Binomial Expansion

Expanding $(x + y)^5$ by multiplying five brackets is slow. Binomial expansion does it in one line, using a row of Pascal’s triangle for the coefficients. It’s the reason the triangle is so useful in algebra.

Key ideas

The pattern

Look at the first few powers of $a + b$:

$$\begin{aligned} (a + b)^1 &= a + b \\ (a + b)^2 &= a^2 + 2ab + b^2 \\ (a + b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \\ (a + b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \end{aligned}$$

For $(a + b)^n$:

• The coefficients are row $n$ of Pascal’s triangle.
• The powers of $a$ go down from $n$ to $0$, while the powers of $b$ go up from $0$ to $n$.
• In every term, the exponents add up to $n$.
• There are $n + 1$ terms.

When the terms have coefficients or negatives

Treat each part of the binomial as a whole, in brackets. For $(2x - 3)^4$, use $a = 2x$ and $b = -3$. Then $(2x)^3 = 8x^3$, and the powers of $-3$ alternate in sign.

Finding one term

Term number $r + 1$ (counting from the left, starting at $1$) uses position $r$ of the row, with $a^{n - r}b^r$. For example, the third term of $(a + b)^6$ uses position $2$: $15a^4b^2$.

Worked examples

Example 1: A basic expansion

Expand $(x + y)^4$.

Solution. Row 4 is $1, 4, 6, 4, 1$:

$$(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$$

Example 2: A negative term

Expand $(x - 2)^5$.

Solution. Row 5 is $1, 5, 10, 10, 5, 1$. Use $a = x$ and $b = -2$:

$$\begin{aligned} (x - 2)^5 &= x^5 + 5x^4(-2) + 10x^3(-2)^2 + 10x^2(-2)^3 + 5x(-2)^4 + (-2)^5 \\ &= x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32 \end{aligned}$$

The signs alternate because the odd powers of $-2$ are negative.

Example 3: Coefficients inside the brackets

Expand $(2x + 3)^4$.

Solution. Row 4 is $1, 4, 6, 4, 1$. Use $a = 2x$ and $b = 3$:

$$\begin{aligned} (2x + 3)^4 &= (2x)^4 + 4(2x)^3(3) + 6(2x)^2(3)^2 + 4(2x)(3)^3 + 3^4 \\ &= 16x^4 + 4(8x^3)(3) + 6(4x^2)(9) + 4(2x)(27) + 81 \\ &= 16x^4 + 96x^3 + 216x^2 + 216x + 81 \end{aligned}$$

Example 4: One term only

Find the term containing $x^3$ in the expansion of $(x + 2)^6$.

Solution. In $(x + 2)^6$, the power of $x$ is $6 - r$, so $x^3$ needs $r = 3$. Position $3$ of row 6 is $20$:

$$20\,x^3(2)^3 = 20(8)x^3 = 160x^3$$

Common mistakes

Thinking $(a + b)^n = a^n + b^n$. $(x + 2)^2 = x^2 + 4x + 4$, not $x^2 + 4$. All the middle terms matter.

Not raising the coefficient. $(2x)^3 = 8x^3$, not $2x^3$. Keep each part of the binomial in brackets.

Losing the negative signs. In $(x - 2)^5$, $b = -2$, so its odd powers are negative and the signs alternate.

Using the wrong row. $(a + b)^5$ uses row 5, $1, 5, 10, 10, 5, 1$. Remember the triangle starts at row 0.

Skipping a term. $(a + b)^n$ always has $n + 1$ terms. Count them.

Practice

1. (Warm-up) Expand $(a + b)^3$.

Solution

$$a^3 + 3a^2b + 3ab^2 + b^3$$

2. (Warm-up) How many terms are in the expansion of $(x + y)^9$?

Solution

$9 + 1 = 10$ terms.

3. (Warm-up) Expand $(x + 1)^4$.

Solution

$$x^4 + 4x^3 + 6x^2 + 4x + 1$$

4. (Core) Expand $(y - 3)^4$.

Solution

$$\begin{aligned} &y^4 + 4y^3(-3) + 6y^2(-3)^2 + 4y(-3)^3 + (-3)^4 \\ &= y^4 - 12y^3 + 54y^2 - 108y + 81 \end{aligned}$$

5. (Core) Expand $(3x + 1)^3$.

Solution

$$(3x)^3 + 3(3x)^2(1) + 3(3x)(1)^2 + 1 = 27x^3 + 27x^2 + 9x + 1$$

6. (Core) Expand $(2x - y)^4$.

Solution

$$\begin{aligned} &(2x)^4 + 4(2x)^3(-y) + 6(2x)^2(-y)^2 + 4(2x)(-y)^3 + (-y)^4 \\ &= 16x^4 - 32x^3y + 24x^2y^2 - 8xy^3 + y^4 \end{aligned}$$

7. (Core) Find the third term in the expansion of $(x + 2)^7$.

Solution

Row 7 begins $1, 7, 21, \dots$ The third term uses position $2$:

$$21\,x^5(2)^2 = 84x^5$$

8. (Challenge) Find the coefficient of $x^2$ in the expansion of $(1 - 2x)^5$.

Solution

Use $a = 1$ and $b = -2x$. The $x^2$ term needs $b^2$, so position $2$ of row 5, which is $10$:

$$10(1)^3(-2x)^2 = 10(4x^2) = 40x^2$$

The coefficient is $40$.

9. (Challenge) Use the expansion of $(1 + 0.1)^4$ to find $1.1^4$ without a calculator.

Solution

$$\begin{aligned} (1 + 0.1)^4 &= 1 + 4(0.1) + 6(0.1)^2 + 4(0.1)^3 + (0.1)^4 \\ &= 1 + 0.4 + 0.06 + 0.004 + 0.0001 \\ &= 1.4641 \end{aligned}$$