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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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==Solution 2==
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Using the remainder theorem, we see that the remainder upon dividing <math>x^8</math> by <math>x+\frac{1}{2}</math> is equal to <math>\left(-\frac{1}{2}\right)^8 = \frac{1}{256}.</math> Thus, we have that
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<cmath>x^8 = \left(x+\frac{1}{2}\right)Q(x) + \frac{1}{256}.</cmath>
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Isolating <math>Q(x),</math> we see
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<cmath>\frac{x^8-\frac{1}{256}}{x+\frac{1}{2}} = Q(x).</cmath>
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Using the difference of squares factorization repeatedly, we get
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<cmath>\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).</cmath>
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Finally, plugging in <math>x=-\frac{1}{2}</math> again, the final answer is
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<cmath>(-1)\left(\frac{1}{2}\right)\left(\frac{1}{8}\right)</cmath>
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<cmath>=\boxed{-\frac{1}{16}}.</cmath>
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~anduran
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== See also ==
 +
{{AHSME box|year=1979|num-b=24|num-a=26}} 
 +
 
 +
[[Category: Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:34, 29 August 2024

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

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