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

##### 1 Answer

Mar 1, 2017

#### Answer:

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

That means that the only possible *rational* zeros are:

#+-1, +-3#

None of these work. Hence

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

I suspect there may be a missing term

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