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Polynomial Division Practice Problems

Every problem here is built backwards from a divisor, a quotient and a remainder that were chosen first and multiplied out, so the division always comes out exactly as the prompt describes. Work the long division on paper, then type the one number the question asks for.

  1. 1.

    Divide by . In the quotient, what is the coefficient of ?

  2. 2.

    Divide by . What is the remainder?

  3. 3.

    Divide by . What is the remainder?

  4. 4.

    Divide by . What is the remainder?

  5. 5.

    Divide by . In the quotient, what is the coefficient of ?

  6. 6.

    Divide by . What is the remainder?

  7. 7.

    Divide by . In the quotient, what is the coefficient of ?

  8. 8.

    Divide by . In the quotient, what is the coefficient of ?

  9. 9.

    Divide by . In the quotient, what is the coefficient of ?

  10. 10.

    Divide by . What is the remainder?
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MPhil in Mathematics, Founder of Mathovia
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Polynomial Division Practice Problems

The polynomial division lesson walks through the long division algorithm one column at a time. This set drills it, with every problem constructed from a known divisor, quotient and remainder so the arithmetic is guaranteed exact.

How the difficulty levels work

DifficultyDivisorDividendWhat you are asked for
Easylinearquadraticthe remainder
Mediumlinearcubicthe remainder, or the quotient’s coefficient
Hardquadraticcubicone coefficient of the linear remainder

Easy gives a quadratic over a linear divisor, so the quotient is linear and the remainder is a single number. Two rounds of the algorithm and you are done.

Medium steps up to a cubic dividend, which means three rounds and a quadratic quotient. Half the problems ask for the remainder and half for a specific coefficient inside the quotient, so you cannot get through the set by only ever evaluating .

Hard divides by a quadratic. Synthetic division does not work here, the remainder is linear rather than constant, and you have to track two terms of the divisor through every subtraction.

The identity every problem satisfies

The dividend equals divisor times quotient plus remainder, and . Every problem in this set was built by choosing , and first and multiplying out to get — which is also the fastest way to check your own answer. Multiply what you got by the divisor, add the remainder, and see whether the original dividend comes back.

A worked problem

Divide by .

Step 1 — divide the leading terms. . Multiply back: . Subtract:

Step 2 — bring down and repeat. , and :

Step 3 — last term. , and :

So the quotient is and the remainder is . Asked for the remainder you would type 9; asked for the quotient’s coefficient you would type 3.

Where people lose marks

  • Missing-degree placeholders. has no or term. Write it as or the columns misalign.
  • Subtracting only the first term. The whole product comes off, and every sign in it flips. Writing the subtraction as an addition of the negated row prevents this.
  • Stopping one step early. Keep going while the running remainder’s degree is at least the divisor’s.
  • Reading the quotient off the wrong row. The quotient sits above the bar; the remainder is what is left at the bottom.

Once this is fluent, synthetic division does the same job in a fraction of the writing for linear divisors, and the remainder and factor theorem skips the division entirely when the remainder is all you need.

Frequently Asked Questions

Why does each problem ask for one number instead of the whole quotient?+

A full quotient like 2x² − 3x + 5 is an expression, and expressions cannot be checked reliably as a single typed value. Asking for the remainder or one named coefficient tests exactly the same long division, but gives an answer that can be marked instantly and unambiguously.

Do I still have to do the whole division?+

For a quotient coefficient, yes — you have to carry the division far enough to reach that column. For a remainder you can shortcut a linear divisor with the remainder theorem and just evaluate P(a), which is a good habit and a useful cross-check.

Why are some divisors quadratic?+

Because a quadratic divisor is the case synthetic division cannot handle, and it is where long division earns its place. It also produces a linear remainder rather than a constant, which is a genuinely different situation worth practising.

What does 'the remainder has the form dx + e' mean?+

When you divide by a quadratic, the remainder can be any polynomial of degree less than 2 — so a linear expression. The problem asks for one of its two coefficients so the answer is a single number.

How should I lay the division out?+

Write the dividend in descending powers with a zero placeholder for any missing degree. A missing term is the most common cause of a wrong answer, because it silently shifts every later column by one.

Is there a limit on how many sets I can generate?+

No. Problems are produced on demand, free, with no sign-up. Reload for a fresh set as often as you like.

More practice problems

Synthetic Division Practice Problems Remainder and Factor Theorem Practice Problems Partial Fractions Practice Problems