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

2 Answers
Nov 24, 2015

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) .

Nov 24, 2015

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 4th roots of 1 except for 1 itself.

In the Complex plane the 4th 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)