D.E.V. The Dawning of great things.

1.Simplify the following equation
(x3+7x2-5x-25) 
(x+5)(x-8)
(x-2)
x2-10x+16


(x3+7x2-5x-25)  .   x2-10x+16
(x+5)(x-8)               x+2


(x3+7x2-5x-25)  .   (x-8)(x-2)
(x+5)(x-8)               x-2


(x3+7x2-5x-25)  .    1 
        (x+5)                1


  x2(x+7)-5(x-5)
         (x+5)

    (x+7)(x2-7)
x3+7x2-7x-49

We start out with a relatively decent sized equation. First thing we want to do is turned this multi-layered fraction into a multiplication. After that we try to maybe figure out some steps toward an easier equation. now we see that -8 could be a factor of 16. With division we could take out that entire part of the equation bringing this giant down to size. Now for our final step we factor out our last numerator and see if there is any moves left. There seems to be so we divide out the rest and here is our simplified equation.

2.Find the domain,range, x, and y values of the given function.
sqrt((x+5)(x+1)(x-8))

D:[-5,-1]U[8,infinity]
R:[0,infinity)
x:-5,-1,8
y:N/A

We have figured our domain from simply taking the x intercepts of each factor and placing them on our graph. We then plot the graph based on our power, x intercepts, and end values. From there we take only the positive values because a square root cannot be negative. Since our values can only be positive this gives us our range.

3.Simplify using completing the square
3x2+24x-30=2
               +30  +30
3(x2+8x)=32

3(x2+8x+16)=80

3(x+4)2=80

3(x+4)2-80=y

-4=y

Here we have a regular run of the mill completing the square. So, we start out with adding 30 to both sides. Then we divide our B value by 2 and square it to get 16. Now we have to be careful because we need to add the multiplied version to both sides. Now our value is 80. We now factor down our left side for the y-intercept formula and boom! put everything to one side and there it is.

4. Find the perimeter in terms of A.
Given
A*B=100
2a+2b=P

(a*b)/A=100/A

b=100/A

2a+2(100/a)=P

This was a simple substitution problem where we can take values of one problem and substitute it to a problem working on a relatively same basis. So all we need to do is divide the total area by a to find it in terms of b. Using that we substitute our new equation for the old one. Finding the perimeter.



I choose these problems to work with because I was scared. A big part of this class intimidated me because of the very first unit and the uncertainty that followed. However I did start to pick up much of the class after settling into a routine. I think that these problems were my understanding of many of the units that we covered. Like in problem 1 when we solved rather complex fractions. With that in mind I feel this projected helped me in the way of reviewing. I had the chance to look over many of my papers and remember what I might have forgotten over time. Other then that I don't think I learned any significantly new knowledge while making this project. A suggestion that I might make though would be to offer different banks of problems ranging from the said expertise to a novice level. Just so that groups can gauge where they are at a little better. So people as in me don't know what to do when given a versatile assignment like this. - Noah P.