Use the Remainder Theorem to determine whether $x+1$ is a divisor of \[ P(x)=2 x^{3}+4 x^{2}-2 x+12 \] (1 point) $x+1$ is a divisor of $P(x)$ $x+1$ is not a divisor of $P(x)$ $x+1$ is a partial divisor of $P(x)$ Not enough information has been provided.

Step 1 ：The Remainder Theorem states that if a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). So, to check if \(x+1\) is a divisor of \(P(x)\), we need to substitute \(x=-1\) into \(P(x)\) and check if the result is zero. If the result is zero, then \(x+1\) is a divisor of \(P(x)\). If the result is not zero, then \(x+1\) is not a divisor of \(P(x)\).

Step 2 ：Substitute \(x=-1\) into \(P(x)\) to get \(P(-1) = 2(-1)^3 + 4(-1)^2 - 2(-1) + 12 = 16\).

Step 3 ：Since \(P(-1)\) is not equal to zero, \(x+1\) is not a divisor of \(P(x)\).

Step 4 ：Final Answer: \(x+1\) is \(\boxed{\text{not a divisor of } P(x)}\).