Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1979 AHSME Problems/Problem 25"

(Created page with "== Problem 25 == If <math>q_1 ( x )</math> and <math>r_ 1</math> are the quotient and remainder, respectively, when the polynomial <math>x^ 8</math> is divided by <math>x +...")
 
Line 14: Line 14:
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
−
First, we divide <math>x^8</math> by <math>x+\frac{1}{2}</math> using synthetic division or some other method. The quotient is <math>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}</math>, and the remainder is <math>\frac{1}{128}</math>. Then we plug the solution to <math>x+\frac{1}{2} = 0</math> into the quotient to find the remainder. Notice that every term in the quotient, when <math>x=-\frac{1}{2}</math>, evaluates to <math>-\frac{1}{128}</math>. Thus <math>r_2 =-\frac{8}{128} = \boxed{\textbf{(B) } -\frac{1}{16}}</math>.
+
First, we divide <math>x^8</math> by <math>x+\frac{1}{2}</math> using synthetic division or some other method. The quotient is <math>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}</math>, and the remainder is <math>\frac{1}{128}</math>. Then we plug the solution to <math>x+\frac{1}{2} = 0</math> into the quotient to find the remainder. Notice that every term in the quotient, when <math>x=-\frac{1}{2}</math>, evaluates to <math>-\frac{1}{128}</math>. Thus <math>r_2 =-\frac{8}{128} = \boxed{\textbf{(B) } -\frac{1}{16}}</math>.

Revision as of 12:58, 10 January 2017

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 $\frac{1}{128}$. 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 $r_2 =-\frac{8}{128} = \boxed{\textbf{(B) } -\frac{1}{16}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_25&oldid=82274"