1962 AHSME Problems/Problem 9: Difference between revisions
No edit summary |
|||
| Line 5: | Line 5: | ||
==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>. | |||
Revision as of 15:00, 16 April 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
.
