Hello, my name is Helen Kolman and I am a

mathematics instructor at Central Piedmont Community College. In this video, we will discuss the AccuPlacer

review, specifically for elementary algebra. The area we are considering is algebraic expressions. The topic we will focus on is division of

polynomials. In this example, we will divide the polynomial

6 X5 plus 2 X4 minus 8 X3 divided by 2 X4. In this problem, we are going to approach

the problem by realizing that we can use the property of fractions to rewrite the fraction

in a more convenient format, so let’s move to that. So, we’re going to take the first term in

the numerator and write it over the common denominator 6X5 over 2X4, then we take a second

term in the numerator write it over the common denominator 2X4 over 2X4, and then we take

the third term in the numerator minus 8X3 and write it over the common denominator 2X4. If you look at the problem in this format,

I believe you will recognize that we have three fractions a common denominator, so the

three numerators were originally expressed together and now we’re breaking it down

in a less common process. The next problem will be to simplify each

of these terms. And simplifying each of those terms we will

need to remember the rule for exponents. That is, we are going to have A in the numerator

raised to the N, value of A in the denominator raised to the M, which will give us A to the

N minus M as a result. Let’s go forward applying that rule in our

simplified process. In the first fraction we have a 6 over 2 in

the numerical piece that gives us 6 divided by 2 or 3, we have X5 over X4 going back to

our rule of exponents. We have X5 minus 4 or X to the 1, so our first

term is now a 3X, our second term is very nice because our second term is a common value

in the numerator and denominator. Two divided by two is one, X4 divided by X4

is one. So our second term is simply a one. Moving on to the third term, we have 8 divided

by 2 which is 4. However, the power of X in the numerator is

smaller than the power of X in the denominator. If I do the rule, I’ll end up with X3 minus

4 or X to the minus 1. So I have an X then in the denominator or

I could simply say since the larger power is in the denominator, I’m going to use

the rule. However I’m going to end up with X4 minus

3 or X1 in the denominator.