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 $(2)$?

To solve $x^4+6x^3−9x^2−162^x−243=0$,
first try and factor, if possible: $(x^2−3x−9)(x^2+9x+27)=0$
So since this factors into to quadratics, we can use quadratic formula and some other simplification methods such as completing the square.
The roots of $x^2-3x-9$ are the same as the roots of $x^2-x-1$, multiplied by three since the absolute value of coefficients strictly increases by a power of $3$. This all amounts to $(3/2)(1±√5)$ being the roots of $x^2-3x-9$, and as for $x^2+9x+27$, one notices the increasing powers of $9$, and therefore a root of $x^2+9x+27$ is equivalent to a root of $x^2+x+1$, multiplied by nine. This amounts to $(9/2)(-1±i√3)$ being the roots of $x^2+9x+27$. 
The complete list of solutions to 
$x^4+6x^3−9x^2−162^x−243=0$
is
$(9/2)(-1+i√3)$
$(9/2)(-1-i√3)$
$(3/2)(1+√5)$
$(3/2)(1-√5)$
