Algebra

$(x^2 + 2x + 5 - x - 4)(3x^2 - x - 4 + 2x + 8) \ge 0\\ (x^2 + x + 1)(3x^2 + x +4) \ge 0\\ \qquad \text{consider the roots of}\; x^2+x+1\\ \qquad \triangle=b^2-4ac=1-4=-3<0 \qquad\text{ no real roots}\\ \qquad \text{consider the roots of}\; 3x^2+x+4\\ \qquad \triangle=b^2-4ac=1-48=-47<0 \qquad\text{ no real roots}\\ $

Since the coefficiant of x^4 is positive, and there are not real roots. 

this funtion will always be positve.

x is in the set of real numbers (no restrictions)