Final answer:
The polynomial equation 6(x-3)(x^2+4)(x+1)=0 has three solutions, including a complex pair. Its real solutions are x=3 and x=-1. The x^2+4 factor is responsible for the complex solutions.
Step-by-step explanation:
The equation given is 6(x-3)(x^2+4)(x+1)=0. To find the solutions of this polynomial, we need to set each factor equal to zero and solve for x.
The first factor is (x-3): setting it to zero gives x=3. The second factor is (x^2+4): since this is a sum of a square and a positive number, it cannot equal zero because the square of a real number is always non-negative and 4 is positive, meaning the sum is always positive. The third factor is (x+1): setting it to zero gives x=-1.
Thus, the solutions are x=3 and x=-1. The equation has three solutions including a complex pair (from the x^2+4 factor), but only two are real solutions.
Remember that the factor x^2+4 will give complex solutions since it can be rewritten as x^2=-4, whose solutions are 2i and -2i (where i is the imaginary unit).