3 or 1 positive

1 negative

2 or 0 imaginary

We found 1 positive solution and we know there is 1 negative solution. We have 2 choices of solutions \(\displaystyle \pm 1\).

If we take our new polynomial, \(\displaystyle x^3+1\), we can do synthetic division again to obtain a quadratic polynomial.

If \(\displaystyle x=-1\), our division looks like:

\(\displaystyle \begin{matrix}

1 & 0 & 0 & 1\\

\vdots & -1 & 1 & -1\\

1 & -1 & 1 & 0

\end{matrix}\)

Now we have \(\displaystyle (x-1)(x+1)(x^2-x+1)\)

We can now use the quadratic equation, \(\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\), to solve the quadratic.