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

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

Solution 2

Using the remainder theorem, we see that the remainder upon dividing $x^8$ by $x+\frac{1}{2}$ is equal to $\left(-\frac{1}{2}\right)^8 = \frac{1}{256}.$ Thus, we have that \[x^8 = \left(x+\frac{1}{2}\right)Q(x) + \frac{1}{256}.\] Isolating $Q(x),$ we see \[\frac{x^8-\frac{1}{256}}{x+\frac{1}{2}} = Q(x).\] Using the difference of squares factorization repeatedly, we get \[\frac{\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(x^2+\frac{1}{2^2}\right)\left(x^4+\frac{1}{2^4}\right)}{x+\frac{1}{2}}=\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2^2}\right)\left(x^4+\frac{1}{2^4}\right)=Q(x).\] Finally, plugging in $x=-\frac{1}{2}$ again, the final answer is \[(-1)\left(\frac{1}{2}\right)\left(\frac{1}{8}\right)\] \[=\boxed{-\frac{1}{16}}.\]

~anduran

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_25&oldid=225998"