Special product formulas

The formulas for special products (abridged multiplication) are the "times tables" of algebra. They give the result of multiplying binomials in ready-made form. Being able to recognise them quickly in both directions (expanding brackets and factorising) makes simplifying expressions, solving equations, cancelling fractions and mental arithmetic several times faster. This lesson explains where each formula comes from, how to use it and where mistakes are most common.

§1. Second-degree formulas

The three basic formulas:

Why are they true? Just multiply out the brackets: (a + b)² = (a + b)(a + b) = a² + ab + ba + b². The two equal terms ab add up to the "middle term" 2ab. Geometrically, a square of side a + b splits into squares of sides a and b plus two a × b rectangles.
In the difference of two squares the middle terms cancel: (a − b)(a + b) = a² + ab − ab − b² = a² − b².
The most common mistake: (a + b)² ≠ a² + b² — the middle term is forgotten. It is easy to check: with a = b = 1, (1 + 1)² = 4, but 1² + 1² = 2.
Any expression may stand in place of a and b: (3x − 2y)² = 9x² − 12xy + 4y². Here a = 3x, b = 2y, so a² = 9x² and .

Worked example

Simplify (3x − 2)² − (3x + 2)(3x − 2).

1. First term: (3x − 2)² = 9x² − 12x + 4.
2. Second term is a difference of squares: (3x + 2)(3x − 2) = 9x² − 4.
3. Subtract: 9x² − 12x + 4 − 9x² + 4 = 8 − 12x.

§2. Third-degree formulas

The cube formula comes from (a + b)³ = (a + b)²(a + b). The coefficients 1, 3, 3, 1 are a row of Pascal’s triangle. In (a − b)³ the signs alternate: +, −, +, −.
The second factor in the sum and difference of cubes, a² ∓ ab + b², is an incomplete square: there is no 2 in front of ab. To remember the signs: the first bracket has the same sign as the original expression, and the sign before ab is the opposite one.
Check: (a − b)(a² + ab + b²) = a³ + a²b + ab² − a²b − ab² − b³ = a³ − b³.

Worked example

Factorise 27x³ + 1.

1. 27x³ = (3x)³ and 1 = 1³, so a = 3x, b = 1.
2. Sum of cubes: (a + b)(a² − ab + b²).
3. Answer: (3x + 1)(9x² − 3x + 1).

§3. Mental arithmetic

The formulas make work with large numbers easier. The key idea is to write the number as a "round" number plus a small correction:

• 102² = (100 + 2)² = 10 000 + 400 + 4 = 10 404.
• 49² = (50 − 1)² = 2500 − 100 + 1 = 2401.
• 999² = (1000 − 1)² = 1 000 000 − 2000 + 1 = 998 001.
• 98·102 = (100 − 2)(100 + 2) = 10 000 − 4 = 9 996.
• 67² − 33² = (67 − 33)(67 + 33) = 34·100 = 3 400.
• 41² − 39² = 2 · 80 = 160.

Rule of thumb: if the number halfway between two factors is round, use the difference of squares; if a number is close to a round one, use the square formula.

§4. Factorising (the reverse direction)

In exams a formula often has to be read "backwards": from a given polynomial back to brackets.

• If a trinomial has two terms that are perfect squares, test for a perfect square: in 4x² − 12x + 9 we have 4x² = (2x)², 9 = 3², 2 · 2x · 3 = 12x — so it is (2x − 3)².
• A difference of two squares always factorises: x⁴ − 16 = (x² − 4)(x² + 4) = (x − 2)(x + 2)(x² + 4).
• A sum of squares such as x² + 4 does not factorise over the real numbers.
• Sometimes grouping comes first: x² − y² + 2y − 1 = x² − (y − 1)² = (x − y + 1)(x + y − 1).

Worked example

Simplify the fraction (x² − 9)/(x² − 6x + 9).

1. Numerator: x² − 9 = (x − 3)(x + 3).
2. Denominator: x² − 6x + 9 = (x − 3)².
3. Cancel (x − 3): (x + 3)/(x − 3), with x ≠ 3.

§5. Symmetric expressions

If a + b and ab are known, many expressions can be found without finding a and b themselves:

These follow directly from §1 and §2: for example, subtracting 2ab from (a + b)² = a² + 2ab + b² leaves a² + b². They are used constantly with Vieta’s theorem, because for the roots of a quadratic x₁ + x₂ and x₁x₂ are known at once.
In particular, if x + 1/x = t, then x² + 1/x² = t² − 2, since x · 1/x = 1.

Worked example

Find a³ + b³ if a + b = 3 and ab = 2.

1. a³ + b³ = (a + b)³ − 3ab(a + b).
2. = 3³ − 3 · 2 · 3.
3. = 27 − 18 = 9.

§6. Irrational expressions

To see a formula you often have to group terms or see a number as the square of a root: 7 = (√7)².
To rationalise a denominator, multiply by the conjugate expression, because (√a − √b)(√a + √b) = a − b — the roots disappear.

Worked example

Simplify (√7 + 1 − √3)(√7 + √3 − 1).

1. Group: (√7 + (1 − √3))(√7 − (1 − √3)).
2. Difference of squares: 7 − (1 − √3)².
3. (1 − √3)² = 1 − 2√3 + 3 = 4 − 2√3.
4. 7 − 4 + 2√3 = 3 + 2√3.

Worked example

Rationalise the denominator of 2/(√5 − √3).

1. Multiply top and bottom by √5 + √3.
2. Denominator: (√5)² − (√3)² = 5 − 3 = 2.
3. 2(√5 + √3)/2 = √5 + √3.

Check yourself

1. Work out 103² mentally.

2. Work out 1001 · 999.

3. a + b = 5, ab = 6. Find a² + b².

4. Simplify (x − 3)² − (x + 3)².

5. Find x² + 1/x² if x + 1/x = 3.

6. Expand (2x − 1)³.

7. Factorise x³ − 8.

8. Simplify (x + 3)² − (x − 1)(x + 1).

Common mistakes

• Writing (a + b)² = a² + b² (the middle term 2ab is lost).
• Putting a 2 into the incomplete square of a sum of cubes: it is a² − ab + b², not a² − 2ab + b².
• Writing b² with a minus sign in (a − b)² (it is always +b²).
• Treating (2x)² as 2x² (the correct value is 4x²).
• Trying to factorise a sum of squares a² + b².
• Forgetting to change signs when removing a bracket after a minus: −(9x² − 4) = −9x² + 4.