1962 AHSME Problems/Problem 9

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 $\boxed{\textbf{(B) } 5}$ .

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1962 AHSME Problems/Problem 9

Problem

Solution