Difference between revisions of "1979 AHSME Problems/Problem 25"
(Created page with "== Problem 25 == If <math>q_1 ( x )</math> and <math>r_ 1</math> are the quotient and remainder, respectively, when the polynomial <math>x^ 8</math> is divided by <math>x +...") |
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Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
− | 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 | + | 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>. |
Revision as of 11:58, 10 January 2017
Problem 25
If and are the quotient and remainder, respectively, when the polynomial is divided by , and if and are the quotient and remainder, respectively, when is divided by , then equals
Solution
Solution by e_power_pi_times_i
First, we divide by using synthetic division or some other method. The quotient is , and the remainder is . Then we plug the solution to into the quotient to find the remainder. Notice that every term in the quotient, when , evaluates to . Thus .