| \(\pm\) | \(1\) | \(2\) | \(3\) | \(6\) |
| \(1\) | \(1\) | \(2\) | \(3\) | \(6\) |
| \(2\) | \(\frac{1}{2}\) | \(1\) | \(\frac{3}{2}\) | \(3\) |
The possible rational zeros are \(\pm1, \pm2, \pm3, \pm6, \pm \frac{1}{2}\) and \(\pm \frac{3}{2}\).
\(f ( 1 ) = - 6, f ( -1 ) = 12, f ( 2 ) = - 12, f ( - 2 ) = 0\) so \(( x + 2 )\) is a factor.
\(f ( 3 ) = 0\) so \(( x - 3 )\) is a factor.
\(f ( 6 ) = 264, f ( - 6 ) = 468, f ( \frac{1}{2} ) = 0\), so \(( x - \frac{1}{2} )\) is a factor.
A cubic polynomial can have only three first degree factors, so we have found all first degree factors. However, when we multiply these three factors together, we get
\(x^3 - ( \frac{3}{2} ) x^2 - ( \frac{11}{2} ) x + 3\) which is only one-half of \(f ( x )\). Thus, \(f ( x ) = 2 ( x + 2 ) ( x - 3 ) ( x - \frac{1}{2} )\)
We can multiply the \(2\) by the \(( x - \frac{1}{2} )\) to get \(f ( x ) = ( x + 2 ) ( x - 3 ) ( 2 x - 1 )\)
We might have found the zeros of \(- 2\) and \(3\) by trial and error, but not the zero \(\frac{1}{2}\). If it were not for the rational zero theorem, we would not have known to try \(\frac{1}{2}\).