1. ## Help with polynomial

How do you find the solutions to the following equation?

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

2. Originally Posted by Ericonda
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

$\frac{p}{q}=\frac{\pm 1}{\pm 1}=+1, -1$

I don't know how to format synthetic division in latex so here is the best I have at it.

If $x=1$, our division looks like:

$\begin{matrix}
1 & -1 & 0 & 1 & -1\\
\vdots & 1 & 0 & 0 & 1\\
1 & 0 & 0 & 1 & 0
\end{matrix}$

Hence x=1 is a solution.

$(x-1)(x^3+1)$

$x^3=-1$

3. the answer should be 1,-1, 1/2 + or - square root of 3/2i

4. Originally Posted by Ericonda
the answer should be 1,-1, 1/2 + or - square root of 3/2i
Ok and??

I gave you two solutions and all the the tools to find the complex conjugate.

5. thank you, but i think your math skills are to hardcore for me. i just need to know how to factor it out

6. 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 $\pm 1$.

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

If $x=-1$, our division looks like:

$\begin{matrix}
1 & 0 & 0 & 1\\
\vdots & -1 & 1 & -1\\
1 & -1 & 1 & 0
\end{matrix}$

Now we have $(x-1)(x+1)(x^2-x+1)$

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