How do you factor the trinomial #n^2+ 3n - 18#?

Concept of method

If you are not sure about the numbers to use the best thing to do is list all the factor pairs of the constant. In this case (18). Look at those factors one set at a time and see which pair has a difference or sum matching the coefficient of the middle term.

As 18 is negative one factor has to be positive and the other negative (negative times positive = negative).

Consequently when we put them together to make 3n we will be looking at difference. Not only that; as we have +3n we know that the larger one has to be positive to give a positive 3n

Not that if the coefficient of #x^2# is not 1 then more experimentation is needed.
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All the factors of 18: {1, 18}, {2, 9}, {3, 6}
The difference #color(white)(xxxx)#17#color(white)(xxx)#7#color(white)(xxx)#3

The last difference is 3 as required. The larger one is 6 so we have (+6) and (-3)

Let us put them into the standard factor form and see what we get!

#(n + 6)(n-3) = n^2 +6n -3n -18#

Which becomes #n^2+3n-18 # as required.