If you mean "factor over the rational numbers", then this cannot be factored.
Here's why:
The given expression is in the form ax^2+bx+c. We have
a = 3
b = 19
c = 15
Computing the discriminant gives us
d = b^2 - 4ac
d = 19^2 - 4*3*15
d = 181
Note how this discriminant d value is not a perfect square
This directly leads to the original expression not factorable
We can say that the quadratic is prime
If you were to use the quadratic formula, then you should find that the equation 3x^2+19x+15 = 0 leads to two different roots such that each root is not a rational number. This is another path to show that the original quadratic cannot be factored over the rational numbers.
