Difference between revisions of "1979 AHSME Problems/Problem 25"
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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}{ | + | 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>. |
== See also == | == See also == | ||
− | {{AHSME box|year=1979|num-b=24|num-a= | + | {{AHSME box|year=1979|num-b=24|num-a=26}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:20, 19 June 2021
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 .
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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