Difference between revisions of "1962 AHSME Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | {{ | + | Obviously, we can factor out an <math>x</math> first to get <math>x(x^8-1)</math>. |
| + | Next, we repeatedly factor differences of squares: | ||
| + | <cmath>x(x^4+1)(x^4-1)</cmath> | ||
| + | <cmath>x(x^4+1)(x^2+1)(x^2-1)</cmath> | ||
| + | <cmath>x(x^4+1)(x^2+1)(x+1)(x-1)</cmath> | ||
| + | None of these 5 factors can be factored further, so the answer is | ||
| + | <math>\boxed{\textbf{(B) } 5}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 8|num-a=10}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:14, 3 October 2014
Problem
When 
is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
Solution
Obviously, we can factor out an 
first to get 
.
Next, we repeatedly factor differences of squares:
![\[x(x^4+1)(x^4-1)\]](http://latex.artofproblemsolving.com/7/1/a/71a5be89c8c9837ba82eaad17c15b410d8b34c0e.png)
![\[x(x^4+1)(x^2+1)(x^2-1)\]](http://latex.artofproblemsolving.com/1/a/2/1a263179eedcc1e11ca73f64968ea3928fdc72fa.png)
![\[x(x^4+1)(x^2+1)(x+1)(x-1)\]](http://latex.artofproblemsolving.com/7/0/c/70cb7aa001ef9be463423705f51e5f8d399c166b.png)
None of these 5 factors can be factored further, so the answer is
.
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
