Adding and Subtracting Rational Expressions

 Today in class, we focused on adding and subtracting rational expressions. Here are a few rules that apply to solving these expressions:
-1) you must find a common denominator before adding or subtracting
-2) when trying to find a common denominator, you must do the same to the top as you did to the bottom.  
Applying these rules: here are some examples :
1/2 + 1/5: Common denominator is lowest number that terms in the denominator both go in to (in this case, the common denominator is 10). If the common denominator is 10, then in the first term, the 2 in the denominator of 1/2 must be multiplied by 5 to reach ten. Then, rule number 2 must be applied, and the top must also be multiplied by 5, making the fraction equal 5/10. With the second term, the 5 in the denominator must be multiplied by 2 to reach ten, meaning that the top must also be multiplied by 2, making the new fraction 2/10. Now, we are able to do the equation 5/10 + 2/10= 7/10

Sometimes, a common denominator cannot be found, so the terms in the denominators must be multiplied together to form the common denominator.

1/x - 1/x+4 = x+4-x/x(x+4) = 4/x(x+4)

Here, x and (x+4) did not have a common denominator, so they were multiplied together to form one.


There are also cases where the denominator of one fraction must be multiplied by a number to make it equal the other denominator.

x/x-1 + 1/1-x = x/x-1 + 1/-1(-1+x) = x/x-1 + -1/x-1 = x-1/x-1 = 1

Here, the denominator of the second fraction was multiplied by -1 so that it equaled the denominator of the first fraction.

Many examples and practice problems can be found on page 227

Here are some examples from page 227:

13. 1/2xy^4 + 1/x^3y^2 = x^2+2y^2/2x^3y^4

19. 1/4x^2 - 1/xy + 1/y^2 = y^2 - 4xy + 4x^2 / 4x^2y^2

Here is a video link that may help you better understand this: http://www.youtube.com/watch?v=FZdt73khrxA

Simplifying Rational Expressions

Hello class,

-For the past two days we have been learning and gathering information on Simplifying Rational Expressions. Simplifying Rational Expressions is basically finding the easiest form of the equation by simplifying it. And Rational expressions are usual in their simplest form, if it's numerator and denominator have no common polynomial factors. We've also been going over how to find the domain and zeros of certain rational functions.



First of all:

What is a rational #?
A rational number would be numbers such as .4=4/10=2/5. It can be written as fractions w/ integers in the numerator & denominator.

It would not be a number for example like: √5/2


Rational Expressions: ---> polynomial/polynomial or polynomial over polynomial


Ex:

x^3-4x/x^3-4x^2+4x = x(x^2-4)/x(x^2-4x+4) = x(x+2)(x-2)/x(x-2)(x-2) = x+2/x-2

Basically what you do is you simplify to the lowest possible terms possible. Then once you are done simplifying you cancel out the same values (that's what the crossed out black values stand for). And then you end with your final answer which will be in the simplest form.



Domain: Is the set of all x values that work in the expression.

Ex:

x^3-4x/x^3-4x^2+4x = x(x^2-4)/x(x^2-4x+4) = x(x+2)(x-2)/x(x-2)(x-2) = x+2/x-2

Domain: All Real #'s - {2}

Which basically means All numbers work in the place of x to make the equation set equal to 0 except for 2.




Zeros: All x value that makes the whole expression 0.

Equation:

x^3-4x/x^3-4x^2+4x = x(x^2-4)/x(x^2-4x+4) = x(x+2)(x-2)/x(x-2)(x-2) = x+2/x-2

x=-2--> zero  -2+2/-2-2 = 0/-4 = 0

Basically what you're trying to find is the number you can substitute the x with to make the whole fraction equivalent to 0.



Now here are some example problems:

Equation: (x-1)(x+1)^2/(x+1)(x^2-1)


Solving:

(x-1)(x+1)(x+1)/(x+1)(x-1)(x+1) = 1

Domain: All Real #'s

So basically what happened within this equation when you simplified was that you had all of the same common value as denominators and numerators. So therefore in result they canceled each other out and you end up with a result of 1.



Equation: 3x^2/6x^2-9x


Solving:

x(3x)/3x(2x-3) = x(3x)/3x(2x-3) = x/2x-3

What happened with this problem was it was simplified to simplest form then once there we canceled out the 3x from both of them leaving us with an answer of x/2x-3.

Tips:

• Be Careful with your simplifying (formulas on page 172 and 173)
• Make sure to cancel out the same integers
• Don't always suspect certain equations are simplified completely if it seems a bit harder then most problems

-http://www.youtube.com/watch?v=B4bVlDgHF5I

-http://www.youtube.com/watch?v=7Uos1ED3KHI

-If you didn't understand this concept before I hope this helped with the gist of it! For more explanation on this go to: Pages 218-219 in your book.

Dividing Monomials and Negative Exponents

Dividing Monomials:                        More examples on page 205 of our textbook

Ex. 15x3/5x = 3x2

To divide these two monomials, first divide the coefficients just like usual. 15/5 = 3. For the exponents with the same base, you subtract the exponents when dividing so 3 (from x3) minus 1 (from x1) equals 2 so your variable is x2. This gives you an answer of 3x2.

Ex. 24a3c6/16a4c3 = 3c3/2a

First divide the coefficients, they both divide by 8 and so we are left with 3 on the top and 2 on the bottom. There is an a3 on the top which cancels out with the a4 on the bottom but we are still left with one a on the bottom. There is a c6 on the top which cancels out the c3 on the bottom and we are left with c3 on the top. This gives us our final answer of 3c3/2a.

Video:
http://www.youtube.com/watch?v=Mn4WuvIGUgI

Negative Exponents: More examples on page 210 of our textbook
x2/x5 = x2-5 =x-3

If we follow our exponent rules and subtract the exponents when we divide then the answer we get is x-3

x2/x5 = (x)(x)/(x)(x)(x)(x)(x) = 1/x3

This is the same problem but solved slightly differently and with a different answer. From this we can assume that 

x-3 is equal to 1/x3

From this we can see that if you bring a negative exponent from the denominator to the numerator, and vice versa, that it then becomes a positive exponent.

Ex.

3a2x-2/axy-1 = 3a2y/ax3 = 3ay/x3

By bringing the bases with negative exponents from the numerator to the denominator, or from the denominator to the numerator we are left with all positive exponents. We moved the x-2 from the numerator to the denominator so it became 

x2

and we brought the y-1 from the denominator to the numerator making it just y. From here all we needed to do was cancel the a's and we got 3ay/x3 as our answer.

Video:

http://www.youtube.com/watch?v=c4aiYf3fzVQ

Tips:

• When Dividing subtract exponents
• When Multiplying add exponents
• When raising a monomial to a power multiply exponents
• When raising monomial don't forget to raise the coefficient

Test Review

We have been covering different kinds of factoring, finding the LCM, finding the GCF, word problems and setting equal to zero in class, here is a quick review for our test hope it helps. 

Factoring by grouping:

ax - bx + ay - by    =      x(a-b) + y(a-b)   =   (x+y)(a-b)

We use this when there is four terms. You group with two or sometimes even three terms and try to get a common factor which in this problem was (a-b). 

Difference of Two Squares:

x2 - 4 = (a+2)(a-2)

The formula for this is a2-b2 = (a+b)(a-b).

Trinomial Squares:

x2 + 2xy + y2 = (x+y)2

The formula for this is a2 + 2ab + b2 = (a+b)2 or a2 - 2ab + b2 = (a-b)2 depending if the middle term is positive or negative.

Sum and Difference of Cubes

z3 + y6 =  (z+y2)(z2- zy+b2) Sum of Cubes

z3 - y6 =  (z-y2)(z2+zy+b2) Difference of Cubes

The formula for the sum of cubes is a3+b3 = (a+b)(a2-ab+b2)

While the formula for the difference of cubes is a3-b3 = (a-b)(a2+ab+b2)

Tips for factoring:

• Always look for GCF (greatest common factor before starting)
• When you have finished make sure your answer is completely factored

Finding the LCM and GCF

To find the GCF make a factor tree for each of the numbers and then create a ven diagram. Put all the factors they all have in common in the middle. Multiply all the numbers in the middle and this is your GCF coefficient. Next look at the variables and take out the variables they all have in common. In this case they all only have a Y in common. To find the LCM you must multiply all of the numbers in the ven diagram, but make sure not to multiply the numbers in the middle more than once. This number is your coefficient. Now look at the variables and take the all the variables from all of the original numbers and if some of the numbers have the same variable, pick the one with the highest exponent.




Word problems/Setting Equal to zero
Find two numbers differing by 3 and whose product is 88

x2+3x=88

x2+3x-88=0

(x+11)(x-8)=0

x+11=0     x-8=0

x=-11        x=8

x=-11, 8

First factor the equation and then set both factored sets equal to zero and then solve.

Zero-Product Property and Consecutive Numbers

Hi Class!

-On wednesday we went over polynomial equations and the zero-product property. A polynomial is an equation that is equivalent to an equation with a polynomial as one side of the = sign, and 0 on the other side. Examples of this type of equation are x^2 + 4x + 4 = 0, and x^4 + x= = 0. To solve these types of equations, you must factor the polynomial into linear factors. If you cannot factor the polynomial, then it is prime. 


-Explanation of the zero-product property in book: p183-185


-Visual example: △ x □  = 0

          ~Either △ = 0, □ = 0, or □ and △ = 0
          ~This can also be shown as: ab = 0
                                   where a = 0 and or b = 0
                 
           *There can be more than one value for X

-The steps to solving polynomial equations with the zero-product property are:
          
          1. Write the equation with one side as zero (set the equation equal to zero)
          2. Factor the non-zero side of the equation
          3. Set both of the factors equal to zero and solve

-Example problem:

          1. One side as zero:  x^2 - 49 = 0
          2. Factor:  (x+2) (x-2) = 0
          3. Set factors equal to zero:  x+2 = 0, x-2 = 0
                                                       x = 2, -2 ---> {2, -2}

-Alternate ways to solve: In class we also learned another way to solve certain problems:

          ~9x^2 - 81 = 0: Instead of factoring, you could also solve for x normally.
            9x^2 = 81
            x^2 = 9
            x = 3, -3

                                *Do not forget that the square root can be positive AND negative

          ~Here is the same problem (^), but using the zero-product property:
            9x^2 - 81 = 0
            (3x-9) (3x+9) = 0
            3x-9 = 0, 3x+9 = 0
            3x = 9, 3x = -9
            x = 3, x= -3
            {3, -3}


-Another example of zero-product property:

          ~x^2 + 10x + 25 = 0
            (x-5) (x-5) = 0 
            x = 5  
            ^When the factor occurs twice (same factor more than once) it's called a 'Double Root'

- Consecutive Numbers: In class we also learned how to express consecutive integers, or consecutive even integers and consecutive odd integers.

          ~For example: 2 consecutive odd integers numbers that multiply to 99
                                    x (x+2) = 99
                                           *x represents the first integer and x+2 represents the second integer
                                   x^2 + 2x = 99
                                   x^2 + 2x - 99 = 0
                                  (x+11) (x-9) = 0
                                   x = -11, x = 9 {-11, 9)
                                            *-11 and 9 are the two possibilities for x NOT the answers to the problem.
                                             To find the answer  ---> plug -11 and 9 into x+2 (add two)

         ~Another example: 2 consecutive even integers  ---> the difference of who's squares is 68
                                         x^2 - (x+2)^2 = 68
                                         (x+2)^2 -x^2 - 68 = 0
                                         x^2 + 4x + 4 - x^2  - 68 = 0
                                         4x - 64 = 0
                                         4x = 64
                                          x = 16     *Now plug into original equation
                                                           (16 + 2)^2 - 16^2 = 68
                                                            {16,18} {-16,-18} 



- http://www.youtube.com/watch?v=1Iay8rFBQ6o


                                                           Hope this helped!
                                           Problems are on pages 185-191 in the book!

Quadratic Factoring and More

Through these classes, there has been a collection of vocab words:
Factoring Quadratics: finds the roots  of a quadratic equation. Factoring quadratic equations in standard form, , can often be accomplished by finding two numbers that add to give b, and multiply to give ac
Factoring by Grouping: means that you will group terms with common factors before factoring as well as common factors inside of a parenthesis. 

Over the past few classes, D block covered and reviewed a couple key topics. First we looked another way or secret to factor an expression. By adding and subtracting a specific factor (which is only adding zero to the expression) you can absorb one of the added factors and continue to solve and factor.

Example #1:

X^4+X^2+1 hint: add and subtract X^2

X^4+X^2+1+X^2-X^2

Now you have to absorb one of the factors that you just added to the expression.

X^4+2X^2+1-X^2

(X^2+1)^2-X^2

Now this is just difference of two squares

A=X^2+1

B=X

=(X^2+1+X)(X^2+1-X)

Example #2:

X^4+4 hint: add and subtract 4X^2

X^4+4+4X^2-4X^2

A=X^2

B=4

(X^2+4)^2-4X^2

(X^2+4-2X)(X^2+4+2X)

Another topic was introduced to the class called Factoring Quadratics.

Here is a video on how to factor a quadratic a basic but helpful way

Click Here to watch video 

There are a couple of ways that you can speed the process up and the way that was taught in class is in a table form. But first here are three example that I think give a good range as to how to factor quadratics with using the table method to help.


Example #1: This is a simple and great star as were get into factoring quadratics
X^2+8X+12
In this case a= (X^2), b=(8X), and c=(12) but we are only going to use the actual consistence of each term listed and bring the variables back later.
These terms finally become a=1, b=8, and c=12
Now is how you use those terms:
Make a table like the one below, substitute the letters (a,b and c) for the terms that equal them, and multiply as shown.
a*c=12
b=8

As you can see, for the first column, I looked for all the factors of 12 and saw if any could be added up to 8. 
So now here comes the math part :)
(X+6)(X+2)
And if you are unsure about your solution, you can FOIL out the problem

Example #2: This is a great one that gets a little more challenging from switching up the negitives and positives, but also just makes sure that you are doing the steps correctly
X^2-12X+20

Now, if you look at the chart that I made, you can look for the once that match up. And in this case the factors will be 2*10. But, as you can see there can be different answers when adding or subtracting these numbers. If you look at the original equation, you can see that (+20) and that means that both of the signs will be the same so the numbers will be -2*-10
(X-2)(X-10)

Example #3: As you can see, there is a coefficient before the first term, this does complicate things just a bit but also can give you reassurance on what you need to work on and what you are good at with these steps.
12X^2+X-1
 This is exactly the same procedure as the previous two Examples. So let's first check out the signs. (-1) this means that the signs will be different. 

So using these numbers we can figure out the factoring (3x-1)(4X+1)

You can also do something called factoring by grouping.

I will continue to use this specific example.

So from the step of using the chart. We will be using -3,4 as because that will give us 1. 

=12X^2-3X+4X-1

now we are going to use factoring by grouping

3X(4X-1)+1(4X-1)

=(4X-1)(3X+1)

This, comes out to the same answer but if you feel more comfortable using this method so you can see what is going on and why.

Some of these, you will need to be guessing and checking the factors that you think would work. But using this method, you will always get your answer.
Here is a version of the examples but may take a bit more time to complete or find the correct factors.

 This might be a little hard to figure out what is exactly going on in this picture due to the lack of quality. This photo is merely the same steps that we were using in example #3 but with possibly a little more complicated numbers.

If you would like to have some practice problems here are some great problems to review. Hint: There will be problems most likely similar to these following on a test. 

A huge hint that I got from these classes  and I think it plays a major role in why some of these might not seem to work is that if numbers are factors in a specific parenthesis, you can take out numbers so that each parenthesis is the same and you can factor by grouping. 

 

Just follow the step for any of the examples and remember to make a table. I think that the table helps a lot.

For the Review in the Book Look Through Page 176-178
For some more information on factoring quadratics and why Click HERE
For some more information for factoring quadratic equations Click HERE
And for my personal favorite site that has great information and helps a lot for reviewing, check this out and Click HERE!!!!

I hope that this summarized the first two classes of this week

Thanks guys, for reading and I hope you liked it :) 

Remember to comment!!!

Have a good weekend (hopefully with snow)