Examples:


1) Factor x3 + 1

SOLUTION: 
A factor of the form (x - k) exists only if k divides (+1) exactly.  Therefore,
k must equal +1, or -1.  This means that (x - 1) and/or (x + 1) are factors.
See if (x - 1) is a factor by long division (remember that if it is, the remainder
must equal zero):
**Note: when using long division, you must include x terms whose coefficients
are zero!!
            x2 + x + 1
           -----------------
   (x - 1)| x3 + 0x2 + 0x + 1
            x3 - x2
            -------
                 x2 + 0x
                 x2 -  x
                 -------
                       x + 1
                       x - 1
                       -----
                         2

The remainder is NOT zero and thus (x - 1) is NOT a factor.  This means that 
(x + 1) probably is (if there is any factor of the form (x - k)) but we will 
use long division to see if it actually is.


            x2 - x + 1
           -----------------
   (x + 1)| x3 + 0x2 + 0x + 1
            x3 + x2
            -------
                 -x2 + 0x
                 -x2 -  x
                 -------
                       x + 1
                       x + 1
                       -----
                         0

The remainder is zero, thus (x + 1) is a factor and (x2 - x + 1)
is the other factor.  That is:
         x3 + 1 = (x + 1)((x2 - x + 1)



2) Factor 2x3 - x2 - 13x + 5

SOLUTION:
Since the coefficient of the leading term (highest order term) is 2, we need to 
find a factor of the form (hx - k) such that h divides exactly into 2, and k 
divides exactly into 5.
Factors of 5 are: 1, -1, 5, and -5  (possible k values)
Factors of 2 are: 1, -1, 2, and -2  (possible h values)
Thus, possible factors of the original polynomial needing to be factored are:
  if h = 1: (x - 1), (x + 1), (x - 5), (x + 5)
  if h = -1: (-x -1), (-x + 1), (-x - 5), (-x + 5)
  if h = 2: (2x - 1), (2x + 1), (2x - 5), (2x + 5)
  if h = -2  (-2x -1), (-2x + 1), (-2x - 5), (-2x + 5)


Note that: (x - 1) = -(-x + 1), and (2x - 1) = -(-2x + 1).  Continue on with 
factoring out a negative from every possiblity in the 1st and 3rd lines above
and see that you get the factors listed in rows 2 and 4 above.  This means that
every factor above appears twice (because the negative of a factor is also a 
factor), and so you only have to test half of these to see which of them are 
factors.  Thus, we need to check:
(x - 1), (x + 1), (x - 5), (x + 5), (2x - 1), (2x + 1), (2x - 5), (2x + 5)

**The best way to get only those possible factors that you want to check is
to always only use the positive values for h but use both positive and negative
values of k. 

It is a lot of work but if you perform long division on all these factors, you
will see that all of them result in remainders not equal to zero, except 
(2x + 5).

              x2 - 3x + 1
             ---------------------
    (2x + 5)| 2x3 - x2 - 13x + 5
              2x3 + 5x2
              --------
                  -6x2 - 13x
                  -6x2 - 15x
                  -----------
                       2x + 5
                       2x + 5
                       ------
                         0

Thus, the factors of  2x3 - x2 - 13x + 5 are (2x + 5) and (x2 - 3x + 1)
Return to the tutorial