Monday, April 25, 2011
More Factoring
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
Tuesday, April 19, 2011
Multiplying Polynomials
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
Sunday, April 10, 2011
Friday, April 8 2011 Post
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:
FINDING OUR POINTS
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.
FINDING OUR DEGREES
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!!!!
FINDING OUR RADIANS
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!
http://www.youtube.com/watch?v=ao4EJzNWmK8
ENJOY!