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 #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))#