Rational Root Theorem

Hello my name is Hardeep, Today I will be explaining how to perform the rational root theorem. We learned this Wednesday October 16. I will explain this through the example in our notes. Pictures of the examples are at the bottom of the post, I didn't know how to post cropped pictures.

 The first step is finding the a and k values.

 a is the leading coefficient of the highest degree term, which in this case is 1. K is simply the constant or the single number at the end of the polynomial, which is 6.

 The next step is to find the roots. This can be done by dividing the factors of k by the factors of a. The factors of a are just 1. The factors of k are 1,2,3,and 6.

Now we will divide 1/1, 2/1, 3/1, and 6/1. We end up with 1,2,3,6 as our roots. These are the numbers we would have plugged into our equation to solve for zero, but we don't have to do that yet, so onto example 2.

Part a) asks us to find the possible roots of 3x3-4x2-5x+2, we will do what we did previously. First we find a and k, in this instance it is a=3 and k=2.

 Next we will find the factors of a and k. The factors of a are 1 and 3. The factors of k are 1 and 2. 

From here we will divide the factors of k by the factors of a. 1/1=1, 2/1=2, 1/3=1/3, 2/3=2/3. We end up with our roots which are +-1, +-2, +-1/3, +-2/3.

 The next step is to substitute our roots into the polynomial function. P(1)= 3(1)3 - 4(1)2 - 5(1) + 2 = 3-4-5+2 = -4 This root did not work so it is not a root of this polynomial. We will continue using the next root which is -1, this step is trail and error.

 P(-1)= 3(-1)3 - 4(-1)2 - 5(-1) + 2 = -3-4+5+2=0 This one worked and so it is one of the roots of this polynomial. The polynomial must equal 0.

 Step 4 is to perform division, I will be using synthetic division.

 -1/3 -4 -5 2
        -3  7-2
-------------------
     3 -7  2 0

 We end up with 3x2 - 7x + 2, from here we will factor this and end up with (3x-1)(x-2)

 Step 6 is to combine all of the factors that we used. (x+1)(3x-1)(x-2)=0

We now find the roots by equating the factors to zero. x+1=0 becomes x=-1, 3x-1=0 becomes x=1/3 and x-2=0 becomes x=2. The roots are x=-1 x=1/3 x=2.

This is all there is to the rational root theorem. If you were keeping up you would have noticed that we completed part b) as we found the roots. The answer to b) is (x+1)(3x-1)(x-2).





Example 3 is the same as the previous example except for a few small details, I will elaborate on these but skim through the rest.

Example 3: Factor x5 + 3x4 - x - 3

Find a and k, a=1   k=-3

Find the roots, +-1, +-3

Now we plug in our roots and equate to zero.

(1)5 + 3(1)4 - 1 - 3 = 0

x4 + 4x3 + 4x2 + 4x + 3

Because our polynomial is a quintic function we must bring it down to quadratic so we will find 3 roots in total.

(-1)5 + 3(-1)4 +1 - 3 = 0

x3 + 3x2 + x + 3

(-3)5 + 3(-3)4 + 3 - 3 = 0

x2 + 1

Our 3 roots are (x-1)(x+1)(x+3)(x2 + 1)







Example 2                                                                   Example 3

Synthetic Division and the Remainder Theorem

Hey, it's Breanne. If you don't know who I am I'm that girl that sits at the back of the classroom in the corner, eating. 

Synthetic Division 
So to refresh your memory synthetic division is a method of performing polynomial division, with less writing and fewer calculations.

Synthetic division is actually super easy, but you really need to master these key steps in order to do it correctly. 

1. Arrange the coefficients in descending order according to the powers of x (if there isn't a power, write 0) 
2. After writing the divisor in the form x - a, use "a" to generate the second and third rows of numbers as follows. Bring down the first coefficient of the dividend.  Multiply this sum by "a", and add the product to the third coefficient of the dividend. Repeat the process until a product is added to the constant term of f(x).
3. The last number in the third row of numbers is the remainder; the other numbers in the third row are the coefficients of the quotient, which is of degree 1 less than f(x).

So in english, you get this equation:

You subtract a from x to get the divisor.

You then write the coefficients down and bring down the first number. 

Multiply the divisor by the first number you brought down. When this is done you carry the result to the next column of numbers. 

You then add down the row.

You continue these steps until you've reached the last number.

You write it out and then simplify, and you're done! 

The Remainder Theorem 

So during our last class we learned about the remainder theorem. We started off doing examples using synthetic division, but always ending up with a remainder. We would then plug "a" into the equation and we get the remainder. 

   

After doing a couple of examples we came up with this conclusion:

When a polynomial P(x), is divided by any binomial, (x-a) we can write the statement: P(a) = R

I hope his helped everyone and I hope your thanksgiving weekend was wonderful!!! 

-Breanne Calvadores

Back to Basics: Polynomial Functions (featuring factoring and long division)

Good [insert time of day] to you all!

For those wondering, my name is Wandaly. I will attempt to fulfill several goals in my scribe. First of all, I will try to summarize today’s class. Hopefully, I will be able to help you understand the concepts we’ve tackled a little bit better and clear up any confusion. Lastly, I’ll try to make this scribe as entertaining as possible. 

Factoring is the process of breaking down an equation or a statement or a term. There are different methods:

Using the Greatest Common Factor: the method is self explanatory. You look at the possible factors of the numbers and take the largest one.

Factoring by grouping: you group the terms together in a polynomial and factor our the common factor from each group. From there, you will be able to see that both of the groups will have the same factor. You can lump the ones not in a group together. 

You may have encountered factoring the difference of perfect squares in grade 11. This one may be the easiest to factor out of all. Simply look at the two terms and square root each one. Write the terms with opposing signs. 

But wait!

We may overlook simple things. At first glance this question may seem like a trick question:

If we stick to what we've learned, we need to take the square root of x squared and square root of 10 as well. It isn't a nice whole number, but it's still a number.

This is the same concept with factoring difference of cubes. The same can also be said for factoring sum of cubes. The important thing to remember when doing questions like these would be to watch out for the signs. 

The formula for difference of cubes would be:

Meanwhile, the formula for sum of cubes would be:

Notice that the only thing that varied between these two formulas was the sign.

We factored perfect square trinomials as well. Each one can be factored as a square of a binomial. A quick way to do it would be to look at the square root of the first and last terms. 

Our work can be reduced if we are able to remember these tips and tricks. Using the form x squared + bx + c: 

1.) If the c is positive, then u and v have the same sign. They are both positive if b is positive. They are both negative if b is negative.

2.) If c is negative, then u and v have opposite signs.

3.) The variable b is the sum of u and v.

4.) The product of u and v is equal to the product of ac.

During the afternoon class, I was pleasantly surprised when I saw the papers with long division questions. I’m sure that’s not a sentence you read often. However, it reminded me of my elementary days in the Philippines where my teachers used to make us do long divisions the whole class. Calculators were foreign objects to us back then and I’m a bit thankful now.

Personal anecdote aside, I can promise you that polynomial long division uses the same methods as normal long division you have encountered before. It may look intimidating with the powers and letters thrown in, but if we’ll navigate it in baby steps, we’ll be able to solve it in no time. 

First step:

1.) Arrange all the terms in descending order of power.

2.) Ask yourself: how many times does the first term in the divisor go in the first term of the dividend?

3.) Multiply the quotient by the divisor.

4.) Subtract.

5.) Take the difference, check how many times does it go to to the divisor.

6.) Keep repeating the steps until you have divided everything or a remainder appears. A reminder is a degree lower than the divisor.

Perhaps if you're a visual learner, you might learn better if we look at an example.

P.S, If all of what I’ve mentioned still made as much sense as the Queen of England driving her Range Rover, I’ll hand the keys over to KhanAcademy. 

Don’t worry – it will eventually make sense. Look it’s Queen Elizabeth driving her Range Rover! It happens.

Function Transformations Review Package

Good morning, everyone! (Or afternoon, depending on when we're reading this in class.) 

Kyle Sarte here, alongside my blog-post-partner, Wandaly Gonzales, who will also be writing later on in this post.

I'm not here with a scribe, but:

A. An assignment

B. A worksheet

C. A quiz

D. A test

E. An exam

F. All of the above

Yes, an assignment-worksheet-quiz-test-exam hybrid!

Last night, I made my way through the archived Pre-Calculus provincial exams and compiled an assortment of questions dealing with the topics covered in our most recently completed unit, Function Transformations. 

I began with the links from 2010 and made my way up the calendar back to 2013, going through each exam and copying and pasting into a separate document every question I felt suitable for a test review. From the 8 booklets I reviewed, I gathered a total of 25 questions.

Questions concerning the stretches and inverses of a function, restricted domains, and the graphing of absolute value functions are some examples of the types of questions included in this review package. Because many of the questions were taken from different exams, the numbering of the questions is rather random - disregard this.

Now, enough about the process! Onto the review package! 

Below, I have provided two free download links (though, I could make a decent amount of cash for charging even just $1 for the download...). One of the links contains the review package in extension DOCX; a file type often associated with Microsoft Word. The other contains the package as a PDF. The text and graphics in the PDF version are much sharper than in the DOCX version, but you can take your pick. (If there are 2 file type choices in which a review package can be downloaded, in how many ways can...)

Download: Function Transformations (DOCX)

Download: Function Transformations (PDF)

I will now pass this blog post to Wandaly, who has provided for everyone an answer key for the review package!

Since Kyle had already greeted you good morning and good afternoon, I might as well cover the good evening. 

Good evening, this is blog-post-partner Wandaly taking over.

It is not my turn to scribe either but Kyle and I wanted to share this package to help us prepare for the upcoming test. Even if we have failed to get some of these correct during our quizzes, as long as we practice over and over again, we're bound to get some right. Ha!

I've provided the answer key to the 12 page test.

P.S., Kyle, I believe the answer to your question is now 3c1, if we add my file.

Download: Function Transformations (ANSWER KEY, PDF)

Kyle here again, here to close off our blog post.

This review package is for everyone to download, view, and use. Pre-Calculus will only get more challenging and time-consuming as we progress throughout the semester. It is because of this that I believe any extra help we can all provide for each other is valuable; nobody wants to be left behind.

Wandaly and I can't promise that these review packages will appear consistently before every unit test, but we will try our best to put them together when we can.

This brings us to the end of this (rather lengthy) blog post. I wish everyone good luck on Friday's test!

Thanks for reading,

Kyle Sarte and Wandaly Gonzales

The Inverse of a Relation

Hey guys this is Mathew Dunning, I sit near the back of the class room in the corner. If you don't know which one I am, you can probably look up considering I am pretty tall, but enough about me lets start talking precal.

So first I will show you how to find the Inverse of a Relation.  To find the Inverse of a Relation and graph you switch the X and Y axis.  The steps to find an Inverse of a function are:

1) Replace f(x) with y

2) Switch x and y

3) Solve for y

4) Replace y with f-1(x)

For Example:  f(x)= 2x+2

y=2x+2

x=2y+2

x-2=2y

y=x-2

    ----

      2

y= 1

    ---- x-1

      2

f-1(x)=1

          ---- x-1

            2

Also in class we worked on finding the domain and range!

This is how you find the domain and range,

Domain is left to right on the graph so if you look at Example #1 in your booklets you can see that the farthest point to the left on the x axis is -5, and the farthest point to the right on the x axis is +4 so then the domain of that function would be (-5,4)

The range is almost the same thing but it goes from low to high so if you look at the original graph you can see the lowest point on the y axis is 1, and the highest point on the y axis is 5 so then we know that the graphs range is (1,5)

If we wanted to find the inverse of the domain and range we would simply switch x and y so then the domain would switch with the range. So the inverse domain would be (1,5) and the inverse range would be (-5,4).

Next we had some examples about applying reflections & inverse.

I will show you a few examples of how to do them. For all your questions the coordinates are (1, -1/2).

ex A. -f(x)

Since the f is negative you know that there will be a reflection in the x axis and you make all y values negative. So then -f(x)= (1,1/2)

ex B. f(-x)

Since the x is negative we know that the reflection will be in the y axis, and you make all x values negative. So then f(-x)= (-1,-1/2)

If you look at the graph on Example #2 you can see how -f(x) makes all the y values negative.

the coordinates change to:

(-4,0) -- (-4,0)

(-2,-1) -- (-2,1)

(-1,2) -- (-1,-2)

(1,4) -- (1,-4)

(2,0) -- (2,0)

Then you just graph the new points and connect the dots!

If you look at Example 5, this is where Mr.P played a trick on us and the questions asks if the two different graphs are inverses of one another.  If you coordinate the points without looking at the different graphs your answer would be yes! but he told us since the lines do not connect the same way they would not be considered inverses of eachother. So don't always let the coordinates trick you sometimes just by looking you can use common sense and tell what the answer will be.

Well guys that is all for now , I will see you tomorrow and I hope everyone had a good time at TA day! :)

Additional info on transformation

     Hello class, This is nick who is about to show you how equations on trasnformation works and explain about this stuff. I hope i don't give you a cancer with my terrible grammar(also i don't find any multiply and devide signs on my lap top. i also have no idea how to put fractions in with this.) and let's don't waste more time on reading this and get in to it.


How Trasformation Equations work.



      Hopefully, you should see what i scanned from the booklets on how equations work.  As you see above, a and d affects the y-values and the effect on the graph is said to be the same on like b and c that affects oppsite. But explaining in words wouldn't help you so much with this terrible grammar -_- so i'll just explain this by a made up question.

 

     If f(x)=(-2,4), what ordered pair does the graph -2f(4(x+2))+4 produce?

 

     Well first, we need to know what the steps are, which is listed on the next page of booklet. 



The steps that you guys don't really need to know if you know the idea in my opinion, but never bad to know.

      using this example of steps above, we should multiply the y-value by -2 which makes it......



 -2*4=-8

 

     and now we need to add 4 and it makes......

-8+4=-4

 

     Now we know that our new y value is -4. Now, what about x-value?

    

     Unlike the y-value, all the thing you see is a TRAP! you should devide by 4 instead of multiplying it! which makes the x-value.......

 

-2/4=0.5

 

     and now we subtract this by 2 instead of adding it(btw you only devide or multiply x-value, not the one next to it.), which make it......

 

0.5-2=-1.5

 

     now there is our x-value, which makes our answer (-1.5,-4).

 

     See how it works??? See how Insides(a and d) affects the y-value and how outsides(b and d) affects the x-value?? Well now, we are going to have graphs to help you understand better in this one.



     The graph of a function f(x) is given below. On the same set of axes draw a graph of the function y=2f(x-1)

    

 

    The coordinates of f(x) are (-5,0),(-4,3),(-2,-3),(-1,2) and (2,0), now we should multiply the y-value by 2 and add 1 to our x-value. which makes these new coordinates......

 

(-5,0) to (-4,0)

(-4,3) to (-3,6)

(-2,-3) to (-1,-6)

(-1,2) to (0,4)

(2,0) to (3,0)

 

now if we draw this on the graph.......

 

 



     A new(ugly) line comes out.

 

     I think that should be enough explaining or giving enough cancer to you guys, I hope you get the idea well by watching this and now it's time for me to go.

 

     (P.S. But before i go i want to pick who's gonna be next and that will be number 1 or 10 if 1 is not availiable.)

 



Reflections and Stretches

Hello, my name is Kyle Cuellar. I will be summarizing our class on Reflections and Stretches. Mr. Piatek began the class by starting a new lesson (reflections) and by making us look at the graph below.

We then answered two questions about the graph.

 When you reflect a point over the x-axis, you will need to make the y-values negative and when you reflect a point over the y-axis, you will need to make the x-values negative.

After that, Mr. Piatek answered example 1. He also showed us the rules about reflections.

 Here are the examples that we answered during the class:

Reflection over x-axis

 Reflection over y-axis

 Reflection over x-axis but y-values are multiplied by -2

Reflection of f(x) over both x-axis and y-axis

The next day, Mr. Piatek started another lesson and it's all about stretches. There are two types of stretches, vertical stretches and horizontal stretches.

Mr. Piatek then showed us how to do the first example.

I found a video about the stretches and reflections of functions:

Just a quick note: When performing transformations always perform the horizontal and vertical stretches before the translations (up, down, left, right).

This is what we learned in class the last two days. I hope I was able to help you with this information.

Sorry I'm not a good writer. Thanks for reading this blog though. I look forward on learning with all of you this sem.

Your Classmate,

Kyle