Question

How do you factor `F(x) = x^4+6x^3+2x^2-3` ?

Answer 1

This quartic has no rational or simple irrational factors, but:

`x^4+6x^3+2x^2-6x-3 = (x-1)(x+1)(x+3-sqrt(6))(x+3+sqrt(6))`

Explanation:

Given:

`F(x) = x^4+6x^3+2x^2-3`

By the rational roots theorem, any rational zeros of `F(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-3` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-3`

None of these work. Hence `F(x)` has no rational zeros.

It is a fairly typically nasty quartic with two real irrational zeros and two complex ones.

I suspect there may be a missing term `-6x` in the question, since:

`x^4+6x^3+2x^2-6x-3 = (x-1)(x+1)(x+3-sqrt(6))(x+3+sqrt(6))`