Simplifying terms

  • Thread starter Ry122
  • Start date
  • #1
565
2

Homework Statement


I'm trying to simplify this polynomial
((s+1)/(2+s))/((s+1)+((s+1)/(s+2))(s+1))

It's more readable if you view it here:
http://www.wolframalpha.com/input/?i=((s+1)/(2+s))/((s+1)+((s+1)/(s+2))(s+1))

It simplifies to 1/(2s+3)


The Attempt at a Solution



I'm not sure what steps are necessary to reduce it down to that.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
962
First, it's not a polynomial, it is a rational function.
Is this it?
[tex]\frac{\frac{s+1}{2+s}}{(s+1+ \frac{s+1}{s+2})(s+1)}[/tex]
The first thing I would do is that addition in the denominator:
[tex]s+1+ \frac{s+1}{s+2}= \frac{(s+1)(s+2)+ s+1}{s+2}[/tex]
which we can then write as
[tex](s+1)\frac{s+2+ 1}{s+2}= (s+1)\frac{s+3}{s+2}[/tex]

With that,the full denominator is
[tex](s+1)^2(1+ \frac{s+3}{s+2}= (s+1)^2\frac{2s+ 5}{s+2}[/tex]

Also,dividing by a fraction is the same as multiplying by its reciprocal so the entire expression becomes
[tex]\frac{s+1}{s+2}\frac{s+2}{(s+1)^2(2s+5)}[/tex]

Can you finish that?
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
First, it's not a polynomial, it is a rational function.
Is this it?
[tex]\frac{\frac{s+1}{2+s}}{(s+1+ \frac{s+1}{s+2})(s+1)}[/tex]
The first thing I would do is that addition in the denominator:
[tex]s+1+ \frac{s+1}{s+2}= \frac{(s+1)(s+2)+ s+1}{s+2}[/tex]
which we can then write as
[tex](s+1)\frac{s+2+ 1}{s+2}= (s+1)\frac{s+3}{s+2}[/tex]

With that,the full denominator is
[tex](s+1)^2(1+ \frac{s+3}{s+2}= (s+1)^2\frac{2s+ 5}{s+2}[/tex]

Also,dividing by a fraction is the same as multiplying by its reciprocal so the entire expression becomes
[tex]\frac{s+1}{s+2}\frac{s+2}{(s+1)^2(2s+5)}[/tex]

Can you finish that?
No: the full denominator is just [tex](s+1)^2 \frac{s+3}{s+2} .[/tex]

RGV
 
  • #4
754
1
According to the link the OP posted, the original "polynomial" is this:

[tex]\frac{\frac{s+1}{2+s}}{(s+1)+\frac{s+1}{s+2}(s+1)}[/tex]


I would start by multiplying by
[tex]\frac{s+2}{s+2}[/tex]


Also, note that the numerator can be restated as
[tex]\frac{s+1}{s+2}[/tex]
so that you have
[tex]\frac{\frac{s+1}{s+2}}{(s+1)+\frac{s+1}{s+2}(s+1)}[/tex]
 
  • #5
565
2
sorry, my original equation might have been interpreted differently by wolfram. But the wolfram one is what I meant.

how did you know to multiply by (s+2)/(s+2) zgozvrm?
 
  • #6
34,688
6,394
sorry, my original equation might have been interpreted differently by wolfram. But the wolfram one is what I meant.

how did you know to multiply by (s+1)/(s+2) zgozvrm?
Because both the overall numerator and overall denominator had a denominator of s + 2. What zgozvrm actually did was multiply by 1 (which is always legal, since it doesn't change the value of what's being multiplied), in the form of (s + 2)/(s + 2).
 

Related Threads on Simplifying terms

Replies
1
Views
1K
  • Last Post
Replies
5
Views
357
Replies
10
Views
973
Replies
5
Views
2K
  • Last Post
Replies
2
Views
653
  • Last Post
Replies
1
Views
648
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
989
  • Last Post
Replies
2
Views
2K
Top