Question

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

Answer 1

In `RR`, the answer is `(x^2 + 1) (x + 1)`.

In `CC`, the answer is `(x+i)(x-i)(x+1)`.

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 ( x + 1) + x + 1 = x^2 (x + 1) + 1 (x + 1) = (x^2 + 1) (x + 1)`

As `x^2 + 1` has no roots for `RR`, you can't factor this expression further in `RR`.

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Note: in `CC`, you can indeed factor it further:

`x^3 + x^2 + x + 1 = (x^2+1)(x+1) = (x+i)(x-i)(x+1)`

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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
`(-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0`

Afterwards, perform a polynomial division of `x^3 + x^2 + x + 1` by `(x-"your root")`, so here, `(x-(-1))`.

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.

`color(white)(xxx) (x^3 + x^2 + x + 1) -: (x+1) = x^2 + 1`
`color(white)(x) - (x^3 + x^2)`
`color(white)(xxx) color(white)(xxxxxx)/color(white)(x)`
`color(white)(xxxxxxx) color(lightgrey)(0) color(white)(x) + x + 1`
`color(white)(xxxxxxxx) - (x + 1)`
`color(white)(xxxxxxxxxx) color(white)(xxxxx)/color(white)(x)`
`color(white)(xxxxxxxxxxxxxx) 0`

This means that you can factor your polynomial as follows:

`(x^3 + x^2 + x + 1) = (x+1)(x^2 + 1)`

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 <=> x^2 = -1`

In `RR`, it's not possible since you can't compute a root of `-1`. In this case you are finished with the factorization

`(x+1)(x^2+1)`.

In `CC` however, `i^2 = -1` and `(-i)^2 = -1` hold, so you can factor
`x^2 +1 = (x+i)(x-i)` .

Answer 2

An alternative method included for fun...

Explanation:

Notice that `(x-1)(x^3+x^2+x+1) = 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 `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 = (x-1)(x-i)(x+1)(x+i)`

and `x^3+x^2+x+1 = (x-i)(x+1)(x+i)`