How do you list all possible roots and find all factors and zeroes of #3x^3+9x^2+4x+12#?

1 Answer
Jul 13, 2016

(see below)

Explanation:

Given
#color(white)("XXX")3x^3+9x^2+4x+12#

Noting that the ratio of the constants for the first two terms is the same as the ratio for the last two terms, provides a hint:
#color(white)("XXX")=3x^2(x+3) +4(x+3)#

#color(white)("XXX")=(3x^2+4)(x+3)#

Since #(3x^2+4)>0# for all Real values of #x#
1. there are no Real factors of #(3x^2+4)#
#color(white)("XXX")rarr # there are no Real zeros corresponding to the #(3x^2+4)# term.

  1. The only Real zero comes from #x+3=0 rarr x=-3#

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If Complex values are allowed:
#(3x^2+4)=3(x-2/sqrt(3)i)(x+2/sqrt(3)i)# as further factoring
and
complex zeroes at #x=2/sqrt(3)i# and #x=-2/sqrt(3)i#