Give all the possible rational zeros of f (x) = 2x^3-x^2+4x-2

Group and undistribute

f (x) = (2x^3-x^2) + (4x-2)

f (x) = (x^2) (2x-1) + (2) (2x-1)

the common is (2x-1)

reverse distribute like this: ba+ca=a (b+c)

so

f (x) = (2x-1) (x^2+2)

set to zero

2x-1=0

2x=1

x=1/2

x^2+2=0

x^2=-2

no rational number works (you need complex numbers)

x=1/2