By Remainder theorem, $p(x) = 2x^{3} + 3x - 1$ divided by $(x + 1)$ leaves remainder

By Remainder theorem, $p(x) = 2x^{3} + 3x - 1$ divided by $(x + 1)$ leaves remainder
  1. $-6$
  2. $6$
  3. $0$
  4. $-2$

Solution

$p(-1) = -2 - 3 - 1 = -6$.

Asked in: MH-SSC-9

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