ACCUPLACER – Elementary Algebra

ACCUPLACER – Elementary Algebra


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.

Leave a Reply

Your email address will not be published. Required fields are marked *