Difference between revisions of "1961 AHSME Problems/Problem 22"
Rockmanex3 (talk | contribs) (Solution to Problem 22) |
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<cmath>(3x^2+4)(x-3)</cmath> | <cmath>(3x^2+4)(x-3)</cmath> | ||
Thus, the answer is <math>\boxed{\textbf{(C)}}</math>. | Thus, the answer is <math>\boxed{\textbf{(C)}}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2515 - AMBRIGGS | ||
==See Also== | ==See Also== | ||
Latest revision as of 11:26, 30 July 2022
Contents
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 
.
![\[3 \cdot 3^3 - 9 \cdot 3^2 + 3k - 12 = 0\]](http://latex.artofproblemsolving.com/7/c/0/7c0eb57f2a03ed1f2993a59257c06a51dfb10b1a.png)
Combine like terms to get
![\[3k - 12 = 0\]](http://latex.artofproblemsolving.com/9/d/4/9d47f58aa8ef6e6f7ae18dbff024fb15861db496.png)
Thus, 
. The cubic is 
, and it can be factored (by grouping or synthetic division) into
![\[(3x^2+4)(x-3)\]](http://latex.artofproblemsolving.com/8/2/4/824bdbe8f7221e29381ffc3001e4911791fc4370.png)
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 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
