help

What is the remainder when \(x^3 + x^6 + x^9 + x^{27}\) is divided by \(x^2 - 1\)?

\(= x^{25} + x^{23} + x^{21} + x^{19} + x^{17} + x^{15} + x^{13} + x^{11} + x^9 + 2x^7 + 2x^5 + x^4 + 2x^3 + x^2 + 3x + 1 \text{ remainder } 3x + 1 \) 

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

Let

 \(\displaystyle f(x)=x^{3}+x^{6}+x^{9}+x^{27}\quad \text{and}\quad \frac{f(x)}{x^{2}-1}=Q(x)+\frac{R(x)}{x^{2}-1}\\ \text{where R(x) is linear, i.e.}\; R(x)=ax+ b \quad \text{for some}\; a\; \text{and}\; b.\\ \text{Then}\\ f(x)=Q(x)(x^{2}-1)+R(x)\\ \text{so}\\ f(-1)=-2=R(-1)=-a+b\dots\dots(1)\\ f(1)= 4=R(1)=a+b\dots\dots\dots(2)\)

and from (1) and (2), a = 3 and b = 1.

Wow that was some BIG latex! cool!