Binomial Expansion Calculator

A binomial expansion calculator rewrites a power such as (2x + 3)⁵ as a sum of terms. Instead of multiplying five copies of (2x + 3), the binomial theorem jumps straight to the answer and can produce any single term on its own.

The binomial theorem formula

For any numbers or expressions a and b and any whole number n:

Three facts follow at once. The expansion has n + 1 terms. In every term the exponents of a and b add up to n, with the power of a falling from n to 0 as the power of b rises from 0 to n. And the coefficients are symmetric, C(n, k) = C(n, n − k), so the second half of each row mirrors the first. When a and b carry their own numbers, as in (2x + 3), those numbers are raised to the same powers and multiplied into each coefficient.

Worked example 1: expanding (2x + 3)⁵

Here a = 2x, b = 3 and n = 5. Row 5 of Pascal's triangle supplies the binomial coefficients 1, 5, 10, 10, 5, 1, and each term multiplies one of them by a power of 2 and a power of 3:

So (2x + 3)⁵ = 32x⁵ + 240x⁴ + 720x³ + 1,080x² + 810x + 243. Two quick checks confirm it: the coefficients add up to 3,125, which is (2 + 3)⁵, and the outer terms are simply (2x)⁵ and 3⁵. With n = 5 there are two middle terms, 720x³ and 1,080x², and 1,080 is the largest coefficient.

Worked example 2: (x − 2y)⁴ and alternating signs

A minus sign belongs to the second term, so b = −2y. Its powers alternate in sign, (−2)⁰ = 1, (−2)¹ = −2, then 4, −8 and 16, and they multiply row 4 of Pascal's triangle (1, 4, 6, 4, 1):

Every term keeps total degree 4 (x³y, x²y², xy³), and the signs alternate because only odd powers of −2 are negative. The coefficients add up to (1 − 2)⁴ = 1, while the alternating sum 1 + 8 + 24 + 32 + 16 equals (1 + 2)⁴ = 81. Those two numbers are the fastest way to catch a dropped sign.

Finding one term without expanding everything

Exams often ask for a single term. Use the general term T_r+1 = C(n, r) a^n−r b^r and remember that position k means r = k − 1:

• The 4th term of (3x − 2)⁷: r = 3, so T₄ = C(7, 3)(3x)⁴(−2)³ = 35 × 81 × (−8) × x⁴ = −22,680x⁴.
• The coefficient of x⁵ in (2x + 3)⁸: x⁵ needs 8 − r = 5, so r = 3 and the coefficient is C(8, 3) × 2⁵ × 3³ = 56 × 32 × 27 = 48,384.
• The term independent of x in (x² + 2/x)⁶: the general term holds x^2(6−r) · x^−r = x^12−3r. Setting 12 − 3r = 0 gives r = 4, so the constant term is T₅ = C(6, 4) × 2⁴ = 15 × 16 = 240.

The calculator's term finder covers both question types: By position returns T_k with every factor shown, and By power searches for the term that contains a chosen power of a variable. For the last example, enter coefficient 2, variable x and power −1 as the second term.

Pascal's triangle: where the coefficients come from

Pascal's triangle starts with a 1 at the top; every other entry is the sum of the two entries above it. Row n, counted from 0, holds C(n, 0) through C(n, n), the coefficients of (a + b)ⁿ. These are rows 0 to 10, each with its sum:

Reading the triangle gives expansions instantly: row 6 means (x + 1)⁶ = x⁶ + 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1. The calculator draws rows 0 to 15 as a triangle and lists the full row n for larger exponents.

The largest coefficient and the middle term

Binomial coefficients rise toward the middle of each row and fall again, so the middle term always carries the largest one. As n grows, that central coefficient gets enormous while taking a shrinking share of the row total:

For odd n the largest value appears twice, in the two middle terms (C(15, 7) = C(15, 8) = 6,435). Once a and b differ, the largest coefficient of the full expansion moves toward the side of the bigger number: in (2x + 3)⁵ it is the 4th term, where 3 carries the higher power. At n = 30 the middle coefficient already has nine digits, which is why the calculator uses exact integers instead of ordinary floating-point numbers.

Sum of the coefficients and other quick checks

Substituting 1 for every variable turns any expansion into a single number, which gives free checks on hand work:

• Sum of all coefficients = (a + b)ⁿ. (2x + 3)⁵ gives 3,125; (3x − 2)⁷ gives exactly 1, even though its terms run from 2,187x⁷ down to −128.
• Alternating sum = (a − b)ⁿ. For (3x − 2)⁷ this is 5⁷ = 78,125.
• Sum of row n of Pascal's triangle = 2ⁿ. Row 10 adds up to 1,024.
• First and last terms are always aⁿ and bⁿ.

How to use this calculator

Enter the first term as a coefficient, a variable letter and the power on that variable, then the second term the same way, then the exponent n. Coefficients accept whole numbers, decimals and fractions (3, −2, 0.5, 3/4). Leave a variable blank for a constant term, and use a negative power for terms like 2/x. The result card shows the expansion, the number of terms, the sum of the coefficients, 2ⁿ and the middle term or terms. Below it you can find any term by position or by power, read every term step by step with C(n, r), a^n−r and b^r in separate columns, and see Pascal's triangle with your row highlighted.

Who this calculator is for

Algebra 2 and precalculus students check expansions and term questions, and the step-by-step columns show where a homework answer went wrong. Statistics students meet the same coefficients in the binomial distribution: the number of ways to get 3 heads in 10 coin flips is C(10, 3) = 120. Calculus students use binomial expansions when they approximate powers or derive series, and teachers use the calculator to build answer keys with fractional or negative coefficients.

Key terms

• Binomial: a sum or difference of two terms, such as 2x + 3 or x − 2y.
• Binomial coefficient C(n, k): n! ÷ (k!(n − k)!), read "n choose k" and written nCr on calculators.
• General term: T_r+1 = C(n, r) a^n−r b^r, the formula for any single term.
• Middle term: the central term (n even) or pair of terms (n odd), carrying the largest binomial coefficient.
• Constant term: the term in which the variable's exponent is zero.
• Pascal's triangle: the triangular table of binomial coefficients in which each entry is the sum of the two above it.

What changes the expansion

• The exponent n sets the number of terms (n + 1) and the row of Pascal's triangle.
• The coefficients a and b scale each term by a^n−r b^r, which shifts the largest coefficient toward the bigger of the two.
• A negative b makes the signs alternate; a negative a does the same, starting with the sign of aⁿ.
• Powers on the variables change only the exponents: (x² + 1)⁴ has the coefficients of (x + 1)⁴ on x⁸, x⁶, x⁴, x² and 1.
• Fractions stay exact: (x/2 + 3)⁴ begins with (1/16)x⁴, and its coefficients sum to (7/2)⁴ = 2,401/16.

Tips and limitations

When expanding by hand, step from one coefficient to the next with the ratio T_r+1 ÷ T_r = (n − r + 1)/r × b/a: in (2x + 3)⁵, 32 × 7.5 = 240, 240 × 3 = 720, 720 × 1.5 = 1,080. Use the symmetry of Pascal's triangle to halve the work, and finish with the sum check.

The theorem as used here needs a whole-number exponent: negative and fractional exponents lead to Newton's infinite binomial series, and three or more terms, such as (x + y + z)ⁿ, need the multinomial theorem. Coefficients must be numbers, and n is capped at 30 for readability, not precision.

Related calculators

• For C(n, k) and nPr on their own, use the Permutation & Combination Calculator.
• For the factorials inside every binomial coefficient, use the Factorial Calculator.
• To multiply out polynomials with more than two terms, use the Polynomial Calculator.
• For binomial probabilities such as 3 heads in 10 flips, use the Probability Calculator.

Sources

• National Institute of Standards and Technology (NIST): Digital Library of Mathematical Functions, §1.2 Elementary Algebra: binomial coefficients (eq. 1.2.1), the binomial theorem (eq. 1.2.2) and the row sum 2ⁿ (eq. 1.2.3).
• NIST: Digital Library of Mathematical Functions, §26.3 Lattice Paths: Binomial Coefficients, including the recurrence behind Pascal's triangle (eq. 26.3.5).
• OEIS Foundation: A007318, Pascal's triangle read by rows, and A000984, central binomial coefficients.
• Every expansion, coefficient and table entry on this page was computed with exact integer arithmetic and checked against the calculator.