If we were doing it the long way, we would need to consider all the factors of 6:

1 & 6;

−1 & −6;

2 & 3;

−2 & −3

and also many factors of −12:

1 & −12;

2 & −6;

3 & −4

and all the negatives of these.

We could spend a **long time** finding the correct combination of
factors if we use the long method.

**Factoring by Grouping Method**

Using grouping method, first we find 2 things, the:

(a) Product of the outer 2 terms of the trinomial

(b) Inner number of the trinomial

Then, the only "guess and check" we need to do is to look for 2 numbers whose:

- Product is the result of (a) above
- Sum is the inner term (from (b) above)

So in our 6*x*^{2} + *x* − 12 example, we are looking for 2 terms whose:

(a) Product is 6

x^{2}× −12 = −72x^{2}(multiply outer numbers)

and whose

(b) Sum is

x(inner number).

We try some terms and easily get 9*x* and −8*x*.

These are correct since:

- (9
*x*)(−8*x*) = −72*x* - 9
*x*+ (−8*x*) =*x*

Now write the original expression replacing *x *with (9*x* − 8*x*), as follows:

6

x^{2}+x− 12 = 6x^{2}+ (9x− 8x) − 12

We now re-group the right-hand side:

6

x^{2}+ (9x− 8x) − 12 = (6x^{2}+ 9x) − (8x+ 12)

Now factor each of the bracketed terms:

(6

x^{2}+ 9x) − (8x+ 12) = 3x(2x+ 3) − 4(2x+ 3)

On the right-hand side, we notice that each term in brackets is the same, so we can combine them as follows:

3

x(2x+ 3) − 4(2x+ 3) = (3x− 4)(2x+ 3)

[**What just happened?**

If we have 3*x**A* − 4*A*, we can factor *A* out of each term, and write (3*x* − 4)*A*. This is how grouping method works. You always end up with brackets that have the same terms inside, and these can be factored out.]

So our answer is:

6

x^{2}+x− 12 = (3x− 4)(2x+ 3)

Always **check your answer** by multiplying it out!