12x³ - 32x² + 25x - 6 = 0
It has three sign changes going from left to right,
so it has 3 or 1 positive solutions.
f(-x) = 12(-x)³ - 32(-x)² + 25(-x) - 6 = -12x³ - 32x² - 25x - 6
has no sign changes so there are no negative solutions.
Any rational solutions it may have must have a numerator which
is a factor of 6 and a denominator which is a factor of 12.
Possible numerators: 1,2,3,6
Possible denominators: 1,2,3,4,6,12
Possible roots: 1/1, 1/2, 1/3, 1/4, 1/6, 1/12, 2/1, 2/2, 2/3, 2/6,
2/12, 3/1, 3/2, 3/3, 3/4, 3/6, 3/12, 6/1, 6/2, 6/3, 6/4, 6/6, 6/12
Reducing the fractions: 1, 1/2, 1/3, 1/4, 1/6, 1/12, 2, 1, 2/3, 1/3,
1/6, 3, 3/2, 1, 3/4, 1/2, 1/4, 6, 3, 2, 3/2, 1, 1/2
Eliminating the duplications: 1, 1/2, 1/3, 1/4, 1/6, 1/12, 2, 2/3,
3, 3/2, 3/4, 1/4, 6
Try 1 as a solution by dividing synthetically
by x-1
1 | 12 -32 25 -6
| 12 -20 5
12 -20 5 -1
No it has 1 as a remainder, not 0, so x-1 is not
a factor and therefore x=1 is not a solution
Try 
as a zero by dividing synthetically
by x-
| 12 -32 25 -6
| 6 -13 6
12 -26 12 0
This has a remainder of 0 so x- is a factor and
therefore x= is a solution solution. The
quotient is 12x²-26+12. That means that we have now
factored the left side of
12x³ - 32x² + 25x - 6 = 0
as
(x - )(12x² - 26x - 6) = 0
Now we can factor 2 out of the the second parentheses:
(x - )2(6x² - 13x - 3) = 0
If you like you can multiply the 2 into the first
parentheses:
(2x - 1)(6x² - 13x - 3) = 0
Now factor the second parentheses:
(2x - 1)(3x - 2)(2x - 3) = 0
Setting the first factor = 0, 2x - 1 = 0, gives solution x =
Setting the second factor = 0, 3x - 2 = 0, gives solution x = 
Setting the third factor = 0, 2x - 3 = 0, gives solution x = 
Edwin
