Question

How do you factor `x^5 - 5x^4 - x^3 + x^2 + 4 = 0`?

Answer 1

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

There are no simpler factors with rational coefficients.

Explanation:

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

First note that the sum of the coefficients is zero.

That is `1-5-1+1+4 = 0`

So `f(1) = 0` and `(x-1)` is a factor

Reversing the signs on the coefficients of odd degree we also find:

`-1-5+1+1+4 = 0`

So `f(-1) = 0` and `(x+1)` is a factor

`(x-1)(x+1) = x^2-1`, so let's separate out this factor to find:

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

By the rational root theorem, the only possible rational zeros of the remaining cubic factor `(x^3-5x^2-4)` are `+-1`, `+-2`, `+-4`.

None of these work, so this cubic only has factors with irrational and/or Complex coefficients.

If this question was asked in Precalculus, I might offer a solution of the cubic, but it's probably beyond the scope of your Algebra course, so let's leave the factoring at this point.