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I'm doing some old multiple tests. It seems I'm pretty stuck around the topic off complex numbers, could someone elaborate how to: Show that 1 is a multiple root of 2nd degree in p

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up vote 4 down vote accepted

$$p(x) = x^3-x^2 - x + 1 = x^2(x-1) - (x-1) = (x-1) (x^2-1) = (x-1)^2(x+1)$$ Hence, $1$ is a multiple root (of degree $2$, since $(x-1)^2$ divides $p(x)$).

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In general, $\alpha$ is a root of multiplicity $k$ of $p(x)$ if and only if $p^{(i)}(\alpha) = 0$ for $0 \leq i \leq k-1$ (considering $p^{(0)}$ as $p$), but $p^{(k)}(\alpha) \neq 0.$ This is a well-known consequence of the remainder theorem for polynomials.

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Here, I am using "root of multiplicity $k$ to mean $(x-\alpha)^{k}$ is a factor, but $(x-\alpha)^{k+1}$i not. – Geoff Robinson Oct 31 '12 at 14:52

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