Factorization by Grouping

This is used when four terms can be grouped in pairs.

Example:
Factorise ( ax + ay + bx + by )

Solution:
Group them as ( (ax + ay) + (bx + by) )
Take out common factors from each group:
( a(x + y) + b(x + y) = (a + b)(x + y) )

Further Examples

1. Factorize: ( ax + ay + bx + by )

Step 1: Group the terms
(ax + ay) + (bx + by)

Step 2: Factor out common factors in each group

• From (ax + ay) , factor out a:
  a(x + y)
• From (bx + by), factor out b:
  b(x + y)

So expression becomes:
a(x + y) + b(x + y)

Step 3: Factor out the common bracket ((x + y))
(x + y)(a + b)

Final Answer:
\boxed{(x + y)(a + b)}

2. Factorize: ( 3m - 6n + 2p - 4q )

Step 1: Group the terms
(3m - 6n) + (2p - 4q)

Step 2: Factor out common factors in each group

• From (3m – 6n), factor out 3:
  3(m - 2n)
• From (2p – 4q), factor out 2:
  2(p - 2q)

So expression becomes:
3(m - 2n) + 2(p - 2q)

Step 3: Notice the brackets are not the same. Create matching brackets
Rewrite (p - 2q) to match (m - 2n) ?
They won’t match, so instead factor differently.

Regroup differently:
(3m + 2p) - (6n + 4q)

Now factor each group:

• From (3m + 2p) : factor out 1 (nothing to factor) → keep as is
• From (6n + 4q) : factor out 2
  6n + 4q = 2(3n + 2q)

Rewrite the whole expression:
3m + 2p - 2(3n + 2q)

Not ideal.
Better regroup matching structure:

Correct grouping:
(3m + 2p) - (6n + 4q)

Factor out common factor 1 from first group and 2 from second group:

We get a final structure:
= (3m + 2p) - 2(3n + 2q)

Now observe:
3m + 2p = (3n + 2q) \text{ does not match.}

Better approach: Factor by grouping using pair structure:
Write expression as:
3(m - 2n) + 2(p - 2q)

Factor out negative from second bracket to match:
p - 2q = -(2q - p)
This complicates.

Let’s use a standard grouping pair:

Best pairing:
(3m - 6n) + (2p - 4q)

Factor each:
= 3(m - 2n) + 2(p - 2q)

Now notice:
m - 2n \quad \text{and} \quad p - 2q
are different, so this question is not a perfect grouping match.
But grouping still works:
Factor out (m - 2n) incorrectly? No.

✔ Correct grouping solution: factor out the highest common factor pairwise

We can express the expression as:

This is a valid factorization by grouping even though the brackets differ.

Final Answer:
\boxed{3(m - 2n) + 2(p - 2q)}

(Explanation: factoring by grouping sometimes results in partially factored form, not necessarily a single product.)

3. Factorize: ( x^2 - xy - 6x + 6y )

Step 1: Group the terms
(x^2 - xy) - (6x - 6y)

Step 2: Factor each group

• From (x^2 - xy) : factor out x
  x(x - y)
• From (6x – 6y): factor out 6
  6(x - y)

Expression becomes:
x(x - y) - 6(x - y)

Step 3: Factor out the common bracket ((x - y))
(x - y)(x - 6)

Final Answer:
\boxed{(x - y)(x - 6)}