Why is (a + b)^2 = a^2 + 2ab + b^2 and not just a^2 + b^2?

The Binomial Theorem includes all the cross terms that arise when you multiply. Expanding (a+b)^2 = (a+b)(a+b) gives you a \cdot a + a \cdot b + b \cdot a + b \cdot b = a^2 + 2ab + b^2. The middle term 2ab comes from the two ways to pick one a and one b from the two parentheses, which is exactly what the binomial coefficient \binom{2}{1} = 2 counts.

When should I use the Binomial Theorem instead of just multiplying the binomial out?

For small exponents like n = 2 or n = 3, multiplying by hand may be quicker. But for n \geq 4, or when you only need a few specific terms, the Binomial Theorem is clearer and much less error-prone. It also shows you the structure of the expansion, which helps with factoring and simplification later.

How do I use the Binomial Theorem for (a - b)^n?

Rewrite it as (a + (-b))^n and apply the formula directly; the expansion will have alternating signs because even powers of (-b) are positive and odd powers are negative. If you track the substitution carefully, the signs sort themselves out automatically.

What is a binomial coefficient \binom{n}{k} and how do I compute it?

A binomial coefficient \binom{n}{k} counts the number of ways to choose k items from n items. You can compute it using \binom{n}{k} = \frac{n!}{k!(n-k)!}, or read the values from Pascal's Triangle for small n. For example, \binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6.

Binomial Theorem

The Binomial Theorem expands (a+b)^n into a sum of terms, giving you a formula for computing powers of binomials without multiplying them out repeatedly.

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}

Solve a problem with Binomial Theorem

Type the problem. The solver will use Binomial Theorem where Binomial Theorem is the right tool, and tell you when it is not.

What each symbol means

What Binomial Theorem takes

Symbol	Meaning
$a$	The first term of the binomial; its power in each term of the expansion ranges from $n$ down to $0$ .
$b$	The second term of the binomial; its power in each term ranges from $0$ up to $n$ .
$n$	The exponent on the binomial, which must be a non-negative integer; if negative or fractional, you need a different form like the generalized binomial series.
$k$	The summation index running from $0$ to $n$ ; it determines which term you are looking at and controls the exponents of $a$ and $b$ in that term.

When to use it

Reach for the Binomial Theorem when you need to expand $(a+b)^n$ where $n$ is a non-negative integer.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Expand a binomial with small positive exponent

Problem

Expand

(x + 2)^3

$(x + 2)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} 2^k$
We apply the Binomial Theorem with $a = x$ , $b = 2$ , and $n = 3$ .
$\binom{3}{0} x^3 + \binom{3}{1} x^2 \cdot 2 + \binom{3}{2} x \cdot 4 + \binom{3}{3} \cdot 8$
We expand the sum by writing all four terms ( $k = 0, 1, 2, 3$ ) and compute the constant powers: $2^0 = 1$ , $2^1 = 2$ , $2^2 = 4$ , $2^3 = 8$ .
$1 \cdot x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 4 + 1 \cdot 8$
We compute each binomial coefficient: $\binom{3}{0} = 1$ , $\binom{3}{1} = 3$ , $\binom{3}{2} = 3$ , $\binom{3}{3} = 1$ .
$x^3 + 6x^2 + 12x + 8$
We multiply out: $3 \cdot 2 = 6$ , $3 \cdot 4 = 12$ , giving the final expanded form.

Answer: $x^3 + 6x^2 + 12x + 8$

This straightforward application of the Binomial Theorem shows how the formula eliminates the need to multiply $(x+2)(x+2)(x+2)$ by hand. The four terms correspond to the four coefficients from Pascal's Triangle, and the powers of $x$ decrease while the powers of $2$ increase systematically.

2. Expand a binomial with a negative term

Problem

Expand

(2a - b)^4

$(2a - b)^4 = (2a + (-b))^4$
We rewrite the subtraction as addition of a negative term so the Binomial Theorem formula applies.
$\sum_{k=0}^{4} \binom{4}{k} (2a)^{4-k} (-b)^k$
We apply the Binomial Theorem with $a = 2a$ , $b = -b$ , and $n = 4$ , substituting directly into the formula.
$\binom{4}{0}(2a)^4 + \binom{4}{1}(2a)^3(-b) + \binom{4}{2}(2a)^2 b^2 + \binom{4}{3}(2a)(-b)^3 + \binom{4}{4} b^4$
We expand all five terms ( $k = 0$ to $4$ ), noting that $(-b)^2 = b^2$ (positive) and $(-b)^3 = -b^3$ (negative).
$1 \cdot 16a^4 + 4 \cdot 8a^3 \cdot (-b) + 6 \cdot 4a^2 \cdot b^2 + 4 \cdot 2a \cdot (-b^3) + 1 \cdot b^4$
We compute powers: $(2a)^4 = 16a^4$ , $(2a)^3 = 8a^3$ , $(2a)^2 = 4a^2$ , $(2a)^1 = 2a$ , and the binomial coefficients $\binom{4}{0} = 1$ , $\binom{4}{1} = 4$ , $\binom{4}{2} = 6$ , $\binom{4}{3} = 4$ , $\binom{4}{4} = 1$ .
$16a^4 - 32a^3 b + 24a^2 b^2 - 8ab^3 + b^4$
We multiply each term: $4 \cdot 8 = 32$ , $6 \cdot 4 = 24$ , $4 \cdot 2 = 8$ , and the signs alternate because of the negative second term.

Answer: $16a^4 - 32a^3 b + 24a^2 b^2 - 8ab^3 + b^4$

This example shows why careful attention to negative terms matters: the signs in the expansion alternate. Many students forget that even powers of a negative number are positive (so $(-b)^2 = b^2$ ) and odd powers are negative. The Binomial Theorem handles this automatically if you substitute the negative term correctly.

3. Apply the theorem to a real-world measurement in a lab

Problem

A laboratory measures temperature change in a heating system. The temperature reading (in Kelvin) is modeled as

(T_0 + \Delta T)^2

, where

T_0 = 300\text{ K}

is the baseline temperature and

\Delta T = 0.1\text{ K}

is a small correction. Use the Binomial Theorem to find the exact expanded form of the temperature reading.

$(300 + 0.1)^2 = \sum_{k=0}^{2} \binom{2}{k} 300^{2-k} (0.1)^k$
We set $a = 300$ , $b = 0.1$ , and $n = 2$ , then apply the Binomial Theorem formula.
$\binom{2}{0} \cdot 300^2 + \binom{2}{1} \cdot 300 \cdot 0.1 + \binom{2}{2} \cdot (0.1)^2$
We expand the summation by writing out all three terms for $k = 0, 1, 2$ .
$1 \cdot 90000 + 2 \cdot 300 \cdot 0.1 + 1 \cdot 0.01$
We compute the binomial coefficients and powers: $300^2 = 90000$ , $\binom{2}{0} = 1$ , $\binom{2}{1} = 2$ , $\binom{2}{2} = 1$ , $(0.1)^2 = 0.01$ .
$90000 + 60 + 0.01 = 90060.01$
We multiply and add: $2 \cdot 300 \cdot 0.1 = 60$ , and summing all terms gives $90000 + 60 + 0.01 = 90060.01$ .

Answer: $90060.01\text{ K}$

In scientific work, the Binomial Theorem reveals the structure of how small perturbations affect a system. The dominant term $300^2 = 90000$ is the baseline, and the correction term $60.01$ shows how the small change $\Delta T = 0.1$ contributes to the overall reading. This is why the theorem is invaluable in physics: it breaks complex expressions into interpretable pieces.

Common mistakes

Where Binomial Theorem usually goes wrong

The mistake	Why it is wrong	The fix
Writing $(a + b)^n = a^n + b^n$ and ignoring all the middle terms.	The Binomial Theorem produces $n + 1$ terms (one for each value of $k = 0, 1, \ldots, n$ ), and the middle terms are often the largest contributions.	Always include every term in the sum. For example, $(a + b)^2 = a^2 + 2ab + b^2$ , not $a^2 + b^2$ ; the $2ab$ term is essential and comes from $\binom{2}{1} = 2$ ways to pick one $a$ and one $b$ .
Using the Binomial Theorem formula when $n$ is negative, fractional, or a variable, without modification.	The standard Binomial Theorem only applies when $n$ is a non-negative integer; negative and non-integer exponents require the generalized binomial series, which is an infinite sum with a different form.	For $n < 0$ or non-integer, use the generalized binomial series or rewrite the expression to avoid the invalid exponent. For example, $(1 + x)^{-1} = \frac{1}{1+x}$ is better solved by using the series formula or algebraic manipulation than by trying to apply the finite binomial theorem.
Forgetting to raise a coefficient to the power, such as writing $(2a)^3 = 2a^3$ instead of $8a^3$ .	When a coefficient and variable are grouped (like $2a$ ), the exponent applies to both parts: $(2a)^3 = 2^3 \cdot a^3$ , not $2 \cdot a^3$ .	Always expand the full power on grouped terms: $(2a)^3 = 2^3 a^3 = 8a^3$ . Write out the intermediate step $2^3 = 8$ if needed to avoid errors.

Tips and when to use something else

Use Pascal's Triangle to find binomial coefficients quickly for small values of $n$ —the $n$ -th row gives the coefficients directly, saving you time computing $\binom{n}{k}$ by hand.
When you only need the first few terms of an expansion (not the whole thing), the Binomial Theorem is far faster than multiplying the binomial out; for large $n$ , it is the only practical method.
If the exponent is negative, fractional, or a variable (like $(a+b)^x$ ), do not use this form—instead reach for the Generalized Binomial Series, which is an infinite series.
For expansions involving subtraction like $(a - b)^n$ , rewrite as $(a + (-b))^n$ and track the alternating signs: even powers of $-b$ are positive, odd powers are negative.

Related formulas

Pascal's TrianglePascal's Triangle displays binomial coefficients and generates the coefficients for polynomial expansion, perfect for computing combinations quickly.Perfect Square TrinomialQuickly factor trinomials of the form a^2 \pm 2ab + b^2 by recognizing the perfect square pattern that gives (a \pm b)^2.Distributive PropertyThe Distributive Property lets you multiply a number by a sum or difference by distributing the multiplication to each term separately.

Frequently asked questions

Why is $(a + b)^2 = a^2 + 2ab + b^2$ and not just $a^2 + b^2$ ?: The Binomial Theorem includes all the cross terms that arise when you multiply. Expanding $(a+b)^2 = (a+b)(a+b)$ gives you $a \cdot a + a \cdot b + b \cdot a + b \cdot b = a^2 + 2ab + b^2$ . The middle term $2ab$ comes from the two ways to pick one $a$ and one $b$ from the two parentheses, which is exactly what the binomial coefficient $\binom{2}{1} = 2$ counts.
When should I use the Binomial Theorem instead of just multiplying the binomial out?: For small exponents like $n = 2$ or $n = 3$ , multiplying by hand may be quicker. But for $n \geq 4$ , or when you only need a few specific terms, the Binomial Theorem is clearer and much less error-prone. It also shows you the structure of the expansion, which helps with factoring and simplification later.
How do I use the Binomial Theorem for $(a - b)^n$ ?: Rewrite it as $(a + (-b))^n$ and apply the formula directly; the expansion will have alternating signs because even powers of $(-b)$ are positive and odd powers are negative. If you track the substitution carefully, the signs sort themselves out automatically.
What is a binomial coefficient $\binom{n}{k}$ and how do I compute it?: A binomial coefficient $\binom{n}{k}$ counts the number of ways to choose $k$ items from $n$ items. You can compute it using $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , or read the values from Pascal's Triangle for small $n$ . For example, $\binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6$ .

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18