Math Homework Help DESPERATLY NEEDED!!!

since the degree of the polynomial is 5, there are five solutions. there is one pos, no negs. therefore, there must be 4 imaginary, which makes sense. the graph crosses the x-axis at approx 2.408 which is the 5th root of 81. you're all saying stuff I already've said.

 

Okay, I'm with you now.

x^5=81
assuming x is a complex number, change to polar form, which will give you.

x^5 = 81 cis 0

(cis is just an abbreviation for (cos(theta) + isin(theta))

now, each root w is given by the formula:

w(k) = r^(1/n) cis ((theta + 2*k*pi)/n)

in your case, n=5, r=81, theta =0 and k= the number of the root you want, 0,1,2,3 and 4, which will give you five roots.

The first root, k=0 will just give you.

w(0)=2.480 cis (0)

The second root, k=1, will give you:

w(1)=2.480 cis (2pi/5)

and so on

w(3) = w.480cis(4pi/5)

you might have to change those solutions back to rectangular form.

edit: if that made so sense whatever, tell me, and I'll see what I can do.

 

Yes, Kitt is right. Do you know what cyclotomic polynomials are, Liz? If you do, that will make my job easier, otherwise just try to follow what Kitt said.

An easy way to visualise this is to think of a regular pentagon being drawn, with the center of the pentagon, as the origin of complex plane, and one of the vertices is on the real axis. One of the roots is the real root of 81^(1/5), just use your calculator to find that. The other roots will be the same distance away from the origin, at the other vertices of the pentagon. That's how to visualise it.

Hope this helps!

 

What Kitt said.

But make sure you have studied that in class (or in a previous class) before using it. It would be kind of strange to use it in something like a high school algebra class.

ps: the above is based on the fact that I think I remember learning that in calc II or III.

 

Yes, Kitt, got it, although I already forgot everything about complex numbers from my Algebra class.

 

eh...we haven't had that yet in class,....so I don't follow at all. sorry.

 

I don't remember messing with polar co-ordinates until secend semester Calc.

 

We don't learn De Moivre's theorem until the end of this year at my school. That's what this is about, also "roots of complex numbers" is the unit in my textbook.

 

What math class is this?

 

It's already been stated that she is in Pre-Calc.

I'm in Calculus AP-AB and I haven't learned the thing that Kitt explained. WE're doing Integrals now....fuuun.

Anyway Liz the only way that I can think of to get the other solutions is to use 81^(1/5) in synthetic division...although now that I think of it you'd have to have one of the imaginar roots. :-(

 

Oh, I won't be doing that till I'm a junior. Sorry *leaves thread*

 

Integrals aren't that bad in AP calc. When you do them in second semster calc, they get a bit harder. Like the ones that just keep going, or the ones you have to use the trig functions to substitute, etc.

 

Anyway Liz the only way that I can think of to get the other solutions is to use 81^(1/5) in synthetic division...although now that I think of it you'd have to have one of the imaginar roots. :-(


yeah, I did. I got stuff with all the roots of 81. the big question is: now what? It doesn't facor, it doesn't do anything. I split it into two equations and did the quadratic eqation, so now I have 3 answers. I'm just tempted to take the quadratic answer and say multpilicy of two.

 

I forget how to do synthetic division. Could you remind me?

7

 

here's your equation: x^3+2x^2-3x+4

you take the co-efficents:

1 2 -3 4

and the number you want to see if it goes into. let's use 2 here.

so:

..1...2...-3...4

..____2____8___10____
2:1...4....5...14

since the last number is 14, it doesn't go in perfectly.

first you bring down the one. then you mult and add, mult and add, mult and add.


edit: maybe the ...s will make the psaces actually be there...

 

Hmm, that didn't help a whole lot. Where did the 14 come from?

 

4+10. my lines are screwy.

 

Oh, ok, I see, but where did the 2, 8, and 10 come from?

Edit: Wait, I think I got it now. If it came out 0 instead of 14, 2 goes in ok?

 

the two is an answer you want to check. see if x could equal 2.


steps:

1)
..1...2...-3...4

.._______________
2:1


pull down the one.

2)

..1...2...-3...4

..____2___________
2:1...4

2*1 = 2
2+2 = 4

3)

..1...2...-3...4

..____2____8_______
2:1...4....5

2*4=8
-3+8=5

4)

..1...2...-3...4

..____2____8___10____
2:1...4....5...14

2*5=10
4+10=14

if the last number was 0, then it would be a factor.

 

Yeah, I figured out what you were doing right as you posted that.

Edit: But that only gets you to 81^(1/5), which you already know is an answer.

 

thank god my Pre-calc wasn't that hard...sheesh..sounds like you guys are discussing stereo instructions or something!

 

Ardens: Just curious, do you learn the proof of de Moivre's theorem when learning the theorem itself? Here in Singapore we don't learn the proof. Oh well.

 

Liz, how much work have you done with complex numbers?

Have they taught you how to change between polar and rectangular form, how to add and subtract them, divide and multiply them, etc?
If not, then you can't do that problem, unfortunately. (or fortunately, depending on how you feel about math )

 

All I know is that there is 1 real root and 4 imaginary roots

 

heh, the teacher explained the problem today. seems she left out an "X". yeppers, the prob should have been:

X^5 - 81X = 0


so you just pull out an X and go from there.