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.

With the substitution $x=t-b/na$, you eliminate the second highest order term of a polynomial $ax^n + bx^{n-1} + \cdots .$

If you substitute $x=t-2/4 = t-1/2$ your polynomial becomes

$$t^4 -\frac{3}{2}t^2 -\frac{11}{16}.$$

This is quadratic in $t^2$ so you can solve to get

$$t^2 = \frac{3\pm 2\sqrt{5}}{4}.$$

So the four roots of your polynomial are plus and minus the square roots of those two solutions (plus $1/2$ because of the original substitution.) We were lucky that the first order term also disappeared.

Hint: From your observation $f(x-1)=f(-x)$, it is clear $x=-\frac12$ is a point of symmetry. Hence use $y=x+\frac12$ to get an even quadratic in $y$, which will then admit the substitution $t=y^2$ to solve...