If q_1(x) and r_1 are the quotient and remainder, respectively, when the polynomial x^8 is divided by x+12, and if q_2(x) and r_2 are the quotient and remainder, respectively, when q_1(x) is divided by x+12, then r_2 equals

1979 AMC 12 Problem 25

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25.

q_1(x)

and

r_1

are the quotient and remainder, respectively, when the polynomial

x^8

is divided by

x+\frac12,

and if

q_2(x)

and

r_2

are the quotient and remainder, respectively, when

q_1(x)

is divided by

x+\frac12,

then

r_2

equals

\frac1{256}

-\frac1{16}

1

-16

256

Let

a=-\frac12

and express the first quotient using

x^8-a^8

Apply the remainder theorem to

q_1(x)

x=a

Let

a=-\frac{1}{2}.

Since the first remainder is

a^8,

\begin{aligned} q_1(x)&=\frac{x^8-a^8}{x-a}\\ &=x^7+ax^6+\cdots+a^7. \end{aligned}

Therefore the second remainder is

\begin{aligned} q_1(a)&=8a^7\\ &=8(-\frac{1}{2})^7\\ &=-\frac{1}{16}. \end{aligned}

Therefore, the correct answer is B.