If $x - 1$ is a factor of $p (x) = x^{2} + x - 2 = 0$ , then

p (1) = 1

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p (1) = 0

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p (1) = - 1

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p (1) = 2

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Solution

The correct option is C $p (1) = 0$
We know by Factor Theorem that if $x - a$ is a factor of $p (x)$ , then, $p (a) = 0$ .
So, here also since $x - 1$ is a factor of $p (x)$ , then $p (1) = 0$

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