4b² + 9b - 28 = 4b² + 16b - 7b - 28

= 4b (b + 4) - 7 (b + 4)

= (4b - 7) (b + 4),

which is the desired answer.

This method can be called 'grouping' or 'splitting the middle term'.

* * *

The trick, as you can see, is in the judicious splitting of 9b into 16b - 7b.

After all, there are infinite ways in which 9b may be split.

(Examples: 9b = 14b - 5b; or 9b = 20b - 11b; neither of which gives any help at all. You can try, and see!)

The real question is, which way of splitting 9b will help us factorise the original expression, and how can we find it? Trial and error will find it, of course; but, as in the present example, trial and error can sometimes become rather tedious.

* * *

Here is a very helpful rule, which works whenever the trinomial given is factorisable:

(Since the three co-efficients in the present example are 4, 9, and -28, in that order, I will express the rule in terms of those three co-efficients.)

Find two numbers whose sum is 9 and whose product is equal to

4 × (-28) = -112.

(Note that the sum must equal the co-efficient of the middle term, namely, 9.)

These two numbers will be found to be 16 and -7. Thus, we will choose to represent 9b as 16b - 7b, and factorisation should proceed smoothly.