# Help understanding these concepts

• Jul 30th 2008, 08:49 AM
Cillian3
Help understanding these concepts
Hello everyone! Thank you all for your time. I'm trying to freshen up for the upcoming school year with math stuff (my worst subject), and theres several concepts I just can't get down. If someone could explain how you arrived at the answer that would be tremendous, thanks!

(1)The first concept is seen with the equation 14!/1!13!. I just don't understand what I'm supposed to be doing with it.

(2) The next one seemed at first easy to me, but I keep getting it wrong. It simply shows 5x=2. Now I thought I'd simply divide both sides by 5, and get x=2/5, but the answers all have the letter "e" in it. I have no idea why though :(

and lastly, (3) I don't know how to find all the possible number of imaginary zeros of f(x) = 2x3 - 4x2 + 8x - 3 I think I just don't do well with the more odd numbers (i, !, and so on). Thank you very much, I hope this isn't to much of a bother. Thanks!
-Cillian3
• Jul 30th 2008, 09:02 AM
Chris L T521
Quote:

Originally Posted by Cillian3
Hello everyone! Thank you all for your time. I'm trying to freshen up for the upcoming school year with math stuff (my worst subject), and theres several concepts I just can't get down. If someone could explain how you arrived at the answer that would be tremendous, thanks!

(1)The first concept is seen with the equation 14!/1!13!. I just don't understand what I'm supposed to be doing with it.

(2) The next one seemed at first easy to me, but I keep getting it wrong. It simply shows 5x=2. Now I thought I'd simply divide both sides by 5, and get x=2/5, but the answers all have the letter "e" in it. I have no idea why though :(

and lastly, (3) I don't know how to find all the possible number of imaginary zeros of f(x) = 2x3 - 4x2 + 8x - 3 I think I just don't do well with the more odd numbers (i, !, and so on). Thank you very much, I hope this isn't to much of a bother. Thanks!
-Cillian3

1) To solve this, you need to understand the concept of the factorial

Let's say for example, we are asked to evaluate $5!$. All this is simply telling us is that we need to find $5\times4\times3\times2\times1$. When you evaluate a factorial, you take the particular number and multiply to each number that preceeded it..

Also, we can break up factorials as well

What if we had to evaluate $\frac{5!}{4!}$?? We need to find something that will cancel out with the $4!$

Take note that $5!=5\times4\times3\times2\times1$ and $4!=4\times3\times2\times1$ Thus, we see that $5!=5\times\underbrace{4\times3\times2\times1}_{4!} \implies 5!=5\times4!$

Thus, $\frac{5!}{4!}=\frac{5\times4!}{4!}=\color{red}\box ed{5}$

Does this seem to make sense? Why don't you try the problem applying what I said here.

2) Would you post the exact question? I can't really help out for some information is probably missing.

3) Try to determine a possible zero of the function. Apply long division, and then apply the quadratic forumula to the remainder. You will probably see that you'll get you're complex [imaginary] roots. Keep in mind that real values are complex numbers as well [i.e. $1=1+0i, \ \mathbb{R}\in\mathbb{C}$]

I hope this makes sense! (Sun)

--Chris
• Jul 30th 2008, 09:04 AM
Quick
1. "!" is not a number.

$5! = 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot$

So: $\frac{14!}{1!13!}$ is equal to:

$\frac{14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{1\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

and then you can just cancel things out like this:

$\frac{14\cdot \not 13\cdot \not 12\cdot \not 11\cdot \not 10\cdot \not 9\cdot \not 8\cdot \not 7\cdot \not 6\cdot \not 5\cdot \not 4\cdot \not 3\cdot \not 2\cdot \not 1}{1\cdot \not 13\cdot \not 12\cdot \not 11\cdot \not 10\cdot \not 9\cdot \not 8\cdot \not 7\cdot \not 6\cdot \not 5\cdot \not 4\cdot \not 3\cdot \not 2\cdot \not 1}$

So your left with: $\frac{14}{1}$ which is just plain 14
• Jul 30th 2008, 09:10 AM
Chris L T521
Quote:

Originally Posted by Chris L T521
3) Try to determine a possible zero of the function. Apply long division, and then apply the quadratic forumula to the remainder. You will probably see that you'll get you're complex [imaginary] roots. Keep in mind that real values are complex numbers as well [i.e. $1=1+0i, \ \mathbb{R}\in\mathbb{C}$]

I attempted it and got ugly looking zeros. You may need some sort of CAS system to figure it out. If you're looking for the number of possible complex zeros a funtion has, apply the fundamental theorem of algebra: a function of degree $n$ will have $n$ complex zeros.

So in your case, since we have a cubic function, the degree is 3. Thus, there will be 3 complex zeros.

I hope I clarified this.

--Chris
• Jul 30th 2008, 12:38 PM
Cillian3
To Chris:
The exact problem is "solve ln 5x = 2." and the possible answers are as follows: http://kcdistancelearning.blackboard...ges/spacer.gif

http://kcdistancelearning.blackboard...06.p2.q69a.gif

http://kcdistancelearning.blackboard...06.p2.q69b.gif

e2

e5

Again, thank you guys again!
• Jul 30th 2008, 02:58 PM
Quick
Quote:

Originally Posted by Cillian3
To Chris:
The exact problem is "solve ln 5x = 2."

Well, that's your problem. "ln" is a part of the equation, it means "natural log" if you know what a log is, then you should know that the natural log is really $\log_e$

If you don't well then you need to study up on logs, there's a good deal to explain (Speechless)

Anyway, here's how to solve the problem:

$\ln {5x} = 2$

$\ln {5} + \ln{x} = 2$

$\ln{x} = 2 - \ln{5}$

$x = e^{2 - \ln{5}}$

$x = \frac{e^2}{e^{\ln{5}}}$

$x = \frac{e^2}{5}$
• Jul 31st 2008, 08:42 AM
Cillian3
Oh! Okay, sorry, somehow I missed the "In" part! I can't thank you enough!
-Cillian3