Substituting c gives the remainder when a polynomial is divided by x − c because the division identity is f(x) = (x − c)q(x) + r. At x = c, the product becomes zero, leaving f(c) = r. The remainder is a constant because the divisor has degree one.
This shortcut has a reason behind it, not just a rule to memorize. The chosen input makes the divisor zero, which removes the entire divisor-times-quotient part of the identity. You can then find the leftover number without calculating the quotient first.
We’ll prove that idea, check it with a cubic, and explain where it stops working. The examples also separate a remainder from a quotient and show why signs matter. For related lessons and tools, start with the algebra hub.
What Does The Division Identity Mean?
Polynomial division writes the dividend as the divisor times a quotient, plus a remainder. The letters name those parts: f(x) is the dividend, q(x) is the quotient, and r is the remainder. For division by x − c, the identity is f(x) = (x − c)q(x) + r.
An identity is an equation that holds for every allowed value of x. It’s stronger than an equation that happens to work at one input. Once you have correctly divided the polynomial, you can replace x with any real number and both sides will still match.
The remainder must have lower degree than the divisor. Degree means the highest power of x with a nonzero coefficient. Since x − c has degree one, a nonzero remainder has degree zero, which makes it a constant; a zero remainder is allowed too.
Why can’t you leave a term such as 3x in the remainder? You could still divide that term by the linear divisor and continue the calculation. Division finishes only after no term of the remainder has a power as high as the divisor’s highest power.
The OpenStax explanation of the remainder theorem uses this division identity and the lower-degree condition to establish the result. The proof below follows those ideas. The numerical examples are separate examples you can check directly.
Why Does Substituting c Leave Only The Remainder?
Replacing x with c makes x − c equal zero. That turns the entire product (x − c)q(x) into zero, regardless of the quotient’s value there. The constant remainder does not change when you change x, so the identity leaves f(c) = r.
Write the substitution in full before simplifying: f(c) = (c − c)q(c) + r. Next, c − c = 0, so f(c) = 0 · q(c) + r. Finally, zero times any finite number is zero, giving f(c) = r.
Notice what the proof does not need: the actual formula for q(x). You know a polynomial quotient exists after division, and that is enough. Every polynomial has a finite value at a real input, so q(c) presents no division-by-zero problem.
This is why substitution can save work when a question asks only for the remainder. If the question asks for the quotient too, evaluating f(c) will not supply it. The shortcut extracts one part of the division result, rather than replacing every step of division.
How Does The Proof Work With A Cubic?
Take f(x) = x³ − 2x² + 3x + 4 and divide by x − 2. The theorem says the remainder is f(2). Substitution gives 2³ − 2(2²) + 3(2) + 4 = 8 − 8 + 6 + 4 = 10.
The full division identity is f(x) = (x − 2)(x² + 3) + 10. Here the quotient is x² + 3, and the remainder is 10. Multiplying the divisor and quotient gives x³ − 2x² + 3x − 6; adding 10 restores the original constant term, 4.
Now use the identity at x = 2: f(2) = (2 − 2)(2² + 3) + 10. The quotient’s value is 7, but its multiplier is zero. That leaves 10, which agrees with the direct calculation from the original cubic.
| Part | Expression or value | Meaning |
|---|---|---|
| Divisor | x − 2 | The polynomial used to divide |
| Quotient | x² + 3 | The polynomial multiplied by the divisor |
| Remainder | 10 | The constant left after division |
| Quotient at 2 | 7 | A value of the quotient, not the remainder |
Could the number 7 be the remainder because it comes from the quotient? No: plugging in the chosen input removes the quotient’s contribution entirely. Keeping the names attached to each expression prevents this common mix-up between q(2) and f(2).
You can use the polynomial division calculator to compare a full quotient and remainder with your substitution result. The useful check is agreement between the two routes. A remainder alone does not reveal all the coefficients of the quotient.
Which Number Do You Substitute?
Substitute the number that makes the divisor zero. For x − 2, use 2; for x + 1, use −1. The sign is not a separate trick to memorize, because solving the small equation x + 1 = 0 already tells you the correct input.
Using the same cubic, f(−1) = (−1)³ − 2(−1)² + 3(−1) + 4. Evaluate the powers first: −1 − 2 − 3 + 4 = −2. The remainder when dividing by x + 1 is therefore −2.
Its division identity is f(x) = (x + 1)(x² − 3x + 6) − 2. Expand the product to get x³ − 2x² + 3x + 6, then subtract 2. This gives the same original polynomial, which checks both the quotient and the negative remainder.
A polynomial remainder can be negative. It does not follow the usual nonnegative remainder rule for whole-number division, such as a remainder between zero and one less than the divisor. Polynomial division restricts the remainder’s degree, rather than its numerical size or sign.
What If The Divisor Is ax − b?
For a linear divisor ax − b with real a ≠ 0, use x = b/a. That input makes ax − b zero, so the remainder is f(b/a). The same proof applies, since the divisor still has degree one and the remainder is still constant.
For example, dividing our cubic by 2x − 4 gives the remainder f(2) = 10. Its quotient is (x² + 3)/2, because 2x − 4 is twice x − 2. Doubling the divisor halves the quotient and leaves the same constant remainder.
Standard synthetic division directly handles the form x − c, whose x coefficient is one. Using c = b/a in that procedure gives the quotient for x − b/a. Divide that quotient by a to get the quotient for ax − b; the remainder needs no such adjustment.
What Does A Zero Remainder Tell You?
A zero remainder means x − c divides the polynomial exactly, so x − c is a factor. For g(x) = x² − 5x + 6, g(2) = 4 − 10 + 6 = 0. The identity is g(x) = (x − 2)(x − 3).
This connection works in both directions. If you already know g(x) contains the factor x − 2, substituting 2 makes the product zero. If substitution gives zero first, the division identity has no leftover constant, so it shows that factor directly.
A nonzero remainder means the divisor is not a factor of that polynomial. Our cubic has remainder 10 at x − 2, even though the quadratic has remainder zero there. You must name both the polynomial and the divisor when reporting the test.
Finding one zero does not automatically finish a factoring problem. It tells you one linear factor; the quotient may need more work. The factoring polynomials lesson explains how to continue when your task is a full factorization.
Why Doesn’t One Value Give Every Polynomial Remainder?
A nonlinear divisor can leave a remainder that still contains x. For a divisor of degree two, the remainder may have degree one. Substitution at a zero of that divisor gives one value of the remainder, rather than the entire remainder polynomial.
For example, x³ divided by x² − 1 has identity x³ = (x² − 1)x + x. The quotient is x, and the remainder is also x. Substituting x = 1 gives 1 = 0 + 1, which checks r(1) but doesn’t turn r(x) into the constant 1.
The same remainder has r(−1) = −1. Those two different values show why one constant cannot replace it. The linear-divisor theorem works because the lower-degree rule has already forced the remainder to be constant before you substitute anything.
Why Not Substitute Into The Fraction?
Use the multiplication identity, not the original fraction f(x)/(x − c), at x = c. That fraction has a zero denominator there and is undefined. The polynomial quotient q(x) from division is a different expression and can be evaluated normally.
When the remainder is nonzero, the fraction has an extra term r/(x − c) for x ≠ c. When the remainder is zero, cancellation may produce a polynomial extension, but the original fraction still has a missing input at c. Neither case requires evaluating the original fraction there.
How Can You Check Your Reasoning?
Start by solving the divisor-equals-zero equation, then substitute that input into the dividend. Use parentheses around negative inputs and calculate powers before multiplying. Label the result as the remainder so you don’t accidentally report a quotient value instead.
If you also have a quotient, multiply it by the divisor and add the remainder. The result should match the original polynomial term by term. This check tests the whole division, while evaluating at c checks the leftover constant.
The reason to remember is simple: the divisor disappears at its own zero, leaving the remainder visible. The shortcut gives the complete remainder for a linear divisor because that remainder is constant. For a higher-degree divisor, keep track of the remainder as a polynomial.