Polynomial Division: Long and Synthetic

Rather than explaining in lots of detail how to divide polynomials either using the long method or the synthetic method, I am sending my students to the following site. This PowerPoint is very brief, but it is very informative and shows every step of each process. Once the presentation is open, just press F5 to begin the slide show. A space bar or enter will move the slide show forward. The backspace key will back up if you need to see something again.

Simplifying Radical Expressions

My Algebra I students are having lots of problems with simplifying radical expressions. It seems that they have run across a concept that eludes them. I am posting about this topic to provide several more examples.

√18

First, we find the prime factorization of 18. If you need help with this concept, go here.

√18 = √(2•3•3)

Since the 3 is repeated twice, we can pull it out from under the square root symbol. Therefore,

√18 = 3√2

Now my students would be asking what happens if you have variables?

√24x³y²

√24x³y² = √(2•2•2•3•x•x•x•y•y)

Pull out everything that is repeated twice - 2, x, y. Leave everything else under the radical symbol.

2xy√(6x)

Some other places to visit for more details on this concept are here and here and here.

If you are one of my students and would like some extra credit, please visit those sites and comment here on whether or not they were helpful.

Solving Quadratic Equations

Here are a few sites that may help you when solving quadratic equations:

• by factoring
• by finding square roots
• by completing the square
• by Quadratic Formula
• by graphing
• extra examples

Solving Systems of Equations Using Elimination

I have found several websites that might help you with this concept.

Purple Math has a great site that shows examples here.

Check out the Unusual Outcomes in this PDF file. The other information there is also very good. The explanations are detailed.

Use this PDF file to help you determine when to add and when to multiply and then add.

Here are some problems that you can try to determine if you are getting it. Their answers are provided as well.

Polynomial Operations and Factoring

Here are some websites that might help clarify polynomial operations and factoring.

Polynomial addition and subtraction - http://math.about.com/library/blpoly.htm

Adding polynomials - http://www.regentsprep.org/rEGENTS/math/polyadd/sp_add.htm

Subtracting polynomials - http://www.regentsprep.org/rEGENTS/math/polyadd/sp_subt.htm

Multiplying polynomials - http://www.purplemath.com/modules/polymult.htm and http://www.jamesbrennan.org/algebra/polynomials/multiplication_of_polynomials.htm

Factoring polynomials - http://www.jamesbrennan.org/algebra/polynomials/factoring_polynomials.htm and http://www.purplemath.com/modules/simpfact.htm

In order for my students to earn extra credit, they must visit at least 2 of the sites and comment on both.

Pythagorean Theorem

The Pythagorean Theorem is used to find a missing side length of a right triangle. It only works for right triangles (one with a right angle).

The formula to use is a² + b² = c². A really good explanation of the formula and its derivation can be found here.

In order to use the theorem, you must know the parts of the triangle. Parts a and b of the theorem are the legs of the triangle. These are the side lengths that make up the right angle. The third side (longest side) is noted as c in the theorem.

Now substitute the values given into the theorem (formula) and solve the equation.

Example: A triangle has a side length of 6 cm and a side length of 8 cm. What is the length of the hypotenuse?

6² + 8² = c²
36 + 64 = c²
100 = c²
√100 = √c²
10 = c

Example: A triangle has a side length of 6 cm and a hypotenuse length of 12 cm. What is the length of the missing side?

a² + 6² = 12²
a² + 36 = 144
a² + 36 – 36 = 144 – 36
a² = 108
√a² = √108
a = √108
a = 6√3

Picture of triangle above from http://en.wikibooks.org/wiki/File:Right_triangle_shows_hyp_legs.PNG

Rate-Time-Distance Word Problems Part 2

Now for the second part of the rate-time-distance problems. This example refers to opposite direction travel.

Let’s look at the following example:

Train 1 leaves the train station travelling south at a rate of speed that is 10 mph more than train 2 which is travelling north. After 3 hours, they are 450 miles apart. Find the rate of train 2.

We are going to use another table to organize the rate, time and distance information.

Train	Rate	Time	Distance
1	r +10	3	3 (r + 10)
2	r	3	3r

Since the trains are moving in opposite directions, we are going to add their distances together to get 450 miles.

3 (r + 10) + 3r = 450
3r + 30 + 3r = 450
6r + 30 = 450
6r = 420
r = 70

Therefore the rate of train 2 is 70 mph.