We have, f(x)=2x^3-4x+8=2(x^3-2x+4)=2g(x), say, where, g(x)=x^3-2x+4f(x)=2x3−4x+8=2(x3−2x+4)=2g(x),say,where,g(x)=x3−2x+4.
Clearly, (x+-1)(x±1) are not the factors of g(x)g(x).
"Now, the Leading Co-eff. of "g" is "1", and the Const. Term is "4Now, the Leading Co-eff. of g is 1, and the Const. Term is 4,
( factors 1,2,41,2,4), we can guess the probable factors of gg as
(x+-1), (x+-2), &, (x+-4).(x±1),(x±2),&,(x±4).
We have already checked that (x+-1)(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."(x−2)∣g(x), then, g(2) must be 0, but, g(2)=8−4+4≠0,so,(x−2) is not a factor.
g(-2)=-8+4+4=0 rArr (x+2)" is a factor of "g(x)g(−2)=−8+4+4=0⇒(x+2) is a factor of g(x).
g(x)=x^3-2x+4g(x)=x3−2x+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!