Audited 05 Aug 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Multiplying Polynomials Calculator

Multiply two polynomials from their coefficient lists, constant term first: get the expanded product coefficients and its value at any chosen x.

Multiplying Polynomials Calculator

Product evaluated at x
55
The product polynomial's value at the chosen x, which always equals A(x) × B(x).
Product coefficients
3, 10, 8

Background.

Expanding (2x + 1)(4x + 3) by hand is a matter of FOIL: four products, collect like terms, done. Expanding a quartic times a cubic means twenty products and a real chance of dropping a sign or filing a term under the wrong power. The bookkeeping, not the algebra, is where the mistakes happen.

The rule behind every expansion is distribution. Each term of the first polynomial multiplies each term of the second, and a term in x^i times a term in x^j lands on x^(i+j). Gather everything that lands on the same power and you have the product. Written with coefficients alone, that is the convolution c_k = Σ a_i × b_j over all pairs with i + j = k, the same operation used to combine probability distributions, smooth digital signals and, minus the carrying, multiply whole numbers digit by digit.

Polynomials go in here as lists of coefficients with the constant first, so 1 + 2x is typed 1, 2 and 5 − x + 3x³ is typed 5, -1, 0, 3. The zero matters: it holds the place of the missing x² term. The product comes back in the same order, together with its value at an x you choose. That second number is a built-in check, because the product evaluated at any x must equal A(x) × B(x), which you can confirm on a pocket calculator.

Only single-variable polynomials with numeric coefficients are handled. Coefficients may be negative or decimal, but factoring, division and coefficients containing other letters, such as a or y, are outside the page's reach.

What is multiplying polynomials calculator?

Multiplying polynomials means expanding the product of two polynomials into a single polynomial in standard form. If A has degree m and B has degree n, the product has degree m + n, and each of its coefficients is the sum of every product of a coefficient from A with one from B whose powers add up to that exponent. The result is given as a coefficient list, constant term first, plus the product's numerical value at a chosen x.

How to use this calculator.

  1. Type the coefficients of polynomial A from the constant term upward, separated by commas, spaces or semicolons. For 4x² + 3 enter 3, 0, 4, with a zero standing in for the missing x term.
  2. Do the same for polynomial B. Each list needs at least two numbers, so a plain constant such as 5 goes in as 5, 0. Use an ordinary hyphen for negative signs.
  3. Choose a value of x at which to evaluate the product, and use it as a check: the result should equal A(x) × B(x).
  4. Read the product coefficients in the same constant-first order. A list of 3, 10, 8 means 8x² + 10x + 3.

The formula.

c_k = sum from i+j=k of a_i b_j

If A(x) = a₀ + a₁x + … + a_m x^m and B(x) = b₀ + b₁x + … + b_n x^n, the product is C(x) = c₀ + c₁x + … + c_(m+n) x^(m+n), where c_k = Σ a_i × b_j over every pair of indices with i + j = k.

In practice each coefficient of A is multiplied by each coefficient of B, (m + 1) × (n + 1) multiplications in all, and each product is added into the slot for the sum of the two powers. The lowest coefficient is always a₀b₀ and the highest a_m × b_n; the ones in between are sums. The page then evaluates C at your x by adding c_k × x^k over every k.

Two properties are worth knowing. The degree of the product is the sum of the degrees, so the coefficient list always has one entry fewer than the two input lists combined. And the product's roots are exactly the roots of A together with those of B, since C(x) is zero precisely when one of the factors is.

A worked example.

Example

Polynomial A is entered as 1, 2, meaning 1 + 2x, and polynomial B as 3, 4, meaning 3 + 4x. The constant term is 1 × 3 = 3. The x term collects two cross products, 1 × 4 + 2 × 3 = 4 + 6 = 10. The x² term is 2 × 4 = 8. So the product is 3 + 10x + 8x², listed as 3, 10, 8. At x = 2 it is worth 3 + 10 × 2 + 8 × 2² = 3 + 20 + 32 = 55. Checking against the factors, A(2) = 5 and B(2) = 11, and 5 × 11 = 55. Evaluate at x = −0.5 instead and the product gives 3 − 5 + 2 = 0, because −0.5 is the root of 1 + 2x.

polynomial B3,4
polynomial A1,2
evaluate At2

Frequently asked questions.

Why are the coefficients entered constant first?
Because each number's position in the list then equals its power of x: the first multiplies x⁰, the second x¹, and so on. That keeps the product rule simple, since position i times position j lands in position i + j. Textbooks usually write the highest power first, so reverse the order when copying.
How do I enter a polynomial with missing powers?
Put a zero in each missing position. x³ − 2 has no x² or x term, so it goes in as -2, 0, 0, 1. Leaving the zeros out would slide the 1 down to the x position and turn the polynomial into x − 2, giving a completely different product.
Can I multiply three or more polynomials?
Not in one step, but multiplication is associative, so you can chain it. Multiply the first two, copy the product's coefficient list into polynomial A, then multiply by the third. Cubing x + 1 takes two passes: 1, 1 times 1, 1 gives 1, 2, 1, and that times 1, 1 gives 1, 3, 3, 1.
How is this related to ordinary long multiplication?
Long multiplication is polynomial multiplication at x = 10, followed by carrying. Treat 12 as 2 + 1x and 34 as 4 + 3x; their product coefficients are 8, 10, 3, and evaluating at x = 10 gives 8 + 100 + 300 = 408, which is 12 × 34. Carrying is what turns the middle 10 into a single digit.
Can polynomial multiplication count dice outcomes?
Yes. Enter 0, 1, 1, 1, 1, 1, 1 for both polynomials, one coefficient for each face from 1 to 6 after a zero for the impossible 0. The product, 0, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, counts the ways two dice reach each total from 0 to 12, and evaluating at x = 1 gives all 36 combinations.
Does matching the value at one x prove my hand expansion is right?
It is a good sanity check but not a proof, because a wrong expansion can happen to agree at a single point. Two different polynomials of degree d can agree at no more than d values of x, so if your expansion matches A(x) × B(x) at d + 1 different points, it is correct.

How this page was produced

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Quanta Calculator
Primary sources
3 cited below
Method
c_k = sum from i+j=k of a_i b_j
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