On Thursday, we learned how to divide factors. First, we did long division, and then we did synthetic division.

Long division:
For both methods of division, we used the example of:
divided by x + 3

First, since there is no term to the third degree, you have to put in a placeholder for that term.

After you have put placeholders in for any missing degrees, you can then start solving the problem. The first step in solving is to take the divisor and see how many times it can go into the first term of the quotient.

Since it goes in x^3 times, you would multiply (x+3) by x^3, giving you an answer of (x^4 - 3x^3). You then subtract that answer from the first two terms in the quotient (x^4+0x^3). The fourth degree terms will cancel out, leaving an answer of -3x^3. You then drop down the -10x^2 and repeat the process until all terms are used.

When you are done, your answer should be:

In the last term of the answer, the 1 is the remainder. Instead of using the remainder like we've always learned (as in stating the answer then saying "remainder 1" at the end,) you must take the remainder and divide it by the divisor. The remainder will always go over the divisor.

Synthetic division is a shorter process that only works when the x term in the divisor is linear and has a coefficient of 1. Using the same equation as above, here is an example of synthetic division:

To start, you must take the coefficients of all the terms in the quotient (including placeholder terms) and line them up.

1  0  -10  -2  4

After doing so, drop down the first term- in this case, 1.

1  0  -10  -2  4

1

Next, multiply 1 by the divisor. Since the divisor is (x+3), you would multiply 1 by -3. This is because x = -3 when x + 3 = 0. Since 1 * -3 is -3, you would add that to the second term, or in this case 0.Your result would be -3, and you would drop that number below the 0.

1  0  -10  -2  4

1  -3

Next, you multiply the -3 under 0 by the -3 in the divisor to give you 9. You add 9 to the next term, -10, and you get -1. You place -1 under the negative term, and repeat this process until all terms have run out. Your result should be:

Again, the 1 is the remainder, which goes over the divisor, (x+3).

Another thing to remember is the rational root theorem.

f(x) = px^n + ... +q

The only possible rational roots are:
factors of p / factors of q

Therefore, if p = 3 and q = 7, then the only possible rational roots would be 1 or 3 divided by 1 or 7.