Question 46540
Well, there'e a bit of trial-and-error processing that occurs when you try to facto a trinomial whose {{{x^2}}} coefficient is greater than 1.

Here's the way I go about it:

I know that the factors of {{{4x^2}}} are:
{{{(4x  )(x  )}}} or {{{(2x  )(2x  )}}}

The factors of 18 are: (1)(18) or (2)(9) or (3)(6)

I also notice that the middle term of the trinomial is negative and the last term is positive.  This tells me that the contants part of the factors are both negative numbers.
 Why? Because a negative times a negative is a positive (last term is positive) and a negative plus a negative is a negative (middle term is negative)

Try the first pair of factors with the x's

{{{(4x - 1)(x - 18) = 4x^2 -72x - x + 18}}} = {{{4x^2 - 73x + 18}}}No go! The middle term is wrong.

Try:
{{{(4x - 2)(x - 9) = 4x^2 -36x -2x + 18}}} = {{{4x^2 - 38x + 18}}} No go! The middle term is wrong.

Try:
{{{(4x - 3)(x - 6) = 4x^2 - 24x - 3x + 18}}} = {{{4x^2 - 27x + 18}}} Bingo!

I hope this helps.