Solve for $x$: $x^4+2x^3-x-1=0$

Question

Solve for $x$:

$$x^4+2x^3-x-1=0$$

My try ;

$$x^4+2x^3-x-1=(ax^2+bx+c)(dx^2+px+q)=0$$ But I cannot go further!!!

Answer 1

$$x^4+2x^3-x-1=\left(x^2+x-\frac{1}{2}\right)^2-\frac{5}{4}$$ and the rest is smooth.

We can get it by the following way. $$x^4+2x^3-x-1=(x^2+x+k)^2-((2k+1)x^2+(2k+1)x+k^2+1)$$ for all value of $k$ and for $k=-\frac{1}{2}$ we'll get a difference of squares.

Answer 2

hint

Let $f (x)=x^4+2x^3-x-1$. $f (0)=-1,f (1)=1,f (-2)=1$ there is at least one root in $(0,1) $ and

one in $ (-2,0)$.

$$f''(x)=12x^2+12x=12x (1+x)\geq 0$$

there are only two real roots.

Answer 3

$(ax^2+bx+c)(dx^2+px+q)=adx^4+(ap+bd)x^3+(aq+bp+cd)x^2+(bq+pc)x+cq$

Try to solve a,b,c,d,p,q

as $$ \left\{ \begin{array}{c} ad=1\\ ap+bd=2\\ aq+bp+cd=0\\ bq+pc=-1\\ cq=-1\\ \end{array} \right. $$