Polynomial Long Division With Remainders: Complete Guide

Polynomial long division divides one polynomial by another using the same divide, multiply, subtract, bring down cycle as whole-number long division, working from the highest power downward. Dividing x³ + 2x² − 4x + 9 by x + 3 gives a quotient of x² − x − 1 with a remainder of 12.

Unlike the shortcuts, it has no restriction on the divisor. It handles x − 2, it handles x² − x + 2, and it handles 2x − 1 with no rescaling step afterwards. That generality is why it is worth learning even though it takes longer.

This guide covers the setup, the four-step cycle in detail, linear and quadratic divisors, missing terms, the case where the divisor is bigger than the dividend, how to verify any answer, and eight practice problems.

Setting Up the Division

Two things must be true before the first step.

1. Both polynomials are written in descending order of power. The highest power comes first, then the next, down to the constant. Rewrite 5 − 2x + x³ as x³ − 2x + 5 before starting.

2. Every missing power has a zero placeholder. The polynomial 2x⁴ − 5x + 3 has no x³ term and no x² term, so it is written:

2x⁴ + 0x³ + 0x² − 5x + 3

Skipping those placeholders is the most common cause of a wrong answer, because the columns stop lining up and terms get subtracted from the wrong place.

The layout is the familiar bracket, with the divisor outside and the dividend inside:

              x² −  x −  1
        ┌──────────────────────
x + 3   │ x³ + 2x² − 4x + 9

The Four-Step Cycle

Each round removes exactly one term from the front of what remains.

1. Divide the leading term of what is left by the leading term of the divisor. That result is the next term of the quotient.
2. Multiply the entire divisor by that quotient term.
3. Subtract the product from the current polynomial. The leading terms cancel by construction.
4. Bring down the next term and repeat, until the remaining polynomial has a lower degree than the divisor.

Only the leading terms are used in step 1. The rest of the divisor plays no part in choosing the quotient term; it only matters during the multiply.

Step 3 is where errors cluster, because subtracting a polynomial flips the sign of every term in it, not just the first.

Example 1: A Linear Divisor

Problem. Divide x³ + 2x² − 4x + 9 by x + 3.

Round 1. Divide x³ by x to get x². Multiply: x²(x + 3) = x³ + 3x². Subtract:

 (x³ + 2x²) − (x³ + 3x²) = −x²

Bring down −4x, leaving −x² − 4x.

Round 2. Divide −x² by x to get −x. Multiply: −x(x + 3) = −x² − 3x. Subtract:

 (−x² − 4x) − (−x² − 3x) = −x

Bring down +9, leaving −x + 9.

Round 3. Divide −x by x to get −1. Multiply: −1(x + 3) = −x − 3. Subtract:

 (−x + 9) − (−x − 3) = 12

The remaining polynomial 12 has degree 0, which is lower than the divisor’s degree 1, so the division stops.

Answer. Quotient x² − x − 1, remainder 12, written as:

x³ + 2x² − 4x + 9 = (x + 3)(x² − x − 1) + 12

Verification. Expand the right side:

(x + 3)(x² − x − 1) = x³ − x² − x + 3x² − 3x − 3
                    = x³ + 2x² − 4x − 3

x³ + 2x² − 4x − 3 + 12 = x³ + 2x² − 4x + 9      ✓

A free second check. Because the divisor is linear, the Remainder Theorem applies: the remainder must equal f(−3). Evaluating, f(−3) = −27 + 18 + 12 + 9 = 12. ✓

Example 2: A Quadratic Divisor

Here is where long division earns its place. Neither the Remainder Theorem nor synthetic division can touch this one.

Problem. Divide x⁴ − 3x³ + 2x² + 5x − 7 by x² − x + 2.

Round 1. Divide x⁴ by x² to get x². Multiply: x²(x² − x + 2) = x⁴ − x³ + 2x². Subtract:

 (x⁴ − 3x³ + 2x²) − (x⁴ − x³ + 2x²) = −2x³ + 0x²

Bring down 5x, leaving −2x³ + 0x² + 5x.

Round 2. Divide −2x³ by x² to get −2x. Multiply: −2x(x² − x + 2) = −2x³ + 2x² − 4x. Subtract:

 (−2x³ + 0x² + 5x) − (−2x³ + 2x² − 4x) = −2x² + 9x

Bring down −7, leaving −2x² + 9x − 7.

Round 3. Divide −2x² by x² to get −2. Multiply: −2(x² − x + 2) = −2x² + 2x − 4. Subtract:

 (−2x² + 9x − 7) − (−2x² + 2x − 4) = 7x − 3

The remainder 7x − 3 has degree 1, which is lower than the divisor’s degree 2, so the division stops.

Answer. Quotient x² − 2x − 2, remainder 7x − 3.

x⁴ − 3x³ + 2x² + 5x − 7 = (x² − x + 2)(x² − 2x − 2) + 7x − 3

Notice that the remainder is a polynomial, not a number. That is the general rule.

The Degree Rule for Remainders

The remainder must have a lower degree than the divisor.

This is the polynomial version of “the remainder must be smaller than the divisor” in whole-number division. Size is replaced by degree.

That is also why the Remainder Theorem only works for linear divisors: a degree-1 divisor forces a constant remainder, which one substitution can find. A quadratic divisor leaves bx + c, which needs two unknowns.

Stopping too early or too late both show up here. If your remainder still has degree at least that of the divisor, another round is due.

Example 3: Missing Terms

Problem. Divide 2x⁴ − 5x + 3 by x² + 2.

Set up with placeholders. The dividend becomes 2x⁴ + 0x³ + 0x² − 5x + 3.

Round 1. 2x⁴ ÷ x² = 2x². Multiply: 2x²(x² + 2) = 2x⁴ + 4x². Subtract:

 (2x⁴ + 0x³ + 0x²) − (2x⁴ + 0x³ + 4x²) = −4x²

Bring down −5x, leaving −4x² − 5x. Note that the x³ column contributed 0, exactly as the placeholder promised.

Round 2. −4x² ÷ x² = −4. Multiply: −4(x² + 2) = −4x² − 8. Subtract:

 (−4x² − 5x + 3) − (−4x² + 0x − 8) = −5x + 11

Degree 1 is below degree 2, so stop.

Answer. Quotient 2x² − 4, remainder −5x + 11.

The quotient has no x term at all, which is a legitimate outcome. Its coefficients are 2, 0, −4.

Example 4: A Non-Monic Divisor

Long division handles a leading coefficient other than 1 with no special treatment and no rescaling afterwards.

Problem. Divide 6x³ + x² − 4x + 5 by 2x − 1.

Round 1. 6x³ ÷ 2x = 3x². Multiply: 3x²(2x − 1) = 6x³ − 3x². Subtract:

 (6x³ + x²) − (6x³ − 3x²) = 4x²

Bring down −4x.

Round 2. 4x² ÷ 2x = 2x. Multiply: 2x(2x − 1) = 4x² − 2x. Subtract:

 (4x² − 4x) − (4x² − 2x) = −2x

Bring down +5.

Round 3. −2x ÷ 2x = −1. Multiply: −1(2x − 1) = −2x + 1. Subtract:

 (−2x + 5) − (−2x + 1) = 4

Answer. Quotient 3x² + 2x − 1, remainder 4.

Verification. (2x − 1)(3x² + 2x − 1) + 4 = 6x³ + 4x² − 2x − 3x² − 2x + 1 + 4 = 6x³ + x² − 4x + 5. ✓

This is the practical advantage over the shortcut method. Synthetic division can handle 2x − 1, but only by dividing by x − 1/2 and then halving the quotient afterwards. Long division just produces the right answer.

When the Divisor Has the Higher Degree

Problem. Divide 3x + 5 by x² + 1.

The first step asks for 3x ÷ x², which would give 3x⁻¹, not a polynomial term. That signals the division is over before it starts.

Answer. Quotient 0, remainder 3x + 5.

The degree rule confirms it: 3x + 5 already has a lower degree than x² + 1, so nothing needs removing.

This is the polynomial equivalent of dividing 3 by 7 and getting 0 remainder 3. Written as a single expression, the result is the proper rational function (3x + 5)/(x² + 1), which cannot be simplified further.

Verifying Any Answer

Every polynomial division can be checked with the same identity used for whole numbers:

Dividend = Divisor × Quotient + Remainder

Expand the right side and compare it term by term with the original dividend. If they match, the answer is right.

Two faster partial checks are worth knowing:

• Degree check. The quotient’s degree must equal dividend degree minus divisor degree. Dividing a quartic by a quadratic must give a quadratic quotient.
• Substitution check. Pick any convenient value, say x = 1, and evaluate both sides. For Example 2 at x = 1: the dividend gives 1 − 3 + 2 + 5 − 7 = −2, and the right side gives (1 − 1 + 2)(1 − 2 − 2) + 7 − 3 = 2 × (−3) + 4 = −2. ✓ This catches most sign errors in seconds without a full expansion.

Long Division vs Synthetic Division

Use synthetic division when the divisor is linear and you want the answer quickly. Use long division for everything else, or when you want a written record that a marker can follow.

Whole-number long division follows the same rhythm on digits rather than terms, and is covered in long division with remainders.

Common Mistakes

• Forgetting placeholder zeros. 2x⁴ − 5x + 3 must be written with 0x³ and 0x².
• Subtracting only the first term. Subtraction flips the sign of every term in the product. Writing the product with its signs already flipped and then adding avoids this entirely.
• Dividing the whole divisor instead of its leading term. Step 1 uses leading terms only.
• Stopping while the remainder still has a high enough degree. Compare degrees, not sizes.
• Continuing past the stopping point and producing negative powers of x.
• Losing terms out of order. Keep columns aligned by power, exactly as place value is kept aligned in number division.
• Skipping the check. One substitution catches most errors.

Practice Problems

Divide, and state the quotient and the remainder.

1. (x³ + 5x² + 2x − 8) ÷ (x + 2)
2. (2x³ − 7x² + 4x + 1) ÷ (x − 3)
3. (x⁴ − 1) ÷ (x² + 1)
4. (3x⁴ + 2x³ − x + 6) ÷ (x² + x − 1)
5. (x⁵ − 2x² + 7) ÷ (x² − 3)
6. (6x³ + 5x² − 2x + 3) ÷ (2x + 3)
7. (x³ + 8) ÷ (x + 2)
8. (9x⁴ − 3x³ + 6x − 4) ÷ (3x² + 1)

Answers

1. Quotient x² + 3x − 4, remainder 0. The zero remainder means x + 2 is a factor, and the quotient factors further as (x + 4)(x − 1).
2. Quotient 2x² − x + 1, remainder 4. Check with the Remainder Theorem: f(3) = 54 − 63 + 12 + 1 = 4. ✓
3. Write the dividend as x⁴ + 0x³ + 0x² + 0x − 1. Quotient x² − 1, remainder 0, since x⁴ − 1 = (x² + 1)(x² − 1).
4. Write the dividend as 3x⁴ + 2x³ + 0x² − x + 6. Quotient 3x² − x + 4, remainder −6x + 10.
5. Write the dividend as x⁵ + 0x⁴ + 0x³ − 2x² + 0x + 7. Quotient x³ + 3x − 2, remainder 9x + 1.
6. Quotient 3x² − 2x + 2, remainder −3. Check: (2x + 3)(3x² − 2x + 2) − 3 = 6x³ + 5x² − 2x + 6 − 3 = 6x³ + 5x² − 2x + 3. ✓
7. Write the dividend as x³ + 0x² + 0x + 8. Quotient x² − 2x + 4, remainder 0. This is the sum of cubes factorisation.
8. Write the dividend as 9x⁴ − 3x³ + 0x² + 6x − 4. Quotient 3x² − x − 1, remainder 7x − 3.

Polynomial Long Division FAQ

How do you do polynomial long division step by step?

Arrange both polynomials in descending powers with zero placeholders, divide the leading term of the dividend by the leading term of the divisor to get a quotient term, multiply the whole divisor by it, subtract, bring down the next term, and repeat until what remains has a lower degree than the divisor.

When do you stop dividing?

When the degree of the remaining polynomial drops below the degree of the divisor. Dividing by a quadratic stops as soon as what is left is linear or constant. Going further produces negative powers, which are not part of a polynomial.

Why do you need zero placeholders?

Because each column stands for one power of x, and a missing power still occupies a column. Leaving it out misaligns everything after it, so terms end up subtracted from the wrong place.

Can the remainder be a polynomial rather than a number?

Yes, and it usually is when the divisor has degree 2 or more. Dividing by a quadratic can leave a remainder of the form bx + c. Only a linear divisor forces the remainder to be a single constant.

What is the difference between long division and synthetic division?

Synthetic division is a shortcut that works only for a linear divisor and manipulates coefficients alone. Long division works for any divisor and writes out full terms. For x − a, synthetic division is faster; for anything else, long division is the only option.

How do I check my answer?

Multiply the divisor by the quotient, add the remainder, and confirm the result matches the original dividend. A quicker partial check is to substitute x = 1 into both sides, which catches most sign errors immediately.

What if the divisor has a bigger degree than the dividend?

The quotient is 0 and the dividend is the remainder. Dividing 3x + 5 by x² + 1 gives quotient 0 and remainder 3x + 5, exactly as dividing 3 by 7 gives 0 remainder 3.

How is this different from dividing whole numbers?

The rhythm is identical: divide, multiply, subtract, bring down. The only change is what makes a remainder acceptable. Whole numbers compare magnitude, as in how to find the remainder, while polynomials compare degree.