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

2 Answers
Nov 24, 2015

Answer:

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

Answer:

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)#