Difference between revisions of "1962 AHSME Problems/Problem 9"

(Created page with "==Problem== When <math>x^9-x</math> is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is: <math> \textbf{(A...")
 
m (Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
"Unsolved"
+
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 22:14, 3 October 2014

Problem

When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:

$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$

Solution

Obviously, we can factor out an $x$ first to get $x(x^8-1)$. Next, we repeatedly factor differences of squares: \[x(x^4+1)(x^4-1)\] \[x(x^4+1)(x^2+1)(x^2-1)\] \[x(x^4+1)(x^2+1)(x+1)(x-1)\] None of these 5 factors can be factored further, so the answer is $\boxed{\textbf{(B) } 5}$.

See Also

1962 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png