1. The problem is to factor the quadratic expression $6x^2 + x + 12$. 2. The general form of a quadratic is $ax^2 + bx + c$. Here, $a=6$, $b=1$, and $c=12$. 3. To factor, we look for two numbers that multiply to $a \times c = 6 \times 12 = 72$ and add to $b=1$. 4. The pairs of factors of 72 are (1,72), (2,36), (3,24), (4,18), (6,12), (8,9). None of these pairs add to 1, so the quadratic does not factor nicely over the integers. 5. We can use the quadratic formula to find the roots: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \times 6 \times 12}}{2 \times 6} = \frac{-1 \pm \sqrt{1 - 288}}{12} = \frac{-1 \pm \sqrt{-287}}{12}$$ 6. Since the discriminant is negative, the roots are complex and the quadratic cannot be factored over the real numbers. Final answer: The quadratic $6x^2 + x + 12$ is prime over the reals and cannot be factored into real linear factors.