Solve degree-four polynomial equation $\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x$ How to solve 
$$\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x?$$
I think there is a trick, because, if $X=x^2-3x-5$, I get 
$$ X^2-3X-5=x.$$
I don't know how to continue. 
I've also tried to factorize but it does not work. Any idea?
 A: We can still do a direct computation, if the trick is not immediately obvious. Then we obtain $f(x)=0$ with
$$
f(x)=x^4 - 6x^3 - 4x^2 + 38x + 35=(x^2 - 2x - 7)(x + 1)(x - 5).
$$
The factorisation is not difficult because we can see the roots $x=-1$ and $x=5$ (by the rational root test, say) and then obtain a quadratic polynomial.
A: Let $f(x) = x^2-3x-5$. What you would really like is to find a fixed point of $f(x)$, since your equation reads $f(f(x)) = x$ so any solution to $f(x)=x$ would satisfy it!
Now you have
$$
f(x) = x \iff x^2-3x-5 = x \iff 0 = x^2-4x-5 = (x+1)(x-5) \iff x \in \{-1,5\}.
$$
The others are a matter of factoring the quartic by the above factors.
A: Factorize as follows,
$$(x^2-3x-5)^2-3(x^2-3x-5)^2-4=x+1$$
$$(x^2-3x-9)(x^2-3x-4)=x+1$$
$$(x^2-3x-9)(x-4)(x+1)-(x+1)=0$$
$$(x+1)(x^3-7x^2+3x+35)=0$$
$$(x+1)[(x^3-5x^2)-(2x^2-3x-35)]=0$$
$$(x+1)[x^2(x-5)-(2x+7)(x-5)]=0$$
$$(x+1)(x-5)(x^2-2x-7)=0$$
$$(x+1)(x-5)(x-1+2\sqrt2)(x-1-2\sqrt2)=0$$
Thus, the roots are $x=-1, 5, 1\pm 2\sqrt2$.
A: Here is  little trick. First set
$$
f(x) = x^2 -3 x -5
$$
Then you see that you want to solve
\begin{align}
y &= f(x) \\
x &= f(y)
\end{align}
On the $x,y$ plane the first equation is a "standard" parabola parametrized by $(t,f(t) )$ while the other-one is the same parabole with $x$ and $y$ exchanged (and it is paramatrized by $(f(t),t)$. It is clear by symmetry that these two parabolae intersect in four points and that two solutions are on the $y=x$ line. 
You find these two solutions by finding the roots of $f(x)=x$ which are $x=-1,5$. At this point you can divide your quartic by $(x+1)(x-5)$ and proceed as Dietrich Burde mentioned. 
