If $\displaystyle \left ( x + p - 1 \right )$ is a factor of the expression $\displaystyle x^{2} + px + 1 - p$, then the roots of the equation $\displaystyle x^{2} + px + 1 = p$ are

Using the factor theorem, if (x + p - 1) is a factor of x^2 + px + 1 - p, then substituting x = 1 - p must yield zero: (1-p)^2 + p(1-p) + 1 - p = 1 - 2p + p^2 + p - p^2 + 1 - p = 2 - 2p = 0, which means p = 1. Substituting p = 1 into the equation x^2 + px + 1 = p gives x^2 + x = 0. Factoring this yields x(x + 1) = 0, so the roots are 0 and -1.

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