Question 173398
You have only considered half of the problem.  What you said is true IF the coefficient on the high order term is 1, but here it is 3.


The part about needing the factors of 10 is almost right.  In fact you need the factors of -10.  So, -2 and 5, 2 and -5, -1 and 10, or 1 and -10.


But the sum required to get the 13 coefficient must consider the 3 coefficient on the high order term.


{{{-2 + 15 = 13}}}, but {{{15=3*5}}}


So

{{{(x+5)(3x-2)=3x^2-2x+15x-10=3x^2+13x-10}}}


Yeah, but how do I know that a quadratic trinomial is factorable in the first place so I don't go rooting (so to speak) around for factors that might not exist? you ask.


Use the discriminant part of the quadratic formula, {{{b^2-4ac}}} where a, b, and c are the coefficients of the 2nd degree, 1st degree, and constant terms respectively.  If the number is a perfect square, then the quadratic is factorable, otherwise not.  For the given problem, {{{13^2-4(3)(-10)=169+120=289=17^2}}}, so your example is factorable.  (If you haven't done the quadratic formula yet, don't worry about it.  Just use the rule above on faith for the time being)