How can information you learned in this course be applied to your life and/or work to increase your future success?

Please include the name of the person or question to which you are replying in the subject line. For example, “Tom’s response to Susan’s comment.”

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Kaitlin:

I am currently working in a bank, so there was a part of this course that I had an understanding on. What I didn’t know was the rest of the material. After struggling through this course, and getting outside help, this course has let me better understand the business outcomes of accounting and my finances. Not only did I understand just a small part of this course, but working hard through the unit assignments and more, I have a better understanding of cash flow in and out rather than just on the banking side of things. This course will help me better understand the finance world, it will help me better understand different ways that cash can come in and out and how to better finance myself that way I don’t go into a slump and I will be able to finance a different way than I have been. 

 

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How can marketers attempt to change beliefs or evaluations directly? How effective do you think these attempts can be? Identify and explain a successful marketing example of attitude change and an unsuccessful marketing example of attitude change.

 

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How can outsourcing create growth opportunities for a new venture?

Venture is used to refer a risky start-up or enterprise company. It is the process ofassessing about the profit potentials of an enterprise. When a new business is started, the riskis the…

 

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How can solar panels be utilized for hospital energy in 2039

 

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Here is a reasonable Pre-Calculus example of to illustrate the concept.

Let’s say you had:

##2x^4 – 3x^3 – 5x^2 + 3x + 8##

Like Joan said, there is a trial and error aspect to this.

Look at all the coefficients, and think about what a common factor might be. If you don’t get a zero remainder, then the factor doesn’t really work and you should try again. If the possible factors are all used up, maybe it isn’t factorable.

Here, factors you might try include the ones that correspond to the fourth order coefficient (2) and the zeroth order coefficient (8).

8 has factors of 1, 2, 4, and 8, and 2 has 1 and 2. So the possible factors might be said to be ##pmp/q##, where ##p## is the factors of the highest degree coefficient, and ##q## is the factors of the zeroth degree coefficient.

You can have:##pm[1, 2, 4, 8, 1/2]##

So you can try all of these (##2/2##, ##4/2##, and ##8/2## are duplicates). Recall that if ##-a## is used as what is written in the synthetic division process on the left corner, it corresponds to ##x+a##. We will use ##-1## here. I tend to try ##1## and ##-1## first, and go up in value, and try the fractions last.

##-1 |## 2 -3 -5 3 8

Drop down the 2, and multiply by the -1 to get -2.

##-1 |## 2 -3 -5 3 8………….-2………………..2

Add -3 and -2, then multiply the resultant -5 by -1 again.##-1 |## 2 -3 -5 3 8………….-2..5…………..2..-5

Repeat until you’re done.

Add -3 and -2, then multiply the resultant -1 by -1 again.##-1 |## 2…-3…-5…3…8…………..-2…..1….0…-3………2…-5…0….3….5

Your answer here happens to be this, where 2 corresponds to ##2x^3##. So, one way to express the result is:

##2x^3 – 5x^2 + 0x + 3 + 5/(x+1)##

where the ##5/(x+1)## was written by saying that the last value below the horizontal bar (below -2, 1, 0, -3), being 5, is divided by the ##x pm a## equation such that ##x pm a = 0##. So, ##x+1## indicates that the factor we have just used is ##-1##.

Naturally, if the remainder is 0, you do not have to divide by the ##x pm a## expression.

 

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