Factoring by Grouping

What this move does

The four terms of $a x + a y + b x + b y$ have no common factor all together. But parts of it do:

both contain a a x + a y + both contain b b x + b y = a (x + y) + b (x + y) = (a + b) (x + y)

The key is just after the second equals sign: once each group has had its common factor taken out, the brackets left over are exactly the same, so that bracket becomes a new common factor.

Group: split the terms into two groups (with four terms, usually two and two).
Factor each group: take out each group's own common factor.
Look at the brackets: if they are the same, take that out too; if not, the grouping is wrong, so try another.

On the workbench above, “the same bracket” is drawn as “two blocks of the same width”: only blocks of the same width stack into a rectangle.

Predict first

There are $3$ ways to split the four terms of $a x + a y + b x + b y$ into two groups of two. How many of them work?

Only

1

2

All

3

Choose an answer first; then you can see the result.

Work through one

Grouping works just as well with only one letter.

Example: factor x³ + 2x² − 9x − 18

Step 1 / 4

A random grouping
First try the first and last terms together and the middle two together. The first group $x^{3} - 18$ has no common factor; the second gives $x$ times $2 x - 9$ . The brackets differ, so they do not fit together.

$(x^{3} - 18) + x (2 x - 9)$
Another grouping
The same bracket
Not finished yet

Grouping the 1st and 3rd terms and the 2nd and 4th terms also works here: $x (x^{2} - 9) + 2 (x^{2} - 9)$ . Try it on the workbench above.

Grouping, then a formula

Sometimes you group not to take out a common factor but to build a formula. Take $x^{2} - y^{2} + 2 y - 1$ : no grouping into pairs works, but put the last three terms together:

x^{2} - (y^{2} - 2 y + 1) = x^{2} - (y - 1)^{2} = (x + y - 1) (x - y + 1) .

The last three terms form a perfect square, and the whole thing becomes a difference of squares. This is called a “three-and-one” grouping.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: Factoring by grouping works

$a x + a y + b x + b y$ : two different groupings make the brackets match.

Practice

Exercise 1Basic

Factor $ab + 3 a + 2 b + 6$ .

Exercise 2Basic

Factor $x^{3} - x^{2} + 4 x - 4$ .

Exercise 3Advanced

Factor $2 a x - 6 a y - b x + 3 b y$ .

Exercise 4Challenge

Factor $x^{2} - 4 y^{2} + 4 y - 1$ .

What this move does

Work through one

Example: factor x³ + 2x² − 9x − 18

Grouping, then a formula

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Putting brackets after a minus sign without changing signs

Stopping halfway and calling it the answer