College Algebra

The Binomial Theorem

Last updated: March 2026 · Advanced

The binomial theorem gives a formula for expanding any power of a binomial (a+b)n(a + b)^n without multiplying it out step by step. It connects algebra with combinatorics through binomial coefficients and Pascal’s Triangle. Beyond pure algebra, the binomial theorem appears in probability theory, compound interest approximations, and numerical analysis.

Factorial Notation

Before stating the theorem, you need factorial notation. The factorial of a non-negative integer nn is:

n!=n×(n−1)×(n−2)×⋯×2×1n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

Special cases:

  • 0!=10! = 1 (by definition)
  • 1!=11! = 1
  • 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120
  • 10!=3,628,80010! = 3{,}628{,}800

Factorials grow extremely fast — 20!20! is already over 2.4×10182.4 \times 10^{18}.

Binomial Coefficients

The binomial coefficient “n choose k” is:

(nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

This counts the number of ways to choose kk items from nn items (order does not matter). You will encounter these again in permutations and combinations.

Computing Binomial Coefficients

(62)=6!2!⋅4!=6×52×1=15\binom{6}{2} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15

Shortcut: Cancel the larger factorial in the denominator. For (nk)\binom{n}{k}, multiply the top kk consecutive integers starting from nn and divide by k!k!:

(83)=8×7×63×2×1=3366=56\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

Key properties:

  • (n0)=1\binom{n}{0} = 1 and (nn)=1\binom{n}{n} = 1 for all nn
  • (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k} (symmetry: choosing what to include is the same as choosing what to exclude)
  • (n1)=n\binom{n}{1} = n

Pascal’s Triangle

Pascal’s Triangle arranges binomial coefficients in a triangular pattern. Each entry is the sum of the two entries directly above it.

\text{Row 0:} \quad 1 \\ \text{Row 1:} \quad 1 \quad 1 \\ \text{Row 2:} \quad 1 \quad 2 \quad 1 \\ \text{Row 3:} \quad 1 \quad 3 \quad 3 \quad 1 \\ \text{Row 4:} \quad 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ \text{Row 5:} \quad 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\ \text{Row 6:} \quad 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \end{array}$$ Row $n$ contains the binomial coefficients $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$. The addition rule that generates Pascal's Triangle is: $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$ ## The Binomial Theorem For any non-negative integer $n$: $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ Written out term by term: $$(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n$$ **Key observations:** - There are $n + 1$ terms in the expansion - The powers of $a$ decrease from $n$ to $0$ - The powers of $b$ increase from $0$ to $n$ - In each term, the exponents of $a$ and $b$ sum to $n$ - The coefficients are the entries of row $n$ in Pascal's Triangle ## Worked Examples ### Example 1: Expand $(x + y)^4$ Using row 4 of Pascal's Triangle: $1, 4, 6, 4, 1$ $$(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$$ ### Example 2: Expand $(2x - 3)^3$ Here $a = 2x$ and $b = -3$, $n = 3$. Row 3: $1, 3, 3, 1$. $$(2x - 3)^3 = \binom{3}{0}(2x)^3(-3)^0 + \binom{3}{1}(2x)^2(-3)^1 + \binom{3}{2}(2x)^1(-3)^2 + \binom{3}{3}(2x)^0(-3)^3$$ $$= 1(8x^3)(1) + 3(4x^2)(-3) + 3(2x)(9) + 1(1)(-27)$$ $$= 8x^3 - 36x^2 + 54x - 27$$ **Watch the signs:** When $b$ is negative, the terms alternate in sign. ### Example 3: Expand $(x + 2)^5$ Row 5: $1, 5, 10, 10, 5, 1$. Here $a = x$, $b = 2$. $$(x + 2)^5 = x^5 + 5x^4(2) + 10x^3(4) + 10x^2(8) + 5x(16) + 32$$ $$= x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32$$ ### Example 4: Finding a Specific Term Find the 4th term of $(3x - 2y)^7$. The $(k+1)$th term (starting from $k = 0$) is: $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$ The 4th term has $k = 3$: $$T_4 = \binom{7}{3}(3x)^{7-3}(-2y)^3$$ $$= 35 \cdot (3x)^4 \cdot (-2y)^3$$ $$= 35 \cdot 81x^4 \cdot (-8y^3)$$ $$= 35 \cdot 81 \cdot (-8) \cdot x^4y^3$$ $$= -22{,}680 x^4 y^3$$ ### Example 5: Finding the Coefficient of a Specific Power In the expansion of $(x + 3)^6$, find the coefficient of $x^4$. We need the term where $x$ has exponent 4. Since $a = x$ and $b = 3$: $$a^{n-k} = x^{6-k} = x^4 \Rightarrow k = 2$$ $$T_3 = \binom{6}{2} x^4 \cdot 3^2 = 15 \cdot x^4 \cdot 9 = 135x^4$$ The coefficient is **135**. ## Special Cases and Identities Setting $a = 1$ and $b = 1$: $$(1 + 1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k}$$ This proves that the sum of all entries in row $n$ of Pascal's Triangle is $2^n$. Setting $a = 1$ and $b = -1$: $$(1 - 1)^n = 0 = \sum_{k=0}^{n} (-1)^k\binom{n}{k}$$ This proves that the alternating sum of entries in any row (except row 0) is zero. ## Real-World Application: Compound Interest Approximation An investment earns 5 percent annual interest. After $n$ years at compound interest, the growth factor is $(1 + 0.05)^n$. For small interest rates, the binomial theorem provides useful approximations. For 3 years: $$(1.05)^3 = \binom{3}{0}(1)^3(0.05)^0 + \binom{3}{1}(1)^2(0.05)^1 + \binom{3}{2}(1)^1(0.05)^2 + \binom{3}{3}(0.05)^3$$ $$= 1 + 0.15 + 0.0075 + 0.000125 = 1.157625$$ The first two terms ($1 + 0.15 = 1.15$) give the "simple interest" approximation. Each additional term adds a smaller correction. ## Common Mistakes 1. **Forgetting $0! = 1$.** This is a definition, not a typo. 2. **Wrong sign pattern with subtraction.** In $(a - b)^n$, substitute $b \to -b$: the signs alternate because $(-b)^k$ is negative when $k$ is odd. 3. **Off-by-one error on term number.** The "4th term" uses $k = 3$ (since the first term has $k = 0$). 4. **Not raising the entire term to the power.** In $(2x)^4$, the coefficient $2$ also gets raised: $(2x)^4 = 16x^4$, not $2x^4$. ## Practice Problems <details> <summary><strong>Problem 1:</strong> Expand $(a + b)^5$ using Pascal's Triangle.</summary> Row 5: $1, 5, 10, 10, 5, 1$ $$(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$ </details> <details> <summary><strong>Problem 2:</strong> Expand $(x - 1)^4$.</summary> Row 4: $1, 4, 6, 4, 1$. With $a = x$, $b = -1$: $$(x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1$$ Signs alternate because $(-1)^k$ alternates. </details> <details> <summary><strong>Problem 3:</strong> Find the coefficient of $x^3$ in $(2x + 5)^5$.</summary> Need $a^{n-k} = (2x)^{5-k}$ with exponent 3, so $5 - k = 3$, $k = 2$. $T_3 = \binom{5}{2}(2x)^3(5)^2 = 10 \cdot 8x^3 \cdot 25 = 2000x^3$ **Coefficient: 2000** </details> <details> <summary><strong>Problem 4:</strong> Find the 5th term of $(x - 3y)^8$.</summary> 5th term: $k = 4$. $T_5 = \binom{8}{4}(x)^4(-3y)^4 = 70 \cdot x^4 \cdot 81y^4 = 5670x^4y^4$ (Positive because $(-3)^4 = 81$ is positive.) </details> <details> <summary><strong>Problem 5:</strong> Compute $\binom{10}{3}$ and verify using the formula.</summary> $$\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$$ **Verification with symmetry:** $\binom{10}{3} = \binom{10}{7} = 120$. </details> ## Key Takeaways - The **binomial theorem** expands $(a + b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ - **Binomial coefficients** $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ count the number of ways to choose $k$ from $n$ - **Pascal's Triangle** gives binomial coefficients visually — each entry is the sum of the two above it - The $(k+1)$th term of the expansion has $k$ as the exponent on $b$ (zero-indexed) - When the binomial involves **subtraction**, signs alternate based on the power of $(-b)$ - The sum of row $n$ of Pascal's Triangle is $2^n$ Return to [College Algebra](/topics/college-algebra/) for more topics in this section.

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Last updated: March 29, 2026