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

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