A process used to find the roots of a

polynomial, usually used with orders of 3 or more since

factoring is usually used with orders of 2. Also known as

synthetic division.

Given an equation, you can determine possible candidates for rational roots by taking the factors of the 0th power coefficient and dividing them by all possible factors of the nth power coefficient (where n is the order of the equation), including both positives and negatives. For example:

x^{3} - 7x^{2} + 6 = 0

possible roots are: x={-6, -3, -2, -1, 1, 2, 3, 6}

2x^{3} - 2x^{2} + 12x + 12 = 0

possible roots are: x={-12, -6, -4, -3, -2, -3/2, -1, -1/2, 1/2, 1, 3/2, 2, 3, 4, 6, 12}

8x^{3} + 40x^{2} + 60x + 5 = 0

possible roots are: x={-5, -5/2, -5/4, -1, -5/8, -1/2, -1/4, -1/8, 1/8, 1/4, 1/2, 5/8, 1, 5/4, 5/2, 5}

So, it's pretty obvious that there can easily be many many roots to try.

The first step is to set the equation up on a grid. We'll use the first example equation for simplicity. Have the top row of the grid be the coefficients (making sure to include a 0 in the event of an equation "missing" a term).
The left column of the grid will hold the numbers that we want to test as roots.

x^{3} - 7x^{2} + 6 = 0

possible roots are: x={-6, -3, -2, -1, 1, 2, 3, 6}

___|____1____-7____0____6
-6 |
-3 |
-2 |
-1 |
1 |
2 |
3 |
6 |

Next, begin with one of the roots (the more problems you do, the more you'll recognize which numbers you should try first based on the coefficients). In this case, I'll begin with -6. The first step is to drop the first coefficient into the first block of the grid.

___|____1____-7____0____6
-6 | 1
-3 |
-2 |
-1 |
1 |
2 |
3 |
6 |

Now, multiply the number in the grid by the test number (1 * -6), and add it to the next coefficient, dropping the result to the next place in the grid.

___|____1____-7____0____6
-6 | 1 -13
-3 |
-2 |
-1 |
1 |
2 |
3 |
6 |

Repeat for each column until the end.

___|____1____-7____0____6
-6 | 1 -13 78 -462
-3 |
-2 |
-1 |
1 |
2 |
3 |
6 |

These results tell us two things. 1) -6 is not a root, because if it were, the last number would be a 0. 2) -6 is a lower boundary, because the sign of the results alternate. This means that no number less than -6 can be a root (which we already knew- but the concept is important).

___|____1____-7____0____6
-6 | 1 -13 78 -462
-3 |
-2 |
-1 | 1 -8 8 -2
1 |
2 |
3 |
6 |

So -1 is also a lower boundary. We've now eliminated -2 and -3 because of that. In this problem there is no upper boundary, unfortunately. An upper boundary occurs when all of the numbers are positive, but in this case, the highest we can try is 6, and the second column will be -1.

Now to examine what happens when a root is found.

___|____1____-7____0____6
1 | 1 -6 -6 0

This means that 1 is a root of the equation, and that we can now factor (x - 1) out of (x^{3} - 7x^{2} + 6). The result will be that we've now changed the equation into

(x - 1)(x^{2} - 6x - 6) = 0.

The second factor can be obtained by lowering the order of the equation by 1, and taking the numbers obtained through synthetic substitution as the coefficients for the terms. The resulting second factor is called the depressed polynomial.

Long story short- now that we know the second factor, we see that it's unfactorable, so now you'd have to use the quadratic formula... but I'd prefer not to go there. If something in this seems unclear, feel free to /msg me and I'll see if I can fix it.

It should also be noted that all of this is just a fancy method of advanced trial and error. Essentially, what you are doing in this is plugging numbers into the equation and seeing if they equal 0. It does, however, have advantages in A) determining which numbers are possible roots, B) eliminating possible roots based on behavior of other possible roots, and C) factoring the polynomial once a root is found.