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 == | ||
| − | {{AHSME box|year=1970|num-b=10|num-a=12}} | + | {{AHSME 35p box|year=1970|num-b=10|num-a=12}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
