Monday, April 25, 2011

More Factoring

Last Thursday we went over the homework and everyone got up and did a problem, from the homework, on the board. After this Mr. Cople gave us a worksheet with some factoring problems and we spent all of class working on the worksheet. The homework was assignment 1103B.

Here are some of the problems from the worksheet:
(they are numbered according to their number on the worksheet)

14. x^2+6x-16

in order to do this problem we first have to list out all the multiples of 16 and find which set adds to 6 and multiplies to -16.

2 and 8: no
1 and 16: no
4 and 4: no
-2 and 8: yes

-2 and 8 add to 6 and multiply to -16.
Now we have to place our parentheses with an x in each one because we have to split up the x^2 into two separate x's.
So it will look like this: (x )(x ).
Next we plug in our -2 and 8 in any order.
So the final answer is: (x-2)(x+8) or (x+8)(x-2).

27. t^4-16

In order to do this problem we first have to find out what numbers will multiply to -16.

4 and 4: no
2 and 8: no
1 and 16: no
-4 and 4: yes

Now we can place our parentheses with our t's in them. However these ones will be t^2 because we started with a t^4.
So it will look like this: (t^2 )(t^2 ).
Now we plug in the -4 and 4.
So it will look like this: (t^2-4)(t^2+4).
But we are not done yet. We have to split up one of the equations in one of the parentheses.
So the final result will be: (t+2)(t-2)(t^2+4)
This works because 2*-2*4=-16.

22. a^2-12a+36

In order to do this problem we have to find multiples of 36 that add up to -12.

4 and 9: no
3 and 12: no
1 and 36: no
6 and 6: no
-6 and -6: yes

This works because -6 and -6 add to -12 and they multiply to 36.
Now we can place our parentheses with our a's in them.
So it will look like (a )(a ).
Now we can plug in -6 and -6.
So it will look like (a-6)(a-6).
Since -6 is in both pairs of parentheses we can combine them.
So the final answer is (a-6)^2.

I hope this was helpful and 1103B is due on April 29, 2011 because of PSAE practice on Monday and Tuesday and no school on Wednesday and very short classes on Thursday.

Owen C.

Wednesday, April 20, 2011

Chapter 11.3 Factoring

At the start of class, Mr. Cope did his daily routine of checking in everyone's homework. After that we began a new section called Factoring. We learned about factoring a polynomial, which is the process of rewriting a polynomial as a product of two or more factors. Mr Cope described to the class that there were 3 special cases of factoring:

1) The Greatest Common Factor (GCF)

Example: Factor 5x^3 - 15x^2

To solve, look for the greatest common monomial factor of the terms. In this case, the GCF of 5 and 15 is 5. X^2 is the highest power of x that divides each term. So 5x^2 is the greatest common monomial factor of 5x^3 and -15x^2. The last step is to apply the Distributive Property.

5x^3 -15x^2= 5x^2(x-3)

to check the answer, you can use the 'expand' command on the NSPIRE calculator by clicking MENU, 3, and then 1.

5x^2(x-3). 5x^2(x-3)= 5x^2 * x - 5x^2 * 3 = 5x^3 -15x^2

2) Difference of Perfect Squares

Example: 121 - y^2
= 11^2 - y^2

= (11+y)(11-y)

To solve a problem like this

3) Perfect Square Trinomial

Example: (8x + 2)^2 = (8x + 2)(8x + 2) =

(8x * 8x) + (8x * 2) + (2 * 2) =

64x^2 +16x +16x + 4

So if you add like terms the answer is 64x^2 +32x+4

After this, we began working on problems on the factoring worksheet he handed out to everybody. Before we began working on the problems, Mr. Cope gave the class the tip to guess and check on problems like x^2 -2x-15.

You want solutions to x^2 -2x-15=(x+____)(x+____)

Because -15 is the product of the the two missing numbers on the right side, you should think of the factors of -15. These factors are either -15 and 1, 15 and -1, 5 and -3, and -5 and 3. Keep filling in the blanks until you get the right answer.

(x-15)(x+1)=x^2 -14x-15 NO!

(x+15)(x-1)=x^2+14x-15 NO!

(x+5)(x-3)=x^2+2x-15 NO!

(x-5)(x+3) = x^2 -2x-15 YEP!

so this means that x^2 -2x-15 = (x-5)(x+3)

On the worksheet, there were many good examples for practice.

For example:

1. X^2 -81

The answer is (x-9)(x+9) because when you factor this using FOIL, you multiply x*x which = x^2. You then multiply x*9 which = 9x. Next you multiply -9x*x which =-9x. Finally you multiply -9*-9 which = 81.

You add like terms and you get x^2 -81

2. x^2 +14x +49

The answer is (x+7)^2 because when you foil this out, you multiply x*x which = x^2. You then mulitply the outsides, which is 7*x=7x. Third, you multiply insides which again is 7*x=7x. The last thing you do is multiply the last terms. For this, you get 7*7 which =49.

You add up the like terms and you get x^2+14x+49

Hopefully this helps you understand polynomial factoring a little bit more, and if it didn't, don't forget Mr. Cope is always willing to help if you go to him!

Tuesday, April 19, 2011

Multiplying Polynomials

Today in class we went over our homework then we went on to 11-2. We were all curious on how to multiply polynomials so we got our 11-2 note sheets and began the new unit.

First we need to learn how to classify polynomials by the number of terms. Here are some examples on how you would classify them.

Monomial: polynomial with one term. Ex: 2, 3x^5

Binomial: Polynomial with two terms. Ex: x+2, X^2-7x

Trinomial: polynomial with three terms. Ex: 2x^4-3x^2+7

Example 1: expand and write in standard form...(5x^2-4x+3)(x-7) so you would multiply the first term in the first parenthesis by the first term in the second parenthesis and then multiply it by the second term in the second parenthesis. Then you would go on to do the same thing again except you would multiply the second term in the first parenthesis by the first and second terms in the second parenthesis. And lastly you would do the same thing again except you would multiply the third term in the first parenthesis by the first and second terms in the second parenthesis, so the answer would come out to 5x^3-35x^2-4x^2+28x+3x-21. But don't forget to combine like terms! After the terms are combined the final answer comes out to, 5x^3-39x^2+31x-21.

Activity: We had to take an 8.5 inch by 11 inch piece of paper and cut out squares of various side lengths from the corners. We had four cut out lengths and we had to find the volume of the resulting box.
For the first box we did 1*9*6.5 which equaled, 58.5in^3.
For the second box we did 2*7*4.5 which equaled, 63in^3.
For the third box we did 3*5*2.5 which equaled, 32.5in^3.
And for the last one we did 4*3*.5 which equaled, 6in^3.
We concluded that box 2 had the largest volume.

Then we made a volume function with the equation, V(x)=x(11-2x)(8.5-2x) and graphed it on our calculator. And that was the end of our activity.

Since we spent lots of the class period doing the activity we did not have time to get to the back side of the note sheet. But we did have enough time for one more example.

Example: Without expanding, we had to find the leading term of the product (5x^2+2)(4x^2+8)(11x-3). It was much easier than I thought. All we had to do was multiply the first term in each set of parenthesis by each other to find the leading coefficient. Then we had to add the exponents of the first terms in each set of the parenthesis. The leading term came out to be 220x^5.

That is all we had time for. Mr. Cope assigned us our homework and before we knew it class was over.

I hope this helped and if it didn't you can always go see Mr. Cope for some more help!

Monday, April 18, 2011

11.1 Intro to Polynomials

Today in class, the first thing we did was go over our tests and our homework quizzes. Then once we finished that, we received the next notes sheet for the next unit. This section is about polynomials. Now you may be wondering what polynomials are so let me explain.

First, in order to learn about polynomials, we need to learn some basic definitions.

Here is the example of a polynomial we were given in class:

Degree of the polynomial: The degree of the polynomial is simply the highest exponent in the equation. So in the example above, the highest exponent is 4, so the degree of the polynomial is 4.

Terms: The terms of the polynomial are each individual piece of the equation that you are adding together. So in the example above, the terms would be , , , -14x and 83.

Standard Form: The standard form of a polynomial is when the polynomial is written with the powers in descending power order. The example is in standard form. It's in standard form because the first term has a power of 4 and then the next has a power of 3 and so on until you get to the last term that has a power of 0.

Coefficients: These are the numbers in front of the variables. In the example, the coefficients are 3,20,-10, -14 and 83.

Leading Coefficient: This is just the number in front of the highest exponent. So in the example the leading coefficient would be 3 because the highest exponent is 4 and 3 is the number in front of it.

Constant: This is the number alone.

Next, we were given names for certain types of polynomial equations. They're classified according to their degree. It would be a good idea to memorize these.

Linear: mx+b

Now, we are going to learn about operations on polynomials and polynomial functions that start on the back side on the notes sheet that you either got in class or you can find on moodle.

First, let and .

1. Evaluate by hand. Give the degree.
a. f(x) + g(x)
In order to solve this, you add the equations together and you simplify. So the answer would be because we added all of the terms together from both equations.
b. f(x) - g(x)
For this one, you just subtract the first equation from the second one which would look like this: and the answer comes out to be
2. Without multiplying the polynomials, predict...
a. the degree of f(x) * g(x)
Make your own prediction of what the answer will be. The class predicted 6.
b. the leading coefficient of f(x) * g(x)
Again make your own prediction of what the leading coefficient will be. The class predicted 10.
3. Now define the function on CAS (menu, 1, 1). Expand f(x) * g(x) and write the result in standard form (menu, 3, 3).
Once you have done these two steps, you should get the answer of
. As you can see here, the prediction for part a of number 2 was wrong. We may have thought that it would be 6 because we were thinking multiplication, but we have to remember what we learned a few chapters ago. We learned that when you are multiplying two terms with exponents, the final answer's exponent will just be the first two exponents added together. So in the example, it is because you just add the 2 in and the 3 in .
4. Generalize to make a rule: The degree of the product of two polynomials is the sum of the degrees of the polynomials. The leading coefficient of the product of two polynomials is the product of the leading coefficients of the polynomials.

5. Continue with f(x) and g(x) in the previous example. Write an unsimplified expression for f(g(x)), predict the degree and predict the leading coefficient.
To write the unsimplified expression, you just plug in g(x) for every x value in f(x) just like we learned in previous chapters. This would be . Then, the degree is going to be 6 because, as we learned in a previous chapter, when you have one exponent raised to another exponent, you multiply them. So since is being squared, the exponent will be 6. Lastly, the leading coefficient is 50 because the number in front of the power of 6 will be 50 once simplified.
6. Expand f(g(x)) on CAS and write it in standard form.
When you do this, you get and we see that we are correct, the degree is 6 and the leading coefficient is 50.
7. Generalize to make another rule: The degree of the composition of 2 polynomials is the product of individual degrees.

This is basically everything that we learned in class as far as taking notes and new material. But, lets try out a few example problems to make sure that you fully understand the new material.

Example 1:
a. list out the coefficients
Think back to what a coefficient is (the number in front of the variables). So based on this, the answer would be 9,24,16.
b. give the degree
When you look back at what a degree is, you will see that it is the highest exponent, so in this case the answer would be 4.
c. give the leading coefficient
Again, look at the definition of a coefficient to see that it is the number in front of the highest exponent. So this number would be 9.

Example 2:
Suppose that f(x) is a polynomial of degree 3, and g(x) is a polynomial of degree 5.
a. What is the degree of f(x) + g(x)?
Between the two polynomials, the highest degree will be so the degree is 5.
b.What is the degree of f(x) * g(x)?
Remember to think back a few chapters ago and think about what we do when we are multiplying terms. When multiplying terms, we must add the exponents together, so 5 + 3 would be 8.
c. What is the degree of f(g(x))?
When raising an exponent to another exponent, don't forget that we have to multiply the two exponents together to get the final power that we are raising to, so 5 * 3 is 15.

That's it! I hope that this explanation make polynomials seem a little less scary and confusing. If this doesn't help, don't forget that you can always go to Mr. Cope for more help.

Sunday, April 10, 2011

Friday, April 8 2011 Post

On Friday, we checked over homework, then started on Day 2 of Unit Circle and Radians. You can find homework answers on moodle.

10-4 (Day 2) More Unit Circle and Radians.
Overall, we learned in this section how to:
1. convert between degree and radian, and between radian and degree
2. fill in our very first unit circle!!

First off, we defined what a radian actually is: the angle created by placing a radius along the circumference of a circle. (If this doesn't make sense, think of a slice of pizza cut in triangular shapes, from the inside out. Take the crust, and that is what the radian is.)

VERY IMPORTANT: for this chapter, and possibly beyond this, you need to switch your calculator mode. Here's how it's done:
1. press home then #5
2. choose #2 for settings
3. choose #1 for general
4. tab down until you reach Angle
5. choose Degree or Radian
6. tab down until you reach Make Default
7. press enter
8. select OK

If you forget to switch modes, you'll notice that you're not going to get your answers in either degree or radian mode.

If we know that 360 degrees around a circle gives us an arc length of 2π, then 180 degrees around will give us half of that..which is just π.
One fourth of that will gives us π divided by two so it's just π/2.
This information will help us convert between degree and radian, and radian and degree.

 \mbox{deg} = \mbox{rad} \cdot \frac {180^\circ} {\pi}

 \mbox{rad} = \mbox{deg} \cdot \frac {\pi} {180^\circ}
Think of it like this. What you want to get rid of it on the bottom, and what you want is on the top.
We then practiced converting some measures.
ex. Rad. to Deg.
1 \mbox{ rad} = 1 \cdot \frac {180^\circ} {\pi} \approx 57.2958^\circ

and from Deg. to Rad.
1 deg. = 1xπ/180 = π/80 (if you're converting to radians, leave in simplest form, so leave the π in)

Unit Circle
It's going to be a bit difficult to explain, but once it's actually filled out, the unit circle does make sense.
Let's start off with a 30-60-90 triangle. trig_30_60_90.gif
The measures are shown, and the way we use this information is by finding
the cos and sin of some angles.
These answers will be later added to the unit circle.


Now, for a 45-45-90 triangle, We will do the same, but because two angles are the same (45) we don't need to do double the work.
cos(45)=√2/2 (remember, we have to multiply by √2 to get the radial out from under and put it on top)

Now that we have these angles, we can use them to help apply them to the unit circle.


This is a completed Unit Circle. It's complicated, but once taken apart, it's essentially just one piece of information being copied four times over. Let's break it down:



1. Let's start with the points. On quadrant I, we have two points that are easy to remember.

At 0 degrees, we have (1,0).

At 90 degrees, we have (0,1).

These could be the first things that you put down to help you remember the unit circle, but first, a trick.

All the angles on the unit circle have a pattern(at least the ones we need to know).

The pattern is up 30 degrees, up 15, up another 15, then up 30. It keeps going around and around.

30, 15, 15, 30 and so on. This should help identify where points are when they're not exactly on the axis.


2. After you have all your degrees filled in, refer back to the triangles. We took the sin and cos of some angles, and now we can use them here.

FOR ALL 45 degree angles (45, 135, 225, and 315) the sin/cos points will be the same because a 45-45-90 triangle only really has that one measure-- √2/2. The positive and negative signs will change because of which quadrant the points will be in, but the numbers are the same.

FOR ALL 30-60 angles, we can use what we found from our triangle above. If you notice on the picture of quad. 1, at 30 degrees, it's coordinates are (√3/2, 1/2). These points are the cos and sin of 30 degrees from our triangle. If you look at 60 degrees, it's also the cos and sin from the triangle. This is how they connect!!!

The same idea from the 45-45 triangle applies to the 30-60 triangle, except you take your points and SLIDE them over. DO NOT FLIP!!!! SLIDE!!!!


3. Finding the radian points are a bit more difficult, but once a simple concept is understood, it becomes less difficult. Just think of each point and a fraction of the whole unit circle. Let's start with what we know--

all the way around, 360 degrees, is equivalent to 2π. From there, we can find the other points. So 90, which is one fourth of 360, or one half of 180 (just π), will be π/2.

Now that we have our two axis points from quad. 1, we can find the ones on the circle.

Take 45 degrees. It's one half of 90 degrees, so take the radian from 90 (which is π/2) and divide that

by two to get the measure for 45.

That is one way of thinking of it.

Another way is by thinking of the top two quadrants. Because 180 is just π, then you can think of 90 and one half of that,

making it π/2.

Then that means that 60 degrees is one third of π, making it π/3.

If that is true, which it is, then 30 is one sixth of π, so π/6.

This process of thinking and each radian point as a fraction can help simplify the process.

If this still doesn't make sense, here's an extra video that helped me. Mr. Cope can also help (considering he is our teacher..)

and the TLC has great tutors!


Monday, April 4, 2011

Chapter 10.1

first day back from spring break we got right down into new material after getting our tests back. we started learning about trigonometric funtions. we learned about the three special ratios for a right triangle. sine, cosine and tangent Sin=opposite/hypotenuse cos=adjacent/hypotenuse tan=opposite/adjacent using these, it is possible to find the lenght of 2 sides of a right traingle given one side and one acute angle. example- a person is standing 10 feet away from a tree, and looking at a 30 degree angle to the top. to find how tall the tree is, you must use one of the ratios. using tangent (as you are looking for the opposite to the angle, and you have the adjacent side) you would put tan30 (the acute angle)=x/10, then just solve for x giving an answer of 5.77 so overall, you have to know what information you have, then know what you're looking for so you use the right ratio. and sorry about my example, I could not get any pictures of a triangle.

Sunday, April 3, 2011

Chapter 8 Review

On the Friday before break we had the chapter 8 test, which covered sections 8.1-8.8.

Here is a review of the main ideas from this chapter:

8.1- Composition of functions:
-review what a function was
-learned how to handle f(g(x)) and g(f(x))
ex: Let f(x)=2x+4 and g(f)=, find f(g(x)) and g(f(x))

so here u will basically plug and chug, and solve for x

8.2-Inverse of Relations:
- in order to find the inverse of a function, you have to switch where x and y, and solve for x
ex: f(x)=3x -17
f(x) is y
switch x and y: x=3y-17
solve for y: -now, this is the inverse of the original function, or

-the inverse graph will be reflected over the y=x line
-the domain of a function is the range of the inverse, and the range on the function is the domain of the inverse.
ex: D of f(x)= R of
R of f(x)= D of

8.3- Property of inverse functions:
- Vertical line test- figure if original function is a function on the graph
-Horizontal line test- figure if inverse is a function on the graph
-Inverse function theorem- basically says that if two functions are inverses of each other, it doesn't matter what u plug in, because they will spit out the thing that was plugged.

8.4- Radical notation for the nth roots:
-= , positive nth root of x
-Geometric mean: multiply all the numbers and square root by the number of values.

8.5 products with radicals:
- jail braking method
9 3
3 3

the answer is

8.6 quotients with radicals:
-wfoo- weird form of one
ex: =1, =1
- use wfoo to get rid of radical in the denominator (rationalize the denominator)
ex: multiply by and simplify

8.7 Powers and roots of negative numbers:
-x is positive: n is 2- positive real , n is odd- positive real, n is even- positive real
-x is negative: n is 2- imaginary, n is odd- negative real, n is even - UNDEF
-x is 0: n is 2- 0, n is odd- 0 , n is even - 0

-x is positive: n is 2- all work , n is odd- all work, n is even- all work
-x is negative: n is 2- UNDEF, n is odd- UNDEF, n is even - UNDEF
-x is 0: n is 2- 0, n is odd- 0 , n is even - 0

8.8 solving equations with radicals:
- look out for no solution
-to solve equation with rdical: isolate radical , raise both sides to the nth ALWAYS CHECK SOLUTIONS!

Hope everyone did well, and had a good break!
that's it, 9 more weeks AND WE ARE DONE!