Rational factors of x3+9x2-108 are found by looking at the factors of 108: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108. Start with plugging in x=1: 1+9-108≠0; x=2: 8+36-108≠0; x=3: 27+81-108=0, so 3 is a zero and x-3 is a factor. Divide by this zero using synthetic division:
3 | 1 9 0 -108
1 3 36 | 108
1 12 36 | 0 = x2+12x+36=(x+6)2.
Therefore (x+6)2(x-3)≥0 is another way of writing the inequality.
(x+6)2 is always positive, so the whole expression is positive provided x-3≥0, that is, x≥3.
So the solution is x≥3, that is, { x≥3 }, the set of all numbers at least as big as 3.