The short answer is that this trinomial is prime, meaning that it is not factorable over the integers. In fact, this one isn't factorable over the real numbers.
This might be getting a little ahead of you, given the way you phrased the question, but it is something that you will be learning in a week or two anyway. What I'm talking about is a little trick to determine the character of the roots of a quadratic equation of the form:
Take the coefficient on the x term and square it:
Multiply 4 times the lead coefficient times the constant term:
Find the difference of these two values:
If the result is zero, then you have a perfect square trinomial, and this would be factorable over the integers..
If the result is positive and a perfect square, then the trinomial is factorable over the integers.
If the result is positive and not a perfect square, the factors are real but irrational numbers, i.e. they involve the square root of a quantity that is not a perfect square. For the time being, you would consider such a trinomial as prime.
If the result is less than zero, then the factors are complex numbers of the form
So, just to illustrate with your problem:
so
and your factors would contain a conjugate pair of complex numbers; they would look like:
Like I said, the trinomial is prime as far as you are concerned.
John