What are the zero(s) for #f(x) = 2x^6 +x^3+3#?

1 Answer
Oct 24, 2015

Answer:

#f(x)# has six Complex zeros which we can find by recognising that #f(x)# is a quadratic in #x^3#.

Explanation:

#f(x) = 2x^6+x^3+3 = 2(x^3)^2 + x^3+3#

Using the quadratic formula we find:

#x^3 = (-1+-sqrt(1^2-4xx2xx3))/(2*2)#

#=(-1+-sqrt(-23))/4 = (-1+-i sqrt(23))/4#

So #f(x)# has zeros:

#x_(1,2) = root(3)((-1+-i sqrt(23))/4)#

#x_(3,4) = omega root(3)((-1+-i sqrt(23))/4)#

#x_(5,6) = omega^2 root(3)((-1+-i sqrt(23))/4)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of unity.