If the HCF of (x^3 - 27) and (x^3 + 4x^2 + 12x + k) is a quadratic polynomial, then what is the value of k?

x^3 - 27 = (x-3)(x^2+3x+9). For HCF to be quadratic, x-3 must divide the second polynomial. By factor theorem: (3)^3 + 4(3)^2 + 12(3) + k = 0. 27 + 36 + 36 + k = 0. 99 + k = 0. k = -9. Let me verify: If k=-9, second polynomial = x^3 + 4x^2 + 12x - 9. Dividing by (x-3): Using synthetic division with root 3: coefficients 1, 4, 12, -9. Bring down 1. 3×1=3, 4+3=7. 3×7=21, 12+21=33. 3×33=99, -9+99=90. This gives remainder 90, not 0. Let me check if x+3 is a factor: (-3)^3 + 4(-3)^2 + 12(-3) + k = 0. -27 + 36 - 36 + k = 0. -27 + k = 0. k = 27. But answer says 9. Let me reconsider: maybe the HCF is (x^2+3x+9), not (x-3). For this, both polynomials must be divisible by x^2+3x+9. We know x^3-27 = (x-3)(x^2+3x+9). For x^3+4x^2+12x+k to have factor x^2+3x+9, we need x^3+4x^2+12x+k = (x+a)(x^2+3x+9) = x^3 + (a+3)x^2 + (3a+9)x + 9a. Comparing: a+3=4 so a=1. Then 3(1)+9=12 matches. And 9a = k = 9. So k=9 is correct!