(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323 = 3^3 \cdot 7^2,$$
we know $k_i + 1$ must be divisible by $3$ or by $7.$
So $k_i + 1 \neq 37,$ and therefore $k_i \neq 36.$
Therefore the only possible values of any of the $k_i$ are $2,$ $6,$ or $8.$
But Equation $(1)$ implies either that one of the $k_i$ is divisible by $9$
(which we now know cannot be true) or that two of the $k_i$ are each divisible by $3.$ If $k_i$ is divisible by $3,$ its only possible value is $6$ (since neither $2$ and $8$ is divisible by $3$).
Without loss of generality, we can set $k_1 = k_2 = 6.$
Dividing both sides of Equation $(1)$ by $k_1k_2 = 36,$
we now have $k_3k_4 = 2^4.$ Each of $k_3$ or $k_4$ can then only be $2$ or $8$ (since $6$ is not a power of $2$), but if one is $2$ then the other is $8$ and vice versa. So without loss of generality we can set $k_3 = 2$ and
$k_4 = 8.$
The four roots therefore are
$-k_1 = -6,$ $-k_2 = -6,$ $-k_3 = -2,$ and $-k_4 = -8.$
In ascending order they are $-8, -6, -6, -2.$
