<s>[INST] Find the roots of the polynomial $x^4+2x^3-x-1=0$. Problem
Find all $4$ roots of the polynomial: $$f(x)=x^4+2x^3-x-1.$$

My Attempt
Observe,
$$f(-2)=1,\; f(-1)=-1,\; f(0)=-1,\; f(1)=1.$$
Therefore, $f(x)$ has a root between $-2$ and $-1$, another root between $0$ and $1$.
Since $f(x)$ does not contain a second degree term, a clever substitution might change $f(x)$ into a polynomial that is easier to deal with. For example $(x\to x-1)$,
$$f(x-1)=x^4-2x^3+x-1=f(-x)$$
Also, I am avoiding computational or graphical assistance.
 [/INST] The "basic" tools for dealing with a polynomial with integer coefficients tell us that the zeroes will not be simple to locate.  The Rational Zeroes Theorem only suggests the candidates $ \ \pm 1 \ \ , $ which you have already shown are not correct.  The Rule of Signs indicates one positive real and three or one negative real zeroes.  We may suspect that two of the zeroes form a "complex conjugate pair", since the function is not positive for values $ \ -1 \ < \ x \ < \ 0 \ \ , $ as    $ \ x^4 - 1 \ $ is more negative in this interval than $ \ 2x^3 - x \ = \ x·(2x^2 - 1) \ $ is ever positive, so there are no further $ \ x-$intercepts.
We could naïvely factor the polynomial as  $ \ x^4 + 2x^3 - x - 1 \ = \ (x^2 + ax + b)·(x^2 + cx + d) \ \ , $ one of these factors being irreducible over real numbers. Following on your observation and lhf's comment, it must be that each of these is the same under the transformations $ \ x \ \rightarrow \ -x \ $ and $ \ x \ \rightarrow \ x - 1 \ \ . $  [I am avoiding the implication of symmetry about $ \ x \ = \ -\frac12 \  , \ $ since Macavity and B. Goddard have already treated it.]  This will mean, for instance, that
$$ x^2 \ - \  ax \ + \ b \ \ = \ \ (x-1)^2 \ - \ a·(x-1) \ + \ b \ \ = \ \ x^2 \ + \ (a-2)·x \ + \ (b - a + 1) \ \ $$
$$ \Rightarrow \ \ -a \ \ = \ \ a \ - \ 2 \ \ \Rightarrow \ \ a \ \ = \ \ 1 \ \ , $$
with $ \ b \ $ remaining unspecified.  The product of the factors is then
$$ (x^2 + x + b)·(x^2 + x + d) \ \ = \ \ x^4 \ + \ 2x^3 \ + \ (b + d + 1)·x^2 \ + \ (b + d)·x \ + \ bd \ \ . $$
This presents us with the coefficient equations $ \ b + d + 1 \ = \ 0 \ $ and $ \ b + d \ =  \ -1 \ \ , $ from the quadratic and linear terms, and $ \ bd \ = \ -1 \ \   $ as the constant term.  Putting these together, we find
$$ ( b \ + \  d )^2 \  = \  1 \ \ , \ \ 4bd \  = \  -4 \ \ \Rightarrow \ \ (b \ - \ d)^2 \  = \  5 $$
$$ \Rightarrow \ \  b \ + \ d  \  = \  -1  \ \ \text{[as already determined]} \ \ , \ \ b \ - \ d \ = \ \pm \sqrt5 $$
$$ \Rightarrow \ \ b \ = \ \frac12 · (-1 \ \mp \ \sqrt5) \ \ , \ \ d \ = \ \frac12 · (-1 \ \pm \ \sqrt5) \ \ . $$
This makes it clear why a straightforward application of the Viete relations, or even an attempt to solve directly for the coefficients in $ \ x^4 + 2x^3 - x - 1 \ = \ (x^2 + ax + b)·(x^2 + cx + d) \ , \  $ is not very enlightening.  The factors may be written in either order, so, for instance,
$$   \left( \ x^2 + x - \frac12 · [1 \ + \ \sqrt5] \ \right) \ · \ \left( \ x^2 + x - \frac12 · [1 \ - \ \sqrt5] \ \right)  \ \ = \  \ 0 \ \ ,   $$
the zeroes of the first factor being the real values $ \ -\frac12 \ \pm \ \frac{\sqrt{3 \ +  \ 2·\sqrt5}}{2} \ \ \approx \ \ 0.8668 \ , \ -1.8668 \ $ and those of the second factor are the "conjugate pair" $ \ -\frac12 \ \pm \ i·\frac{\sqrt{2·\sqrt5 \ - \ 3}}{2} \ \ \approx \ \ -0.5 \ \pm \ i·0.6067 \ \ . $
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