# How do you factor completely x^3+x^2+x+1?

Nov 24, 2015

In $\mathbb{R}$, the answer is $\left({x}^{2} + 1\right) \left(x + 1\right)$.

In $\mathbb{C}$, the answer is $\left(x + i\right) \left(x - i\right) \left(x + 1\right)$.

#### Explanation:

In some cases, you can "see" how to factor such a term with some experience.

Here, for example, you could do the following transformation:
${x}^{3} + {x}^{2} + x + 1 = {x}^{2} \left(x + 1\right) + x + 1 = {x}^{2} \left(x + 1\right) + 1 \left(x + 1\right) = \left({x}^{2} + 1\right) \left(x + 1\right)$

As ${x}^{2} + 1$ has no roots for $\mathbb{R}$, you can't factor this expression further in $\mathbb{R}$.

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Note: in $\mathbb{C}$, you can indeed factor it further:

${x}^{3} + {x}^{2} + x + 1 = \left({x}^{2} + 1\right) \left(x + 1\right) = \left(x + i\right) \left(x - i\right) \left(x + 1\right)$

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Now, how do you do the factorization if you don't "see" my steps from above?

First, find one root.
Generally, you can start with plugging $x = 1$, $x = - 1$, $x = 2$, $x = - 2$ until one of those reveals itself as a root.

Here, it's obvious that $x = - 1$ is a root since
${\left(- 1\right)}^{3} + {\left(- 1\right)}^{2} + \left(- 1\right) + 1 = - 1 + 1 - 1 + 1 = 0$

Afterwards, perform a polynomial division of ${x}^{3} + {x}^{2} + x + 1$ by $\left(x - \text{your root}\right)$, so here, $\left(x - \left(- 1\right)\right)$.

I know that in some countries, the notation for long division is different. I will write it in the notation that I'm familiar with and I hope that it will be easy for you to re-write it in your notation if necessary.

$\textcolor{w h i t e}{\times x} \left({x}^{3} + {x}^{2} + x + 1\right) \div \left(x + 1\right) = {x}^{2} + 1$
$\textcolor{w h i t e}{x} - \left({x}^{3} + {x}^{2}\right)$
$\textcolor{w h i t e}{\times x} \frac{\textcolor{w h i t e}{\times \times \times}}{\textcolor{w h i t e}{x}}$
$\textcolor{w h i t e}{\times \times \times x} \textcolor{l i g h t g r e y}{0} \textcolor{w h i t e}{x} + x + 1$
$\textcolor{w h i t e}{\times \times \times \times} - \left(x + 1\right)$
$\textcolor{w h i t e}{\times \times \times \times \times} \frac{\textcolor{w h i t e}{\times \times x}}{\textcolor{w h i t e}{x}}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times} 0$

This means that you can factor your polynomial as follows:

$\left({x}^{3} + {x}^{2} + x + 1\right) = \left(x + 1\right) \left({x}^{2} + 1\right)$

Finding a factorization for ${x}^{2} + 1$ is easier since this is a quadratic polynomial. All you have to do is search for solutions for

${x}^{2} + 1 = 0 \iff {x}^{2} = - 1$

In $\mathbb{R}$, it's not possible since you can't compute a root of $- 1$. In this case you are finished with the factorization

$\left(x + 1\right) \left({x}^{2} + 1\right)$.

In $\mathbb{C}$ however, ${i}^{2} = - 1$ and ${\left(- i\right)}^{2} = - 1$ hold, so you can factor
${x}^{2} + 1 = \left(x + i\right) \left(x - i\right)$ .

Nov 24, 2015

An alternative method included for fun...

#### Explanation:

Notice that $\left(x - 1\right) \left({x}^{3} + {x}^{2} + x + 1\right) = {x}^{4} - 1$

So the zeros of this cubic are the $4$th roots of $1$ except for $1$ itself.

In the Complex plane the $4$th roots of $1$ lie at $\frac{1}{4}$ rotations around the unit circle:

graph{(x^2+y^2-1)((x+1)^2+y^2-0.004)(x^2+(y-1)^2-0.004)(x^2+(y+1)^2-0.004)((x-1)^2+y^2-0.004) = 0 [-2.812, 2.814, -1.406, 1.406]}

That is $1$, $i$, $- 1$ and $- i$

So ${x}^{4} - 1 = \left(x - 1\right) \left(x - i\right) \left(x + 1\right) \left(x + i\right)$

and ${x}^{3} + {x}^{2} + x + 1 = \left(x - i\right) \left(x + 1\right) \left(x + i\right)$