A polynomial identity holds for every value: (a + b)² = a² + 2ab + b², and the middle term is what people forget. Row n of Pascal’s triangle holds the coefficients of (a + b)ⁿ, the entries are the combinations nCr, and the powers of a fall while the powers of b rise, each term totalling n. Substituting a = b = 1 checks any expansion.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Cubes
a³ − b³ = (a − b)(a² + ab + b²) and a³ + b³ = (a + b)(a² − ab + b²). x³ − 27 = (x − 3)(x² + 3x + 9). Check by expanding: x³ + 3x² + 9x − 3x² − 9x − 27 = x³ − 27.
The triangle
1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1. Each row starts and ends with 1, and every other entry is the sum of the two above. Row 5 is ready for a fifth power.
They are combinations
The entry 6 in row 4 is 4C2 = 4!/(2! 2!) = 6: the number of ways to choose which 2 of the 4 brackets contribute a b. Row 6, position 2: 6C2 = 15, without writing the triangle.
With coefficients inside
(2x − 3)⁴: a = 2x, b = −3. The a-powers are 16x⁴, 8x³, 4x², 2x, 1; the b-powers 1, −3, 9, −27, 81. Multiply each pair by its row-4 coefficient: 16x⁴ − 96x³ + 216x² − 216x + 81.
One term only
The x³ term of (x + 3)⁶ is 6C3 · x³ · 3³ = 20 · 27 x³ = 540x³. The term with bʳ is nCr · aⁿ⁻ʳ · bʳ; no need to expand the other six.
Check with x = 1
(2 − 3)⁴ = 1, and 16 − 96 + 216 − 216 + 81 = 1. One substitution catches a wrong sign or a dropped coefficient anywhere in the expansion.
Step 2: Try It Yourself
Tap and try it out.
Expand (2x − 3)⁴ and check it; then find the x³ term of (x + 3)⁶ without expanding
- 1row 4: 1, 4, 6, 4, 1; a = 2x, b = −3Pascal, and name the parts
Step 0 of 4
4C0
4C1
4C2
4C3
4C2 has the most. It has 5 more than 4C0.
Step 3: In Real Life
Ten years of growth
(1 + r)¹⁰ is what a decade of growth multiplies by. The binomial theorem expands it: 1 + 10r + 45r² + …, a quick estimate of ten years’ interest without a calculator.
Step 4: Watch an Example
One step at a time.
Watch Sana Expand a Power
Sana expands (a + b)⁴ without multiplying four brackets.
- Step 1
Row 4 of Pascal’s triangle is 1, 4, 6, 4, 1. The powers of a fall from 4 to 0 across the terms while the powers of b rise from 0 to 4.
Step 5: Your Turn
Practice makes it stick.
The Terms
Problem 1 of 2
How many terms does (a + b)⁶ have?
The Coefficient
Problem 2 of 2
In (a + b)⁵, what is the coefficient of a³b²?
Expand It
1 of 8
How many terms in (a + b)⁹?
2 of 8
What is 6C2?
3 of 8
In (a + b)⁴, coefficient of a²b²?
4 of 8
In (a + b)², what is the coefficient of ab?
5 of 8
Sum of row 3 of the triangle: 1, 3, 3, 1?
6 of 8
In (a + b)⁷, the exponents of every term total what?
7 of 8
Match each expression with its factorisation.
Tap a card on the left to start.
8 of 8
In (a + b)⁶, coefficient of a⁵b?
Step 6: Quick Check
Show what you know.
Question 1 of 2
How many terms does (a + b)⁸ have?
Question 2 of 2
Where do the binomial coefficients come from?
What You Learned
- Row n of Pascal’s triangle gives the coefficients of (a + b)ⁿ, and its entries are the combinations nCr.
- The powers of a fall while the powers of b rise, always totalling n.
- With coefficients or signs inside, list the a-powers and b-powers first, then multiply.
- The term with bʳ is nCr · aⁿ⁻ʳ · bʳ, and x = 1 checks the whole expansion.