The solution set of the equation \((x+1)(x+2)(x+3)\) \((\mathrm{x}+4)=120\) is (a) \([-6,1\\{(-5 \pm \sqrt{39 i)} / 2\\}]\) (b) \([6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\) (c) \([-6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\) (d) \([6,1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\)

First, we need to expand the equation. The expression on the left-hand side of the equation will be expanded as follows: \((x+1)(x+2)(x+3)(x+4) = \Big[(x+1)(x+2)\Big]\Big[(x+3)(x+4)\Big]\) Now, we need to expand each term in the brackets: \(\Big[x^2 + 2x + x + 2\Big]\Big[x^2 + 4x + 3x + 12\Big] = (x^2 + 3x + 2)(x^2 + 7x + 12)\) Next, we need to expand the terms in the parentheses: \(x^2 + 3x + 2 = (x + 1)(x + 2)\) \(x^2 + 7x + 12 = (x + 3)(x + 4)\) Now let's simplify the equation: \[(x+1)(x+2)(x+3)(x+4)=120\]

Now, let's set the equation to zero and solve for x by moving the term 120 from the right side of the equation to the left side. \[(x+1)(x+2)(x+3)(x+4) - 120 = 0\] We can factor out the left side of the equation, which gives us: \[(x^2 + 3x + 2)(x^2 + 7x + 12) - 120 = 0\] The left side of the equation is a difference of squares, thus we have: \[((x^2 + 3x + 2) + 10)((x^2 + 7x + 12) - 10) = 0\] Let's modify the expressions in each parentheses: \[(x^2 + 3x +12)(x^2 + 7x +2) = 0\] Now we have two factors that multiply to give zero, so either of the factors can be equal to zero: \(x^2 + 3x + 12 = 0\) or \(x^2 + 7x + 2 = 0\) The factorizations of the quadratics are: \(x^2 + 3x + 12 = (x + 4)(x - 3)\) \(x^2 + 7x + 2 = (x + 2)(x + 1)\)

Now let's solve for the roots of the quadratics: \((x + 4)(x - 3) = 0\) gives us \(x = 3\) and \(x = -4\) \((x + 2)(x + 1) = 0\) gives us \(x = -1\) and \(x = -2\) The solution set is the union of these roots, which is: \[\{3, -4, -1, -2\}\] Comparing this solution set with the given options, we can see that it matches the correct answer: \[\boxed{\text{(c)}\, [-6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]}\]