# 1979 AHSME Problems/Problem 25

## Problem 25

If $q_1 ( x )$ and $r_ 1$ are the quotient and remainder, respectively, when the polynomial $x^ 8$ is divided by $x + \tfrac{1}{2}$ , and if $q_ 2 ( x )$ and $r_2$ are the quotient and remainder, respectively, when $q_ 1 ( x )$ is divided by $x + \tfrac{1}{2}$, then $r_2$ equals $\textbf{(A) }\frac{1}{256}\qquad \textbf{(B) }-\frac{1}{16}\qquad \textbf{(C) }1\qquad \textbf{(D) }-16\qquad \textbf{(E) }256$

## Solution

Solution by e_power_pi_times_i

First, we divide $x^8$ by $x+\frac{1}{2}$ using synthetic division or some other method. The quotient is $x^7-\frac{1}{2}x^6+\frac{1}{4}x^5-\frac{1}{8}x^4+\frac{1}{16}x^3-\frac{1}{32}x^2+\frac{1}{64}x-\frac{1}{128}$, and the remainder is $\frac{1}{256}$. Then we plug the solution to $x+\frac{1}{2} = 0$ into the quotient to find the remainder. Notice that every term in the quotient, when $x=-\frac{1}{2}$, evaluates to $-\frac{1}{128}$. Thus $r_2 =-\frac{8}{128} = \boxed{\textbf{(B) } -\frac{1}{16}}$.

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