Given that when P(x) is divided by (x+2), the remainder is 3, we have P(-2) = 3.

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Step 1: Apply the Remainder Theorem. Let the polynomial be $P(x)$. According to the Remainder Theorem, if a polynomial $P(x)$ is divided by $(x-c)$, the remainder is $P(c)$. Given that when $P(x)$ is divided by $(x+2)$, the remainder is $3$, we have $P(-2) = 3$. Given that when $P(x)$ is divided by $(x-1)$, the remainder is $-1$, we have $P(1) = -1$.

Step 2: Express the polynomial in terms of the new divisor and remainder. When $P(x)$ is divided by $(x+2)(x-1)$, the divisor is a quadratic polynomial. Therefore, the remainder must be a linear polynomial of the form $Ax + B$. We can write $P(x)$ as: $P(x) = Q(x)(x+2)(x-1) + (Ax + B)$ where $Q(x)$ is the quotient and $Ax + B$ is the remainder.

Step 3: Use the given conditions to form a system of equations. Substitute $x = -2$ into the equation for $P(x)$: $P(-2) = Q(-2)(-2+2)(-2-1) + (A(-2) + B)$ $3 = Q(-2)(0)(-3) + (-2A + B)$ $3 = -2A + B \quad (*)$ Substitute $x = 1$ into the equation for $P(x)$: $P(1) = Q(1)(1+2)(1-1) + (A(1) + B)$ $-1 = Q(1)(3)(0) + (A + B)$ $-1 = A + B \quad (**)$

Step 4: Solve the system of linear equations for $A$ and $B$. We have the system:

1. $-2A + B = 3$
2. $A + B = -1$ Subtract equation (2) from equation (1): $(-2A + B) - (A + B) = 3 - (-1)$ $-2A - A + B - B = 3 + 1$ $-3A = 4$ $A = -\frac{4}{3}$ Substitute the value of $A$ into equation (2): $-\frac{4}{3} + B = -1$ $B = -1 + \frac{4}{3}$ $B = -\frac{3}{3} + \frac{4}{3}$ $B = \frac{1}{3}$

Step 5: State the remainder. The remainder is $Ax + B$. Substitute the values of $A$ and $B$: $\text{Remainder} = -\frac{4}{3}x + \frac{1}{3}$

The remainder when the polynomial is divided by $(x+2)(x-1)$ is $\boxed{-\frac{4}{3}x + \frac{1}{3}}$.

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