This week I learned a new method of factoring trinomials. It is the box method. You would really only need this when a doesn’t equal one in a    equation.

Example:

For this method you make a for sided box with the highest term in the top left and the lowest one on the bottom right.  You multiply the two to get the number that two numbers have a product a sum of. (The middle term bx)

Then you find two numbers that fit.

Then you go and find what the terms have in common, if they dont have anything in common you put one. And to find something that thet do have in common you find the gcf.

Then you simply right out the factored equation.

Some challenges are; when the positives and the negatives are implemented to the equation, it can be confusing. Another challenge is finding the correct numbers that fit.

Factoring trinomials in this form : 

To do this by inspection you need to find two integers which have a product of c and a sum equal to b.

Example: 

You need to identify what two numbers have a product of 20 and has a sum of 21. In many this case 20 has many possibilities for the sum and a product,, this can be a challenge. For example 4 & 5 equals to 20 as a product, however their sum is only 9 so that wouldn’t work. What would work though would be 1,20, because they make a product of 20 and have a sum of 21. Then what you need to do is factor and write out the equation

 = 

You know that this is correct when you foil the term and get the same equation.

I earned to factor a polynomial by removing the greatest common factor.

In this equation the common factor is .
  Since  is a factor of both the terms you divide it to get it’s simplest form.

You know you have done it correctly if you can go the other way around and get the same equation.

I learned how to use FOIL :

F-First term in each bracket

O- Outside terms

I- Inside terms

L- Last term in each bracket

You apply it when you multiply two binomials. Example:

 First you multiply the first terms of each bracket. (x()x) 

Then you multiply the outside term with the first term in the first bracket. (x)(2) 

Then you multiply the inside term from the first bracket with first term in the second bracket. (2)(x) 

Then you multiply the last term in each bracket. (2)(2)=4

The last step is to add like terns and to simplify.

=

Some challenges are getting the order confused.

This week I learned how to label the sides of a triangle and to find the angles using trigonometry. Using SOH CAH TOA you can find these all without using Pythagorean theorem. A challenge can be correctly labeling the Adj., Opp., and the Hyp. Another one is getting the wrong formula SOH CAH TOA. And another challenge is not doing the right calculations on a calculator. This is applied when you are finding the sides or angles of a triangle.

For this example I’ll show how to find X(Hyp.) and angle A and angle B.

For this you really need to label it correctly on what angle you want to find. First Angle A.

Then Angle B..

With these angles you can now find X(Hyp). As long as don’t use Tan. again you will get the same answer. Also as long as you use the formula correctly.

This week I learn how to convert with this system using a strategy I never used before. A challenge with this strategy is remebering whether or not to have a negative exponent and remebering what goes in the denominator and what goes in the numerator. This is obviously use when converting between the system as well as between the imperial system.

How to apply between SI:

I learned to to solve equations that have a negative exponent.

Expressions such as this:

You apply these sort of expressions when you are solving various equations. This is how you apply it. 

Since it is an exponent that is negative you put it over one because its less then one. And to solve this:    you just have to solve what the exponent is, so 

Some challenges are if the coefficient is a fraction such as: 
 An easy way to get around this is simply switcing the numbers around like this : because it means the same thing.