Question

How do you factor `x^4 + x^3 + x^2 + x + 1`?

Answer 1

`x^4+x^3+x^2+x+1`

`=(x^2+(1/2+sqrt(5)/2)x+1)(x^2+(1/2-sqrt(5)/2)x+1)`

Explanation:

This quartic has four zeros, which are the non-Real Complex `5`th roots of `1`, as we can see from:

`(x-1)(x^4+x^3+x^2+x+1) = x^5-1`

So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:

`x^4+x^3+x^2+x+1`

`=(x-(cos((2pi)/5) + i sin((2pi)/5))) * (x-(cos((4pi)/5) + i sin((4pi)/5))) * (x-(cos((6pi)/5) + i sin((6pi)/5))) * (x-(cos((8pi)/5) + i sin((8pi)/5)))`

A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if `x=r` is a zero of `x^4+x^3+x^2+x+1`, then `x=1/r` is also a zero.

Hence there is a factorisation in the form:

`x^4+x^3+x^2+x+1`

`=(x-r_1)(x-1/r_1)(x-r_2)(x-1/r_2)`

`=(x^2-(r_1+1/r_1)x+1)(x^2-(r_2+1/r_2)x+1)`

So let's look for a factorisation:

`x^4+x^3+x^2+x+1`

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

`=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1`

Equating coefficients we find:

`a+b = 1`

`2+ab=1`, so `ab = -1` and `b=-1/a`

Substituting `b=-1/a` in `a+b=1` we get:

`a-1/a = 1`

Hence:

`a^2-a-1 = 0`

Using the quadratic formula, we can deduce:

`a = 1/2 +- sqrt(5)/2`

Since our derivation was symmetric in `a` and `b`, one of these roots can be used for `a` and the other for `b`, to find:

`x^4+x^3+x^2+x+1`

`=(x^2+(1/2+sqrt(5)/2)x+1)(x^2+(1/2-sqrt(5)/2)x+1)`

If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.