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1979 AHSME Problems/Problem 25

Revision as of 19:34, 29 August 2024 by Anduran (talk | contribs) (→‎Solution 2)
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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
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All AHSME Problems and Solutions

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