Find the real zeros of f (x)= 2x^3 + 3x^2 – 8x + 3
try x = ±1, ± 2, ± 3
First try I found x = 1 to work
so x-1 is a factor.
Using either long algebraic division or synthetic division we get
2x^3 + 3x^2 – 8x + 3 = (x-1)(x+3)2x-1) = 0
x = 1, -3 , 1/2