Mariher presentation

Part 2: Denominator

Part 1: Numerator

First you long divide the polynomial by the factor given, (x-5). We talked about the steps for this in problem 2, you do the same thing here. We do get zero here so we can then take the polynomial we are left with and factor it. DO NOT FORGET TO INCLUDE THE FACTOR WE DIVIDED WITH.

You can take the denominator and simply just factor using grouping. The sign in between the two factors is a subtraction, therefore you have to change the sign in the second factor (seen highlighted) You can get the x-intercepts and graph them (pay attention to the degree the graph and the a-value). Then find the domain.

Here we are factoring by grouping. Remember we are setting this greater than and equal to zero because this is the numerator. We can solve for the x-intercepts and graph the function. Notice that next to the x=5 it says bounce. This is because when a factor is to an even power or there are an even number of the same factor the graph bounces at that x-intercept. Pay attention to the degree and a-value of the graph From this graph we can the domain.

Grouping

We take the polynomial and the factor we used to divide and set them greater than or equal to zero. We do this because the original problem is under a radical. You cannot have a negative number under a radical! After that you factor how you normally would. There are lots of tricks but you can do it however you learned. Here we factored by grouping (Tip: When you group and the sign between the two factors is a subtraction you have to change the sign in the second factor). We learned this concept in problem 1. After we have factored we can find where it is equal to zero; 2, -2, 8, and, 4. We will use these numbers for the next step.

First you have to group the variables into groups of two, leaving the sign in between out. Then you have to factor each group. In this case we factor out a 7x^2 in the first group and a 3 in the second. Then you take that and factor out the common factor ; (2x+11). You end with, (2x+11)(7x^2+3), the final factored form.

Part 1: Numerator

After we divide we are left with two things, a polynomial and a factor. Some times you may need to divide that polynomial again using another factor, but in this case we do not need to. We do still need to continue factoring though. From here you can choose how to factor. I used the guess and check method. I guessed, and oh look it worked out. Now it is completely factored. You have finished but here we found where it equals zero, just for some extra practice. For the first and third factor it was easy. As you can see the second factor was a little different. There are no solutions. This is because when you solve it algebraically you end up with a negative under the radical, that results in an imaginary number, therefore no solutions.

Find the domain

The original problem:

First you must long divide by the factor given, (x-4). To long divide you have to subtract to get rid of the front number, always. So for this we see that the first thing we have to do is multiply the (x-4) by 2x^3. We then get (2x^4-8x^3). Now we have gotten rid of the 2x^4 and we are left with -16x^3. Then just like long division with numbers you drop the next number (if the number you had left over from the subtraction is zero you then drop two numbers). You continue to subtract until you get zero. There is a chance you might not get zero. Check your work. If your work is correct then you have a remainder and it won't factor this way. For this problem we get zero though. The polynomial you get at the top is what you will use to factor

Factor completely

Now we are going to factor using long division, we know how to do that now. What we do not know is why we are using it. There are many options when factoring. When you have more than five variables it can be hard to use other methods, therefore generally we use this method when there are five or more.

To start we take out the numerator and use the factor given to use long division

Heather's

Reflection

Part 2: Denominator

For the denominator you can simply use grouping. The only difference is we only set this one greater to zero, not equal. You cannot have a negative number under a radical and you cannot have a zero in the denominator. From the factored form we can find the x-intercepts. We can also see that the a-value is a 4, so it is positive. The degree is a 3 so the arrows do not point in the same direction, they point in the opposite direction. We then find the domain of this graph. The only thing different is we use parentheses instead of brackets because we are not including zero.

Grouping

I choose these problems for two reasons. One I really to do these kinds of problems. Two I think they all relate and touch on every unit. I think these show an overview of what I know best because they expand to most of the units we covered. I personal did not feel as if this assignment helped me. I don't feel as if I learned much. It was a hard project though. I do understand as to how it could help others, I learn well from teaching, just not this particular time.

First you have to group the variables into groups of two, leaving the sign in between out. Then you have to factor each group. To start we factor out the greatest common factor. In this case we factor out a x^2 in the first group and a 6 in the second. When you are done both factors in parentheses should be the same like in the second line. Then you take that and factor out the common factor ; (3x+8). You end with, (3x+8)(x^2+6). That is the final factored form

Finding the Domian

Marissa's Summary

Those numbers are called x-intercepts, where the graph crosses the x-axis. There are a few things we have to pay attention to when graphing. Where the x-intercepts are, if the a-value is negative or positive, and the degree of the graph. From this picture we can tell where the x-intercepts are, some might be able to tell a little bit about the degree but we want to know everything, right? So if we look back at the original polynomial we see that the highest power degree is 4. Four is even so that tells us something about the graph, the arrows point the same way. The a-value is a 2 so it is positive. This tells us the arrows point the same way. From this information we can sketch a graph that looks like this. The domain is where the graph is above zero or equal to zero. If we include zero then we use brackets.

To find the domain of this problem we take the numerator out and use long division to factor it.

Final Domain

Part 3: The final domain

Finding the final domain is the easiest part. We a number line. First plot where your graphs equal zero (x-intercepts). Remember to pay attention to which numbers we can actually have in our equation. The open circles mean it is not equal to that number, closed means it is equal. For this number line I graphed denominator on top and numerator on the bottom. Then highlight where they overlap. Where the two domains overlap is where the final domain is. In this case the final domain is (3,4] U [8, infinite).

I choose to make problems from units that I had the most trouble understanding, such as long division then finding the domain and grouping, so I would get more practice with them. I tried to make them more complex so I could combine units that we did. The problems involved long division, graphing, grouping and finding individual domains then combing two domains into a final one. This assignment helped me understand how to do the problems more efficiently and it taught me about the silly mistakes I was making. This project is just a lot of stress put on people though because we already have to worry about studying for exams and then on top of that we have to now worry about a project. In the future for this assignment maybe assign less questions we have to make because then it takes some weight off our shoulders and we can also focus on studying for exams. I'm just excited to have this project out of the way so now I can worry about doing good on this exam!

For the final domain we graph the domains from the numerator (bottom) and denominator (top) on a number line. Pay attention to the open and closed circles.The domain of the final is where the two domains overlap, the highlighted area.

Mariher Presentation