<s>[INST] Solving a fourth degree equation Well, I want to find the roots of the equation:
$$x^4+6x^3-9x^2-162x-243=0\tag1$$
And Wolfram Alpha tells me that I can factor this and get:
$$(x^2-3x-9)(x^2+9x+27)=0\tag2$$
Then it will be easy to solve... But my question is, how can I go from $(1)$ to $(2)$?
 [/INST] In case anyone ever looks at this question again...it is best to try things that might give a factorization first. What I notice is the powers of 3 in $x^4 + 6 x^3 - 9 x^2 - 162 x - 243 \; . \;$ This suggests taking $x=3t,$ then dividing through by the resulting constant factor $81,$ resulting in
$$ t^4 + 2 t^3 - t^2 - 6 t - 3 \; . \; $$
The two things to try next are
$$ (t^2 + At + 1) (t^2 + B t - 3)  $$
and
$$ (t^2 + Ct - 1) (t^2 + D t + 3)  $$
The first one fails, you get $A = 8/5$ and $B = 2/5,$ the coefficient of $t^2$ is wrong.
The second one works, $C+D=2,$ then $3C-D= -6.$ Add these to get $4C = -4,$ then $C = -1,$ then $D = 3.$ The $t^2$ term also works out correctly. 
$$ (t^2 -t - 1) (t^2 + 3 t + 3) =   t^4 + 2 t^3 - t^2 - 6 t - 3 \; . \; $$
To get back to $x$ I suggest multiplying by $9 \cdot 9 = 81$ for
$$ (9t^2 -9t - 9) (9t^2 + 27 t + 27) =   81t^4 + 162 t^3 - 81 t^2 - 486 t - 243 \; . \; $$
Finally $t = x/3$ becomes
$$ (x^2 -3x - 9) (x^2 + 9 x + 27) =   x^4 + 6 x^3 - 9 x^2 - 162 x - 243 \; . \; $$
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