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The problem is given as follows:

Let $p(x) = x^{2004} - x^{1901} - 50$. What is the remainder of the division of $p(x)$ by $(x-1)^2$.

The solution is straightforward when using the derivative of $p(x)$. However, considering that I stumbled upon this problem in a high school textbook, I'm assuming that an elegant solution exists without the use of calculus?

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Let $$ x^{2004}-x^{1901}-50=q(x)(x-1)^2+ax+b. $$ Translating by $-1$, we have $$\begin{align*} q(x+1)x^2+ax+(a+b)&=(x+1)^{2004}-(x+1)^{1901}-50\\&=\sum_{j=0}^{2004}\binom{2004}jx^j-\sum_{j=0}^{1901}\binom{1901}jx^j-50 \\&= \text{(higher order terms})+103x-50. \end{align*}$$ This gives $a=103$ and $b=-153$.

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Write: $$x^{2004} - x^{1901} - 50 = (x-1)^2k(x)+ax+b$$

Setting $x=1$ we get $$-50 = a+b\implies b=-50-a$$

so $$x^{2004} - x^{1901} - 50 = (x-1)^2k(x)+ax-a-50$$

so $$x^{1901}(x^{103}-1) = (x-1)^2k(x)+a(x-1)$$

so $$ x^{1901}(x^{102}+ x^{101}+...+x+1)= (x-1)k(x)+a$$

and finally putting $x=1$ again we get $$103 =a\implies r(x)=103x-153$$

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$x\!=\!1\,\Rightarrow\, p = -50 + (x\!-\!1)\color{#c00}g.\ $ $\, g = \dfrac{p+50}{x-1}=x^{1901}\,\dfrac{x^{103}\!-\!1}{x\!-\!1} = x^{1901}(x^{102}+\cdots+x+1)\,$

so $\,x\!=\!1\,\Rightarrow\,\color{#c00}g = 103+(x\!-\!1)h\ $ thus $\ p = -50+103(x\!-\!1)+(x\!-\!1)^2h$

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  • $\begingroup$ We twice applied the Remainder Theorem $\,f(x) = f(1) = (x-1)g(x)\,$ for some polynomial $\,g(x)\ $ See also the Double Root Test in algebraic form. $\endgroup$ – Bill Dubuque Mar 14 at 15:38

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