What are all the possible rational zeros for f(x)=2x^3-4x+8 and how do you find all zeros?

1 Answer
Sep 20, 2016

Thus, all the zeroes of f are, -2, 1+i, and, 1-i; of these, only

-2 is rational.

Explanation:

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!