We have, #f(x)=2x^3-4x+8=2(x^3-2x+4)=2g(x), say, where, g(x)=x^3-2x+4#.
Clearly, #(x+-1)# are not the factors of #g(x)#.
#"Now, the Leading Co-eff. of "g" is "1", and the Const. Term is "4#,
( factors #1,2,4#), we can guess the probable factors of #g# as
#(x+-1), (x+-2), &, (x+-4).#
We have already checked that #(x+-1)# are not factors.
If #(x-2)|g(x)", then, "g(2)" must be 0, but, "g(2)=8-4+4ne0, so, (x-2)" is not a factor."#
#g(-2)=-8+4+4=0 rArr (x+2)" is a factor of "g(x)#.
#g(x)=x^3-2x+4#
#=ul(x^3+2x^2)-ul(2x^2-4x)+ul(2x+4)#
#=x^2(x+2)-2x(x+2)+2(x+2)#
#=(x+2)(x^2-2x+2)#
#:. f(x)=2g(x)=2(x+2)(x^2-2x+2)#
For the Quadr.
#x^2-2x+2, Delta=(-2)^2-4(1)(2)=4-8=-4lt0#.
Hence, that quadr. can not have rational zeroes. Its complex
zeroes are #(2+-2i)/2=(1+-i)#
Thus, all the zeroes of #f# are, #-2, 1+i, and, 1-i#; of these, only
#-2# is rational.
Enjoy Maths!
