Friday, January 16, 2009

Thursday, January 15, 2009

Today's Slides: January 15

Here they are ...



Remainder Theorem and The Rational Roots Theorem

Today in class, we went over one more time how to do synthetic division and how to find the roots of a function using the Remainder Theorem.

After that, we learned how to find the missing coefficient in a polynomial if we already know what the remainder is. Here's how you would do this:

Let's say you have the question:


The first thing you want to do is write it out in a way that is easy to understand:


Then you plug the root of the denominator into the function:


Since we already know the remainder we can rewrite it this way:


Now all we do is isolate K with a little algebra, and solve it:


And there you have it!

Near the end of the class we managed to quickly learn the Rational Roots Theorem. This theorem allows us to find any rational roots of a polynomial function. Here's an example:

So you're given the equation:


The first thing to do is the find all the possible positive and negative factors of the constant term:


Now we find all the positive factors of the leading coefficient:


We then list all the possible rational roots, eliminating any duplicates:


We can then test out these roots by using synthetic division and the factor theorem to turn the function into a quadratic (Remember: If the remainder is 0, then it is a root):


This then gives you:


Now you just factor the equation and find the roots:


And there you have it! That's about all we did for today, tomorrow's scribe will be...Niwatori-san

Wednesday, January 14, 2009

Today's Slides: January 14

Here they are ...



Yesterday we did more examples on synthetic division but we learn't something new.

Here is an example:


From the above we can observe that the remainder is the same with the rational root .

Another example was about finding out if a fraction is a factor or not.

If a fraction is a factor, it's remainder and rational root must be eqaul to o


Here's an example: