Simplifying Polynomial: (1/(2s+3))

[Ry122]

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.

 

[HallsofIvy]

First, it's not a polynomial, it is a rational function.
Is this it?
\frac{\frac{s+1}{2+s}}{(s+1+ \frac{s+1}{s+2})(s+1)}
The first thing I would do is that addition in the denominator:
s+1+ \frac{s+1}{s+2}= \frac{(s+1)(s+2)+ s+1}{s+2}
which we can then write as
(s+1)\frac{s+2+ 1}{s+2}= (s+1)\frac{s+3}{s+2}

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

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

Can you finish that?

 

[Ray Vickson]

First, it's not a polynomial, it is a rational function.
Is this it?
\frac{\frac{s+1}{2+s}}{(s+1+ \frac{s+1}{s+2})(s+1)}
The first thing I would do is that addition in the denominator:
s+1+ \frac{s+1}{s+2}= \frac{(s+1)(s+2)+ s+1}{s+2}
which we can then write as
(s+1)\frac{s+2+ 1}{s+2}= (s+1)\frac{s+3}{s+2}

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

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

Can you finish that?

No: the full denominator is just (s+1)^2 \frac{s+3}{s+2} .

RGV

 

[zgozvrm]

According to the link the OP posted, the original "polynomial" is this:

\frac{\frac{s+1}{2+s}}{(s+1)+\frac{s+1}{s+2}(s+1)}


I would start by multiplying by
\frac{s+2}{s+2}


Also, note that the numerator can be restated as
\frac{s+1}{s+2}
so that you have
\frac{\frac{s+1}{s+2}}{(s+1)+\frac{s+1}{s+2}(s+1)}

 

[Ry122]

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?

 

[Mark44]

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).