<s>[INST] Find all roots of $\,(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$ The question is to find all complex roots of 
$$(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$$
and it is meant to be solved by hand.
Is there any quick way to solve this using some trick that I'm not aware of? The solution given by Wolfram used a lot of multiplying out and factoring that would be very difficult to think of or write. Also, just by observation $x=0$ and $x=-6$ are solutions.
 [/INST] $360 = 2^3 \times3^2 \times 5 = 1 \times 2 \times 3^2 \times 4 \times 5 = (-5) \times (-4) \times (-3)^2 \times (-2) \times (-1)$. We see that $x=0$ and $-6$ are solutions.
Let $(x+3) = y$. Then we get
$$(x+1)(x+2)(x+3)^2(x+4)(x+5) = y^2 (y^2-1)(y^2-4) = 360$$ Note that since $x=0$ and $-6$ are solutions, we have $y = \pm 3$ to be solutions.
Calling $y^2=t$, we then get that
$$t(t-1)(t-4) = 360$$
Since $y=\pm3$ is a solution, $t=9$ is a solution for the above equation.
Hence,
$$t^3 - 5t^2 + 4t = 360 \implies t^3 - 5t^2 + 4t - 360 = (t-9)(t^2+at+b)$$
Hence, $a-9=-5$ and $9b = 360$. This gives us $a = 4$ and $b=40$. Hence, the solutions are
$$t = 9, -2 \pm 6i$$
Hence,
$$y = \pm3 , \pm \sqrt{-2\pm6i}$$
Hence, $$x = 0,-6, -3 \pm \sqrt{-2\pm6i}$$
</s>