Expanding bigger expressions

Distributivity doesn't stop at two terms in a bracket. Whatever is outside multiplies every term inside, however many there are.

Take (3a/2)(a − b + 1/5). Distribute across all three terms: (3a/2 × a) − (3a/2 × b) + (3a/2 × 1/5).

Now simplify each:

• 3a/2 × a = (3/2)(a × a) = (3/2)a²;
• 3a/2 × b = (3/2)ab;
• 3a/2 × 1/5 = (3/2 × 1/5)a = (3/10)a.

So the answer is (3/2)a² − (3/2)ab + (3/10)a. Can any two of those be combined? No: a², ab and a are three different collections of letters. Terms with exactly the same letters, each to the same power, are called like terms, and only like terms can be added into one. (3a²b and −5a²b are like terms; a²b and ab are not.) So this expression is as simple as it gets.

Now expand (a + b)(a + b). Distribute: (a + b)a + (a + b)b = a × a + b × a + ab + b × b = a² + ba + ab + b².

Here two terms are alike: ba and ab are the same thing, since multiplication can be done in either order. So ba + ab = 2ab, and (a + b)(a + b) = a² + 2ab + b².

One step further: expand (a + b)(a² + 2ab + b²). Distribute over three terms: (a + b)a² + (a + b)2ab + (a + b)b², which opens out to a × a² + ba² + a × 2ab + b × 2ab + ab² + b × b².

Simplify each: a × a² = a³, ba² = a²b, a × 2ab = 2a²b, b × 2ab = 2ab², b × b² = b³. Collect the like terms: a²b + 2a²b = 3a²b, and ab² + 2ab² = 3ab². Result: a³ + 3a²b + 3ab² + b³.