Question

f(s) = (s^3-1) / (s^2-1)

How do you divide something like this? Does it factor? If so, how?

Answer 1

yes it factors

Answer 2

difference of cubes/difference of squares

Answer 3

I'm trying to give you a hint

Answer 4

Well so i understand that it ends up being 3/2 but can you show me the steps involved? Because all i get so far is that the denominator can factor to (s+1)(s-1) but what can the numerator factor to?

Answer 5

Everybody wants to help :)

Answer 6

nuemerato = (s-1)(s^2 +s + 1)

Answer 7

- if my memory serves me right!!

Answer 8

Yes try to factor:
f(s) = (s^3 -1)/(s^2 -1)
f(s) = (s -1) (s^2 +s +1) /[(s-1)(s+1)] <--cross out s-1
f(s)= (s^2 +s +1)/ (s +1)

and im not sure how you end up with 3/2 though :D

Answer 9

I think he's thinking about taking the limit with coefficients in the numerator and denominator with polynomials of the same degree.

Answer 10

you can go further to break it up to
f(x)= (s^2 +s )/ (s+1) + 1/(s+1)
f(x) = s(s+1)/(s+1) + 1/(s+1)
f(x) = s + 1/(s+1)

Answer 11

if s =1 then it is 3/2

Answer 12

yeah the limit is 3/2 when x->1 that's what i meant.... but i'm not seeing how (s-1)(s^2 + s +1) = s^3 - 1

Answer 13

oh, so this is a limit question xD

Answer 14

I knew it

Answer 15

yep

Answer 16

\[\color{red}{\frac{s^3-1}{s^2-1}=s + \frac{1}{s-1}}\text{ right?}\]

Answer 17

mighta got a - and + switched

Answer 18

It does?

Answer 19

bovice: that's just a property of difference of cube...
they have formula for it (a^3 - b^3) = (a - b) (a^2 +ab +b^2)

Answer 20

f(x) = X^3 -1 then f(1) = 0 so x-1 is a factor

Answer 21

divide x-1 into x^3 -1 gives x^2 + x+ 1

Answer 22

yeah (s-1) will be cross out instead of x+1

Answer 23

s +1 i meant

Answer 24

I guess so...... still confused on this but i dont really know the steps for diving polynomials

Answer 25

dividing polys are the same as dividing numbers; if you recall how to divide longhand

Answer 26

I knew he would be confused...too many people giving their way of dividing them

Answer 27

hmmm.... x-1 / x^3 -1

x goes into x^3 x^2 times....

Answer 28

that's as far as i can get....dunno what to do next

Answer 29

the one thng you have to watch when u divide polys is the powers must go down by 1
so when dividing into x^3 - 1 u need to write it as

x^3 + 0x^2 + 0x - 1

Answer 30

amistre64 would use an online classroom to show you step by step but he needs an instruction manual to use the whiteboard, lol

Answer 31

\(\color{blue}{\text{s}}\)
\(\color{green}{\text{------------}}\)
\(\color{green}{\text{s^2 - 1| s^3 - 1}}\)
\(\color{red}{\text{-s^3 +s}}\)
\(\color{black}{\text{----------}}\)
\(\color{green}{\text{s - 1}}\) <- remainder

Answer 32

whiteboards are for sissys :)

Answer 33

lol

Answer 34

Wow, how creative

Answer 35

Here is a nice summary of the difference of cubes and difference of squares. :)

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut29_specfact.htm

Answer 36

amistre64 proves he doesn't need a whiteboard

Answer 37

real mathmatickers \(\color{orange}{\textbf{ do it in color}}\) :)

Answer 38

I don't know amistre64....Latex wastes manhours

Answer 39

When you waste manhours, you become a sissy

Answer 40

lol , so true :)

Answer 41

alright i found out how to using that link....thanks

Answer 42

So yeah, should be \[s + {1 \over s+1}\] when all is said and done.

Answer 43

that looked better in preview :)

Answer 44

He used over instead of frac

Answer 45

but everyone's favorite thing to do is use l'hospital's rule

Answer 46

You're so right. I love l'hopital rule

Answer 47

Wait, you spelled hospital wrong...

Answer 48

lol its the thought that counts

Answer 49

l'hopitals rule why didnt i think of that.... x^3 -1 = 3x^2 and x^2-1 = 2x
so 3(1^2) / 2(1) = 3/2 hahahaha woooo

Answer 50

lol

Answer 51

Because you can only use it when you have 0/0 or infinity/infinity

Answer 52

we do have 0/0

Answer 53

lol, you're right...nobody thought to do it. I blame the OP

Answer 54

Well he originally didn't say he was looking for a limit.

Answer 55

right im glad i came late in the conversation

Answer 56

:)

Answer 57

exactly...but he did mention it though, but by the time he did, everyone was already working on the hard way...

Answer 58

Hey Bovice, I have your alternate solution. factor the numerator and the denominator as before but then substitute s = 1 into the function to get 3/2.

Answer 59

lim x --> 1 of ((s-1)(s^2 + s+1))/((s-1)(s+1)

Answer 60

Simplify: lim x --> 1 (s^2+s+1)/(s+1)

Answer 61

substitute 1: (1+1+1)/(1+1) = 3/2