Expanding $(a + b)^5$ by multiplying five factors is slow. The identity $(a + b)^2 = a^2 + 2ab + b^2$ has a pattern that continues for every power, and the coefficients come from Pascal's triangle.
Key idea
$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, \qquad \binom{n}{k} = \frac{n!}{k!\,(n - k)!}.$$ The powers of $a$ go down while the powers of $b$ go up, and the coefficients are row $n$ of Pascal's triangle.
Why it works
Multiplying out $(a + b)(a + b)\cdots(a + b)$, each term is made by choosing $a$ or $b$ from each of the $n$ factors. A term $a^{n-k}b^k$ comes from choosing $b$ in exactly $k$ of the factors, and there are $\binom{n}{k}$ ways to choose which ones. Pascal's rule (each entry is the sum of the two above it) matches this: a choice of $k$ factors from $n$ either includes the last factor or doesn't.
| row | coefficients |
|---|---|
| 0 | 1 |
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
Worked examples
Worked example: A full expansion
Expand $(x + 2)^4$.
- Row 4: $1, 4, 6, 4, 1$.
- $x^4 + 4x^3(2) + 6x^2(2^2) + 4x(2^3) + 2^4$.
Answer: $x^4 + 8x^3 + 24x^2 + 32x + 16$
Worked example: One coefficient
Find the coefficient of $x^3$ in $(2x - 1)^5$.
- The $x^3$ term uses $(2x)^3$ and $(-1)^2$, with $\binom{5}{3} = 10$.
- $10 \cdot (2x)^3 \cdot (-1)^2 = 10 \cdot 8x^3 \cdot 1$ $= 80x^3$.
Answer: $80$
Common mistake
Raising only the variable to the power. In $(2x - 1)^5$ the term is $\binom{5}{3}(2x)^3(-1)^2$, and $(2x)^3 = 8x^3$, not $2x^3$. Keep each term of the binomial in parentheses, signs included, and the negative signs take care of themselves.