How do you find the solutions to the following equation?
x^4-x^3+x-1=0
First, identify what kind of solution we are searching for via Des Cartes Rule of Sign.
3 or 1 positive
1 negative
2 or 0 imaginary
$\displaystyle \frac{p}{q}=\frac{\pm 1}{\pm 1}=+1, -1$
I don't know how to format synthetic division in latex so here is the best I have at it.
If $\displaystyle x=1$, our division looks like:
$\displaystyle \begin{matrix}
1 & -1 & 0 & 1 & -1\\
\vdots & 1 & 0 & 0 & 1\\
1 & 0 & 0 & 1 & 0
\end{matrix}$
Hence x=1 is a solution.
$\displaystyle (x-1)(x^3+1)$
$\displaystyle x^3=-1$
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.