Difference between revisions of "1970 AHSME Problems/Problem 11"
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== Solution == | == Solution == | ||
− | <math>\fbox{E}</math> | + | From the Remainder Theorem, we have <math>2(-2)^3 - h(-2) + k = 0</math> and <math>2(1)^3 - h(1) + k = 0</math>. Simplifying both of those equations gives <math>-16 + 2h + k = 0</math> and <math>2 - h + k = 0</math>. Since <math>k = 16 - 2h</math> and <math>k = h - 2</math>, we set those equal to get: |
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+ | <math>16 - 2h = h - 2</math> | ||
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+ | <math>3h = 18</math> | ||
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+ | <math>h = 6</math> | ||
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+ | This gives <math>k = 4</math> when substituting back into either of the two equations, and <math>|2h - 3k| = 0</math>, which is answer <math>\fbox{E}</math>. | ||
== See also == | == See also == |
Latest revision as of 20:26, 13 July 2019
Problem
If two factors of are and , the value of is
Solution
From the Remainder Theorem, we have and . Simplifying both of those equations gives and . Since and , we set those equal to get:
This gives when substituting back into either of the two equations, and , which is answer .
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |
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