Predetermine the middle term of an quadratic equation when there are two possibilities 
Factorise  completely:

*

*$2x^2+13x+15$

*$3x^2-5x+2$


*

*As i figured, there could be two ways to break the middle term. Either, $15x-2x$ or, $10x+3x$.

If I continue with  $15x-2x$:
$2x^2+13x+15$
$2x^2+15x-2x+15$
$2x(x-1)+15(x+1)$
But If I go down with $10x+3x$:
$2x^2+13x+15$
$2x^2+10x+3x+15$
$2x(x+5)+3(x+5)$
I know the former is incorrect and the later one is correct. But I would really like to know how to predetermine this problem and avoid continuing with the former one.
NB: No. 2 has the exact same approach.
 A: There is a way to factorize a general quadratic trinomial
$$Ax^2 + Bx + C = (Dx + E)(Fx + G),$$
that is, assuming that the trinomial can be factored in the first place.  (This method will also let you determine whether the trinomial cannot be factored at all.)
Set
$$Ax^2 + Bx + C = 0$$
and solve for $x$ using the Quadratic Formula:
$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.$$
Note that, in order for this method to work, then the discriminant
$$d = B^2 - 4AC$$
needs to be a perfect square.

Now, let us apply this method to your first quadratic trinomial:
$$2x^2 + 13x + 15$$
Set $2x^2 + 13x + 15 = 0$, then solve for $x$ using the Quadratic Formula, where $A_1 = 2, B_1 = 13, C_1 = 15$:
$$x = \frac{-{B_1} \pm \sqrt{{B_1}^2 - 4{A_1}{C_1}}}{2{A_1}} = \frac{-13 \pm \sqrt{169 - 4\cdot{2}\cdot{15}}}{4} = \frac{-13 \pm \sqrt{49}}{4}.$$
Note that the discriminant
$$d_1 = {B_1}^2 - 4{A_1}{C_1} = 49$$
is a perfect square.  Hence, this means that the first quadratic trinomial is indeed factorable.
We obtain the $x$-values
$$x = \frac{-13 \pm 7}{4} = -5, -\frac{3}{2}$$
which means that we have the factors
$$(x + 5)(2x + 3) = 2x^2 + 13x + 15.$$

I leave it as an exercise for you to apply the Quadratic Formula Method for factoring your second quadratic trinomial:
$$3x^2 - 5x + 2$$
