Find \( s \) and \( r \) if \( (x+3) \) is a factor of \( s x^{3}+r^{2} x^{2}-28 x+15 \) and has a remainder of -60 when divided by \( (x-3) \).

To find the values of \( s \) and \( r \), we can use the Factor Theorem and the Remainder Theorem. First, since \( (x+3) \) is a factor, we can use the fact that if \( (x+3) \) is a factor of the polynomial, then substituting \( x = -3 \) into the polynomial will yield zero: \[ s(-3)^3 + r^2(-3)^2 - 28(-3) + 15 = 0. \] Calculating this step by step: \[ s(-27) + r^2(9) + 84 + 15 = 0, \] which simplifies to: \[ -27s + 9r^2 + 99 = 0. \] Thus, we have our first equation: \[ -27s + 9r^2 + 99 = 0. \quad \text{(1)} \] Next, we turn to the condition that the polynomial has a remainder of -60 when divided by \( (x - 3) \). According to the Remainder Theorem, we can substitute \( x = 3 \) into the polynomial: \[ s(3)^3 + r^2(3)^2 - 28(3) + 15 = -60. \] Calculating this: \[ s(27) + r^2(9) - 84 + 15 = -60, \] which simplifies to: \[ 27s + 9r^2 - 69 = -60. \] Thus, we have our second equation: \[ 27s + 9r^2 - 69 = -60. \quad \text{(2)} \] Now we can simplify equation (2): \[ 27s + 9r^2 = 9. \] From this, we can divide everything by 9: \[ 3s + r^2 = 1. \quad \text{(3)} \] Now we will solve equations (1) and (3) simultaneously. Starting with (3): \[ r^2 = 1 - 3s. \] We can substitute \( r^2 \) from (3) back into (1): \[ -27s + 9(1 - 3s) + 99 = 0, \] which expands to: \[ -27s + 9 - 27s + 99 = 0. \] This simplifies to: \[ -54s + 108 = 0. \] Solving for \( s \): \[ 54s = 108 \implies s = 2. \] Now substitute \( s = 2 \) back into equation (3) to find \( r^2 \): \[ 3(2) + r^2 = 1, \] which simplifies to: \[ 6 + r^2 = 1 \implies r^2 = 1 - 6 = -5. \] Since \( r^2 = -5 \) yields no real solutions, \( r \) must take on complex values. Therefore: \[ r = i\sqrt{5} \quad \text{or} \quad r = -i\sqrt{5}. \] So the final solution is: \[ s = 2, \quad r = \pm i\sqrt{5}. \]