Help with polynomial

Apr 2008
7
1
How do you find the solutions to the following equation?

x^4-x^3+x-1=0
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.

\(\displaystyle (x-1)(x^3+1)\)

\(\displaystyle x^3=-1\)
 
Last edited:
Apr 2008
7
1
the answer should be 1,-1, 1/2 + or - square root of 3/2i
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.
 
Apr 2008
7
1
thank you, but i think your math skills are to hardcore for me. i just need to know how to factor it out
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.
 
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