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)
```
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