$p(x) = x^{2} + 2x + 1$. $p(-1)$ equals

$p(x) = x^{2} + 2x + 1$. $p(-1)$ equals
  1. $0$
  2. $1$
  3. $-1$
  4. $4$

Solution

$1 - 2 + 1 = 0$. ($-1$ is a double root.)

Asked in: MH-SSC-9

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