Polynomial Calculator

How to use this calculator

1. Write powers with a caret3x^2, not 3x2 and not 3x². Coefficients of 1 can be omitted, so x^3 and 1x^3 both work, and terms may appear in any order.
2. Pick the operationFor division, the first field is the dividend and the second is the divisor. The order matters for subtraction and division, not for addition and multiplication.
3. Use the evaluation pointEntering a value for x shows what each polynomial and the result come to there. It is the quickest way to check an answer — if the result is correct, evaluating it must match doing the operation on the two evaluated values.
4. Read the remainderOn division, a zero remainder means the divisor is a factor. A non-zero remainder is reported separately rather than folded into the quotient.

The four operations, and what changes the degree

Addition and subtraction work power by power. Line the two polynomials up, combine the coefficients of matching powers, and leave untouched any power that appears in only one of them. The degree of the result is at most the larger of the two degrees — and can be smaller, if the leading terms cancel. Subtracting 3x² − 5x from 3x² + 2x gives 7x, a degree-one result from two degree-two inputs.

Multiplication distributes every term across every term. Coefficients multiply and exponents add. The degree of a product is always exactly the sum of the degrees, with no exceptions, which makes it a reliable check: multiply a cubic by a quadratic and the answer must be quintic. If it is not, a term was dropped.

Long division is the one that needs a procedure rather than a formula. Divide the leading term of the dividend by the leading term of the divisor to get the first quotient term. Multiply the entire divisor by that term and subtract, which cancels the leading term and lowers the degree by one. Repeat until what remains has a lower degree than the divisor. That leftover is the remainder.

The relationship that ties it together is dividend = divisor × quotient + remainder, and it is worth using as a check. Multiply the quotient by the divisor, add the remainder, and you must get back the original dividend exactly.

Worked example: (3x² − 5x + 7)(x − 2)

Distribute the first polynomial across both terms of the second.

Against x: 3x² · x = 3x³, −5x · x = −5x², 7 · x = 7x.

Against −2: 3x² · (−2) = −6x², −5x · (−2) = +10x, 7 · (−2) = −14.

Collect like powers. The x³ term stands alone at 3x³. The x² terms are −5x² and −6x², giving −11x². The x terms are 7x and 10x, giving 17x. The constant is −14.

Result: 3x³ − 11x² + 17x − 14.

Degree check: a degree-2 polynomial times a degree-1 polynomial must give degree 3, and it does.

Numerical check at x = 2. The first polynomial gives 3(4) − 5(2) + 7 = 12 − 10 + 7 = 9. The second gives 2 − 2 = 0. Their product must be 0. And the result at x = 2 is 3(8) − 11(4) + 17(2) − 14 = 24 − 44 + 34 − 14 = 0. It matches.

That zero is not a coincidence — it says x = 2 is a root of the product, which it must be, since x − 2 was a factor. Reversing the operation and dividing 3x³ − 11x² + 17x − 14 by x − 2 returns 3x² − 5x + 7 with no remainder.

Degree behaviour at a glance

The last row is the stopping condition for long division, and it is what tells you the algorithm has finished. If the remainder still has degree equal to or greater than the divisor, there is another step to take. Students who stop early usually stop exactly one step early.

The factor and remainder theorems

Long division connects directly to two results that appear constantly in exams, and understanding the link makes both obvious rather than memorised.

The remainder theorem says that dividing a polynomial P(x) by (x − a) leaves a remainder equal to P(a). Not a remainder related to P(a) — exactly P(a). So to find the remainder when 3x³ − 11x² + 17x − 14 is divided by x − 2, evaluate the polynomial at 2. We did that above and got 0.

The factor theorem is the special case where that remainder is zero. If P(a) = 0, then (x − a) divides P(x) exactly, so (x − a) is a factor and a is a root. The two statements — a is a root, and (x − a) is a factor — are the same statement.

This is why finding one root of a cubic collapses the problem. Divide out the corresponding linear factor and what remains is a quadratic, which the quadratic formula handles directly. A cubic that looks intractable becomes routine as soon as one root is spotted.

Spotting that first root usually means trying the rational root candidates: factors of the constant term divided by factors of the leading coefficient. For 3x³ − 11x² + 17x − 14, the constant is −14 and the leading coefficient is 3, so the candidates are ±1, ±2, ±7, ±14, each optionally over 3. Testing 2 gives zero on the first sensible guess.

Synthetic division is a shorthand for dividing by a linear factor that drops the variables and works only with coefficients. It is faster by hand and produces the same quotient and remainder, but it applies only when the divisor is linear — which is why the general long division above is still worth knowing.

Assumptions and limitations

• Single variable only — the parser reads x and no other letter
• Exponents must be whole numbers from 0 to 20
• Does not factorise, find roots symbolically, or simplify algebraic fractions
• Coefficients are handled as decimals, so results may show rounding at many decimal places
• Division reports quotient and remainder separately rather than as a single algebraic fraction
• Does not handle complex coefficients
• Implicit multiplication such as (x+1)(x-2) is not parsed — expand brackets first, or use the multiply operation

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