# Help with polynomial

#### Ericonda

How do you find the solutions to the following equation?

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

#### dwsmith

MHF Hall of Honor
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.

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

$$\displaystyle x^3=-1$$

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#### Ericonda

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

#### dwsmith

MHF Hall of Honor
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.

#### Ericonda

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
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.

Ericonda