When a polynomial \( P(x) \) is divided by a linear factor \( x - r \), the Remainder Theorem states that the remainder is \( P(r) \). In this case, since \( x + 3 \) is a divisor of the polynomial \( P(x) = x^3 - 3x^2 - 13x + 15 \), we can find the corresponding value of \( r \) by setting \( x + 3 = 0 \), which gives \( r = -3 \).
According to the Remainder Theorem, if \( x + 3 \) divides \( P(x) \) evenly, then \( P(-3) \) must equal zero.
Let's compute \( P(-3) \):
\[ P(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15 \] \[ = -27 - 3(9) + 39 + 15 \] \[ = -27 - 27 + 39 + 15 \] \[ = -54 + 54 \] \[ = 0 \]
Thus, since \( P(-3) = 0 \), the remainder when dividing \( P(x) \) by \( x + 3 \) is \( 0 \). This confirms that \( x + 3 \) is indeed a factor of the polynomial \( P(x) \). Therefore, we conclude that the remainder is 0.