Algebraic manipulation and the binomial expansion

Expanding and factorising

• Expand three brackets by multiplying two first: (x + 1)(x + 2)(x + 3) = (x² + 3x + 2)(x + 3) = x³ + 6x² + 11x + 6.
• Factorise quadratics with a ≠ 1 by finding two numbers that multiply to ac and add to b. 6x² + x − 2: ac = −12, and 4 and −3 work, so 6x² + 4x − 3x − 2 = 2x(3x + 2) − (3x + 2) = (3x + 2)(2x − 1).
• Always take out common factors first: x³ − 4x = x(x² − 4) = x(x − 2)(x + 2).

Algebraic fractions

• Simplify by factorising and cancelling: (x² − 9)/(x² + 5x + 6) = (x − 3)(x + 3)/((x + 2)(x + 3)) = (x − 3)/(x + 2).
• Add or subtract using a common denominator: 2/(x + 1) + 3/(x − 2) = (2(x − 2) + 3(x + 1))/((x + 1)(x − 2)) = (5x − 1)/((x + 1)(x − 2)).

Rearranging formulae

• When the subject appears twice, collect those terms on one side and factorise.
• y = (x + 2)/(x − 3) gives y(x − 3) = x + 2, so xy − 3y = x + 2, xy − x = 3y + 2, x(y − 1) = 3y + 2 and x = (3y + 2)/(y − 1).

The binomial expansion

• The coefficients of (a + b)ⁿ come from Pascal's triangle (1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1…) or from nCr = n! ÷ (r!(n − r)!).
• (a + b)ⁿ = aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + … + bⁿ.
• (2 + x)³ = 8 + 12x + 6x² + x³.
• The x² term in (1 + 2x)⁵ is 5C2 × (2x)² = 10 × 4x² = 40x², so the coefficient is 40. Remember to raise the whole term, including its number, to the power.

Key terms

Binomial expansion

Expanding (a + b)ⁿ into a sum of terms.

Pascal's triangle

A triangle of numbers giving binomial coefficients.

nCr

The number of ways to choose r items from n, used for binomial coefficients.

Coefficient

The number in front of a term.

Algebraic fraction

A fraction with algebraic expressions in it.

Subject

The letter a formula is rearranged to find.