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Binomial expansion

Expand binomials using Pascal’s triangle, factorials and combinations. Learn the pattern of powers, handle signs and fractions, and simplify exact surd and integer calculations.

Before you startExpanding brackets and laws of indices

01 / The pattern

Choose one term from each bracket.

A binomial has two terms, such as a + b or 2 − 3x. Raising it to a positive integer power means multiplying that many identical brackets.

(a + b)³ = (a + b)(a + b)(a + b)
= a³ + 3a²b + 3ab² + b³

There are three ways to choose b from one bracket and a from the other two. That produces the coefficient 3 of a²b. The total power of a and b in each term is 3.

For exponent n, the a-power falls from n to 0 while the b-power rises from 0 to n. There are n + 1 terms before any substitution causes terms to combine or vanish.

02 / Pascal’s triangle

Each inside entry is the sum of its two parents.

Start with 1. Each new row begins and ends with 1, and each inside entry is the sum of the two entries above it.

We label a row by its exponent n: the top is n = 0. If you count the top as the first row, the coefficients for power n are in the (n + 1)th row. Check which convention a question uses.

Choose a row and an entry in the model. Entry r counts ways to choose b from r of the n brackets. Here r starts at 0, so it labels the (r + 1)th entry.

Choose a binomial coefficientChoose and compare

Exponent n Entry r

Row n = 4 has coefficients 1, 4, 6, 4, 1. Entry r = 2 is 6, the sum of the two 3s above it. This is the fifth row when counting from one, and its third entry. C(4,2) = 6.

Watch neighbouring entries build Pascal’s triangle

Pause, replay or seek freely. The notes explain the same idea and stay in view.

03 / Factorials

A factorial multiplies consecutive positive integers.

For a positive integer n, n! = n(n − 1)…2 × 1. By convention 0! = 1; this makes counting “choose nothing” work consistently.

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
8!/6! = 8 × 7 = 56

Cancel common factorial factors before multiplying large numbers. (n − 1)! means the factorial of n − 1; it does not mean n! − 1.

Why define 0! as 1?

The recurrence n! = n(n − 1)! gives 1! = 1 × 0!, so 0! must equal 1. There is also exactly one way to choose an empty selection: choose nothing.

04 / Combinations

Count selections without counting their order.

We write C(n,r), also written ⁿCᵣ or “n choose r”, for the number of ways to choose r items from n distinct items without regard to order.

C(n,r) = n! / [r!(n − r)!]
0 ≤ r ≤ n, with n and r integers

Choosing in order gives n(n − 1)…(n − r + 1) possibilities. Every set of r chosen items appears r! times in those orders, so divide by r!.

C(7,2) = 7 × 6 / (2 × 1) = 21

Choosing r items is equivalent to choosing the n − r items to leave out. Thus C(n,r) = C(n,n − r), explaining Pascal’s left-right symmetry. In particular C(n,0) = C(n,n) = 1.

Prove Pascal’s addition rule

For integers n ≥ 1 and 1 ≤ r ≤ n, put the two fractions over the common denominator r!(n − r + 1)!:

C(n,r − 1) + C(n,r)
= [n!r + n!(n − r + 1)] / [r!(n − r + 1)!]
= n!(n + 1) / [r!(n + 1 − r)!]
= C(n + 1,r)

A counting proof gives the same result: choose r people from n ordinary candidates and one distinguished candidate. Either include that person and choose r − 1 of the other n, or exclude them and choose all r from the n. The cases are disjoint and cover every selection.

Evaluate a probability formula

For six independent fair coin tosses, the chance of exactly two heads is C(6,2)(1/2)²(1/2)⁴ = 15/64. The coefficient counts the possible positions of those two heads; each complete sequence has probability 1/64.

More generally, for n independent trials each with the same success probability p, exactly r successes have probability C(n,r)pʳ(1 − p)ⁿ⁻ʳ, with 0 ≤ p ≤ 1. These assumptions matter.

05 / Expand a binomial

Apply the coefficient, then raise both whole terms to their powers.

For a positive integer n, the binomial theorem is a finite identity:

(a + b)ⁿ = aⁿ + naⁿ⁻¹b
+ C(n,2)aⁿ⁻²b² + … + bⁿ

The general term is C(n,r)aⁿ⁻ʳbʳ for r = 0,…,n. If b is negative or contains a numerical factor, raise the entire b-term to the power r.

The n = 0 row is the constant polynomial 1; numerically a non-zero base raised to power zero is 1. Negative and fractional exponents require a different version of the binomial expansion.

Expand (2x − 1)⁴Worked example

Coefficients: 1, 4, 6, 4, 1

Use a = 2x and b = −1.

(2x)⁴ + 4(2x)³(−1)
+ 6(2x)²(−1)²
+ 4(2x)(−1)³ + (−1)⁴

The signs alternate because b is negative.

16x⁴ − 32x³ + 24x² − 8x + 1

The powers of 2 are part of the coefficients.

06 / First few terms

Start with the constant when ascending powers are requested.

“First four terms in ascending powers of x” normally means the constant, x, x² and x³ terms when all occur. Choose a as the constant and b as the x-term.

Writing + … means the remaining terms are omitted, not zero. A partial expansion is not an exact replacement for the original expression.

A fractional x-termWorked example

(3 − x/2)⁶

Use r = 0, 1, 2, 3.

3⁶ + 6 × 3⁵(−x/2)
+ 15 × 3⁴(−x/2)²
+ 20 × 3³(−x/2)³ + …

Square and cube the fraction as well as x.

729 − 729x + (1215/4)x²
− (135/2)x³ + …

These are the first four terms; this finite expansion continues up to x⁶.

Two variables

(2u + v)³
= 8u³ + 12u²v + 6uv² + v³

Each term has total degree 3 in u and v. You can regard this as descending powers of u or ascending powers of v.

07 / Substitute expressions

A binomial term can itself be an expression.

To expand (1 + x − x²)³, first treat x − x² as a single quantity t. Expand (1 + t)³, then substitute and collect equal powers.

(1 + t)³ = 1 + 3t + 3t² + t³
t = x − x²
t² = x² − 2x³ + x⁴
t³ = x³ − 3x⁴ + 3x⁵ − x⁶
(1 + x − x²)³
= 1 + 3x − 5x³ + 3x⁵ − x⁶

The x² and x⁴ terms cancel after substitution. Do not simply apply the two-term formula to three independent terms.

08 / Use symmetry

Adding opposite substitutions keeps the even powers.

In (a + x)ⁿ and (a − x)ⁿ, the even-power terms agree and the odd-power terms have opposite signs. Adding cancels odd powers; subtracting cancels even powers.

(2 + x)⁴ + (2 − x)⁴
= 32 + 48x² + 2x⁴

At x = √5, this is 32 + 240 + 50 = 322. The irrational terms cancel exactly.

Use cancellation to solve an equation

Solve (1 + x)⁵ + (1 − x)⁵ = 82 for real x. Adding the expansions removes all odd powers:

2 + 20x² + 10x⁴ = 82
x⁴ + 2x² − 8 = 0

Put y = x². Then y² + 2y − 8 = (y + 4)(y − 2) = 0. Since y ≥ 0, retain y = 2 and reject y = −4. The real solutions are x = ±√2.

An exact difference containing a surd

(1 + x)⁵ − (1 − x)⁵
= 10x + 20x³ + 2x⁵

Putting x = √2 gives (10 + 40 + 8)√2 = 58√2.

Exact large-integer arithmetic

Find C(9,7) without evaluating 9! in full, and explain the symmetry you use.

Hint

Choosing seven items leaves two out.

Worked solution

C(9,7) = C(9,2) = 9 × 8 / 2 = 36

Each selection of seven corresponds to exactly one pair left out.

10 / Recap

The pattern counts choices from repeated brackets.

For exponent n, Pascal’s row n has n + 1 entries.
C(n,r) counts the choices of r brackets contributing the second term.
The term powers are n − r and r.
Raise every numerical factor and sign to the appropriate power.
Keep omitted terms marked with an ellipsis.
A complete finite expansion is exact; using only some terms is a separate approximation step.

Next: finding binomial coefficients →

Section 1 of 10 · The pattern