Show that $(x - 1)$ is a factor of $(x^{n} - 1)$

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Solution

Let $p (x) = x^{n} - 1$
In order to show that $(x - 1)$ is a factor of $(x^{n} - 1)$ , it is sufficient to show that $p (1) = 0$ .
Now, $p (x) = x^{n} - 1 ∴ p (1) = 1^{n} - 1 = 1 - 1 = 0$
$∴ (x - 1)$ is a factor of $(x^{n} - 1)$

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