MHB Is the Remainder of Polynomial $f(x)$ the Same for Two Different Divisors?

[anemone]

Show that the remainder of the polynomial $f(x)=2008+2007x+2006x^2+\cdots+3x^{2005}+2x^{2006}+x^{2007}$ is the same upon division by $x(x+1)$ as upon division by $x(x+1)^2$.

 

[kaliprasad]

Spoiler

$f(x)=2008+2007x+2006x^2+\cdots+3x^{2005}+2x^{2006}+x^{2007}$1st we provide the premise

this shall have same remainder when divided by $x(x+1)$ and $x(x+1)^2$

provided this shall have same remainder when divided by $(x+1)$ and $(x+1)^2$

when we divide by (x+1)^2 we shall have a linear polynomial say m(x+1) + c and when we divide by (x+1) it shall be d and both are same if m= 0 and m+c = d

so if we convert
f(x) as a polynomial of (x+1) the coefficient of x should be zero

or f(x-1) should have coefficient of x to be zero

now we provide the solution based on premise

$f(x-1)=2008+2007(x-1)+2006(x-1)^2+\cdots+3(x-1)^{2005}+2(x-1)^{2006}+(x-1)^{2007}$

should have coefficient of x to be zero

the coefficient of x
$= 2007 + 2006 * (-2) + 2005 * 3 \cdots + 2 * 2006 - 2007 = 0$

hence proved

 

[anemone]

Spoiler

$f(x)=2008+2007x+2006x^2+\cdots+3x^{2005}+2x^{2006}+x^{2007}$1st we provide the premise

this shall have same remainder when divided by $x(x+1)$ and $x(x+1)^2$

provided this shall have same remainder when divided by $(x+1)$ and $(x+1)^2$

when we divide by (x+1)^2 we shall have a linear polynomial say m(x+1) + c and when we divide by (x+1) it shall be d and both are same if m= 0 and m+c = d

so if we convert
f(x) as a polynomial of (x+1) the coefficient of x should be zero

or f(x-1) should have coefficient of x to be zero

now we provide the solution based on premise

$f(x-1)=2008+2007(x-1)+2006(x-1)^2+\cdots+3(x-1)^{2005}+2(x-1)^{2006}+(x-1)^{2007}$

should have coefficient of x to be zero

the coefficient of x
$= 2007 + 2006 * (-2) + 2005 * 3 \cdots + 2 * 2006 - 2007 = 0$

hence proved

Ops, I want to apologize to kaliprasad for not replying to your solution earlier...sorry, Kali!:(

Thanks for your participation and your great solution by the way.

I have another solution that I want to share here with you and MHB:

Spoiler

We have that

$\begin{align*}f(x)&=2008+1004x+1003(x^3+2x^2+x)+\cdots+3(x^{2003}+2x^{2002}+x^{2001})+2(x^{2005}+2x^{2004}+x^{2003})+(x^{2007}+2x^{2006}+x^{2005})\\&=x(x+1)^2(1003+\cdots+3x^{2000}+2x^{2002}+x^{2004})+(1004x+2008)\end{align*}$

from which the result follows with remainder $1004x+2008$