1961 AHSME Problems/Problem 22
Contents
[hide]Problem
If is divisible by , then it is also divisible by:
Solution
If is divisible by , then by the Remainder Theorem, plugging in in the cubic results in . Combine like terms to get Thus, . The cubic is , and it can be factored (by grouping or synthetic division) into Thus, the answer is .
Video Solution
https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2515 - AMBRIGGS
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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