Breaking Down a Polynomial

Hi everyone!

So last week we were assigned a problem in Dropbox called "Breaking Down a Polynomial" for homework. This blog is a walkthrough of how to do that problem. It also kind of goes over synthetic division. Sorry it's so long...

http://www.youtube.com/watch?v=GyubhcJg7wQ

Hope this was helpful. Good luck with studying for exams!!

-Acadia

Completing the Square

Hi Everyone!
On wednesday, everyone was hastily finishing their radical equation assessments, so I will just be blogging for thursday. In class on thursday we learned how to solve Quadratic Functions by Completing the Square.

Completing the square gets its name because you can use a square as a visual to break down quadratics.

For example: x² + 10x + 25

http://www.helpalgebra.com/articles/completingthesquare.htm


An example of a perfect square equation is (x-1)² = 25 where both sides of the equation are perfect squares so you can square root both sides of the equation.
√(x-1)² = √25 x-1= +/- 5
x= 6 and x= -4
☞note: if you start with a square, you should not have any extraneous solutions.

But if you have an equation like x² -10x - 18 = 0, where both sides of the equation are NOT square roots, and the equation is not factorable, you must follow these steps. 


Here is an example problem:

x² - 10x - 18 = 0
Can this equation be factored? You want to make this equation factorable in ax² + 2ab + b² form.
a = 1
b = -10
c = 18

First, the third term (18) must be brought over to the other side. The mystery number will make the first side factorable. You add it to the other side as well because then you are not actually changing the equation, because the numbers would cancel. As a rule in algebra, what you do to one side, you must do to the other.

x -10x + ? = 18 + ?

To find your mystery number, divide b, which in this case is -10,  by 2, and then square it.  (b/2)² will always give you your mystery number.

(-10/2)² = 25

Now your equation is factorable.

x - 10x + 25 = 18 + 25 
(x-5)² = 43

Square root both sides of the equation.

√(x-5)² = √43
x-5 = +/- √43  ☞note: all square roots are positive and negative
43 is not a square, and is not divisible by any squares, so you leave it inside the square root prison.

Finally, Isolate the x.

x = 5 +/- √43



Here is another example of how to do an equation that is not a perfect square equation:

x² +4x -96 =0

Bring the 96 over to the other side. 

x² + 4x + ? = 96 + ?
(4/2)² = 4
x² + 4x + 4 = 96 + 4 (which conveniently adds to 100 which is a square! Its always nice when math works out in whole numbers☺)

Square root both sides.

√(x+2)² = √100
x+2 = +/- 10   ☞note: both positive and negative 10 are square roots of 100

x = +/- 10 - 2

x= 8 and x= -12

What do you do if the x² has a coefficient?
3x² + 5x -7 = 0

If the x² in your trinomial square has a coefficient, you have to divide the equation by the number of the nasty coefficient to cancel it out.

3x² + 5x- 7 = 0

_____________    

            3

x² + 5/3x - 7/3 = 0   ☞note: 0 divided by anything is 0 (remember 0/K and N/0? 0 divided by anything is 0, but it is impossible to divide anything by 0)
x² + 5/3x + ? = 7/3 + ?
(5/2 • 1/2)² = (5/6)² = 25/36
x² + 5/3x + 25/36 = 7/3 + 25/36
(x + 5/6)² = +/- √109/36
x = (+/-√109/ 6) - 5/6

x =  +/- √109 - 5 
       __________
                  6

The Quadratic Formula and Completing the Square

Does everyone remember the quadratic formula? Well, here's how the quadratic formula ties into completing the square.

ax² + bx + c = 0

Divide the equation by a to get the x² on its own.

ax² + bx + c = 0
_____________
           a

x² + (b/a)x + b²/4a² = -(c/a)x + b²/4a²
(x + b/2a)² = -4ac/4ac² + b²/4a²

Square root both sides.

x + b/2a =  √(b² - 4ac)/4ac  looking familiar?
x + b/2a = +/- √b² - 4ac
                  ____________
                              2a
x = -b +/- √b² - 4ac
      _____________
                   2a
                
yay!!

Here is a video about the Quadratic Formula. You may remember it if you were in Dubuque's math class last year. Beware the end...

           

Here is a khan academy video on completing the square. I often use his videos for review, and find them really helpful.

Problems using these concepts are on page 294 in the book.
Hope my blogging was satisfactory! 
-Grace

Cooper's Post

Hello Class,

- Monday: 6-5 Equations Containing Radicals:

   On Monday we looked over the blog from last week and went over some of the problems that occurred in our homework on Equations Containing Radicals.       
   As the name would suggest, these are equations that contain a square root. The first step in solving on of these equations is to isolate the square root or radical...

     ...with the radical isolated, one squares both sides of the equation...

      ...and set the equation equal to zero...

      The next step is to factor the equation and find the value of x...  

      The final step in solving an equation of this nature is the ever important step of checking the solutions by plugging them back into the original equation...

      Tips:

-Don't forget to check the solutions or you might end up with a solution that doesn't work.

-Be sure to square the whole root and not individual terms as this may lead to an incorrect solution.

-When you have two square roots, it is important to get the roots on different sides or the equal sign.

     Here is a video that illustrates how to solve an equation this two roots.  

For the video click here

    

This material is covered in pages 263-266 of the text book and problems from this section appear in self test 1 on page 267 as well in the chapter review and Chapter test on pages 286-287.

-Tuesday: Overview of Section Six:
      On Tuesday we reviewed our homework which covered the material from section 6 that we have learned so far.

      Some common mistakes that we went over evolved the properties of radicals and the use of conjugates.

      -Properties of Radicals: 
       Here is one of the problems were went over that concerns this subject...

       The first thing to do in solving this type of problem is to multiply it by a term that will that will make the denominator a perfect square, cube, ect...

       With the denominator cleaned up and a square root remaining on the top, solving the rest of the problem is fairly simple...

       -Conjugates:
        When one is faced with a problem with two roots in the denominator, it is necessary to us the conjugate of the denominator. The conjugate is the same as the original equation except the sign is the opposite. For example, the conjugate of 3x-6y would be 3x+6y.

        Here is an equation evolving conjugates that we worked on today:

        The first step is to multiple by the conjugate...        

        At this point, the bottom works out to 8 which cancels with the numbers in the top...

        Leaving us with...

       Tips:

        -Remember, the conjugate is the same as the original equation only the sign is different. 

         Example:  The conjugate of 3x-6y would be 3x+6y

        -To clean up the bottom in a fraction with roots in the numerator and the denominator, multiply the                    whole equation by something that will make the bottom a perfect square, cube, ect.

      In the text book, section six spans from page 248-289 and covers irrational and complex numbers.

       Applications:
       -Irrational numbers such as pi are used to find the circumference  of a circle
       -Irrational numbers are always more exact the physical measurements  
     

         

Four Rules 3/18

So during Monday's class we elaborated on what we learned at the end of last week.. Up on the board we went through 4 major rules and then did lots of practice problems in the book.

I went ahead and copied my notes over to Notability and cleaned them up a bit, so here they are:

I hope this helped clear up any questions you might have had and explain anything you were unsure of.
Also, I hope everyone had a great snow day!
- Izzy

Hi class,

This 3 day weekwas really laid back.  

We went through the project that we created the week before. All of you know that you can refrence the movies and look in the texbook.

On Thursday was pi day, potato chip day and Esme's birthday (by the way happy belated birthday). We celebrated by memorizing Pi (the person who got the most in a group was 68 by christan, esme and anne and individually was 68 again by esme) and creating poems.

Even though we had lots of fun and had a laid back week we did start to learn a new unit which was irrational and complex numbers. We started by staing everything about square roots. from there we learned the different parts of a square root such as the index and the radical

here is a video on rational and irrational square roots

and heres another on simplyfying square roots 

I hope that these will help

Now I found a call me maybe parody (I know the song is so last year) that has a little bit of radicals in it. It might help also but also its just for fun. I am terribly sorry if it makes you an more confused.

-Nadia
Here are some notes from class.

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.