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Difference between revisions of "1979 AHSME Problems/Problem 25"
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Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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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>.
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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}{256}</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>.
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== See also ==
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{{AHSME box|year=1979|num-b=24|num-a=26}} 
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:20, 19 June 2021

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

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

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