Melody

What is the remainder when   x^3 + x^6 + x^9 + x^27   is divided by   x^2 - 1  ?     

Well answered by Heureka and Tiggsy.   Thanks.

Thanks Tiggsy,

That makes good sense :)

Thanks very much Tiggsy.  I have not studied it yet but I will.  

Thanks Heureka,

What is the remainder when x^3 + x^6 + x^9 + x^27 is divided by x^2 - 1?

Heureka has done this the standard way but I just wanted to see if I could figure out how to do it without the 

algebraic division.  I think my method is valid but I am not 100% sure on the divide by 2 bit.  I think it is valid though.

\(x^2-1=(x-1)(x+1)\\ let\;\;f(x)=x^3+x^6+x^9+x^{27}\\ \text{When f(x) is divided by x+1 the remainder will be }f(-1)\\ f(-1)=-1+1-1-1=-2\\ \frac{x^3+x^6+x^9+x^{27}}{(x+1)(x-1)}\\ =\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\ =\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\ =g(x)+\frac{(1+1+1+1+2)/2}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\\qquad \qquad \text{I divided by 2 because I factored out (x+1)}\\ \\ \qquad \qquad if \;\;Q(x)=x+1\;\;then\;\;Q(1)=2\\~\\ =g(x)+\frac{3}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\ =g(x)+\frac{3(x+1)}{(x-1)(x+1)}+\frac{-2}{(x+1)(x-1)}\\ =g(x)+\frac{3x+1}{(x-1)(x+1)}\\ \text{so the remainder will be }3x+1\)

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coding

x^2-1=(x-1)(x+1)\\
let\;\;f(x)=x^3+x^6+x^9+x^{27}\\
\text{When f(x) is divided by x+1 the remainder will be }f(-1)\\
f(-1)=-1+1-1-1=-2\\
\frac{x^3+x^6+x^9+x^{27}}{(x+1)(x-1)}\\
=\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\
=\frac{x^3+x^6+x^9+x^{27}+2}{(x+1)(x-1)}+\frac{-2}{(x+1)(x-1)}\\
=g(x)+\frac{(1+1+1+1+2)/2}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\\qquad \qquad \text{I divided by 2 because I factored out (x+1)}\\
\\ \qquad \qquad if \;\;Q(x)=x+1\;\;then\;\;Q(1)=2\\~\\
=g(x)+\frac{3}{(x-1)}+\frac{-2}{(x+1)(x-1)}\\
=g(x)+\frac{3(x+1)}{(x-1)(x+1)}+\frac{-2}{(x+1)(x-1)}\\
=g(x)+\frac{3x+1}{(x-1)(x+1)}\\
\text{so the remainder will be }3x+1

That looks very impressive Heureka !

Thanks Tiggsy    

Forget your condescending "Hint"

You are, by choice, a nobody.  

You are addressing acknowledge Mathematicians.

If you believe you know the correct answer then spell out your reasons and give the answer, as you see it, in full.

Or if you are incapable of doing that then state why you believe a different answer is correct.

Like you could day. "The answer that the teacher gave me is .... but i do not know how she got it."

You are very welcome :)

How do you know if the graph of a fractional function has asymptotes?

Firstly ou look for values that x or y cannot be.  That will take care of the vertical and horizontal asymptotes.

For instance

You cannot divide by 0 so   

   for instance in       \(y=\frac{3}{x+2}\)

x=-2 would be an asymptote, 

also 3 divided by anything cannot be 0 so   y=0 is also an asymptote.

You do have to be careful though because sometimes these values are holes in the graph and not asymptotes at all.

It takes practice to work out the nuances of graphing.

After you think about what a graph might look like enter it into desmos graphing calculator to see if you are right.

That will help you to learn.