Remainder and Factor Theorem Practice Problems

The remainder and factor theorem lesson shows why the shortcut works. This set makes it automatic, then pushes into the harder direction where you solve for a missing coefficient.

How the difficulty levels work

Difficulty	Polynomial	Task
Easy	quadratic	find the remainder on division by
Medium	cubic	find , or the remainder — same thing, stated both ways
Hard	cubic with an unknown	find the that makes a factor

Easy and Medium are the theorem used forwards: substitute and evaluate. The two phrasings in Medium are deliberate — “find ” and “find the remainder on division by ” are the same question, and recognising that is the whole point of the theorem.

Hard runs it backwards. You are told is a factor, so , and that equation has exactly one unknown in it.

The two statements

The first is the remainder theorem, the second the factor theorem. The second follows from the first: a remainder of zero is precisely what “divides exactly” means.

A worked problem — forwards

Let . Find the remainder on division by .

Step 1 — the divisor is zero at , so evaluate there.

Step 2 — work through the powers.

The remainder is 2. Since it is not zero, is not a factor.

A worked problem — backwards

For what value of is a factor of ?

Step 1 — the factor theorem says .

Step 2 — evaluate everything that is not .

Step 3 — collect and solve.

So . Check it: with , . It works.

Where people lose marks

Substituting the wrong sign. is zero at , not .
Losing a minus inside an even power. , not . Brackets around the substituted value prevent this.
Treating a non-zero remainder as a factor. Only means a factor.
On the problems, forgetting that sits on . Substituting multiplies it by , which is always positive — a sign trap when is negative.
Solving without checking. Putting your back in and confirming takes one line.

Where the remainder does come out zero, you have found a factor — and the polynomial zeros practice picks the story up from there.

Frequently Asked Questions

What does the remainder theorem say?+

When a polynomial P(x) is divided by x − a, the remainder is exactly P(a). So instead of running the whole division you substitute a into the polynomial and evaluate — one line instead of several rows of arithmetic.

How is the factor theorem different?+

It is the special case where the remainder is zero. If P(a) = 0 then x − a divides P(x) exactly, and conversely. The remainder theorem gives you a number; the factor theorem reads a zero of that number as a statement about factors.

Why do the hard problems ask for k?+

Because that is how the factor theorem is normally used in practice. You are told x − a is a factor, which means P(a) = 0, and that single equation is enough to solve for one unknown coefficient. It tests whether you understand the theorem rather than just quoting it.

Do I substitute a or negative a?+

Substitute a, the value that makes the divisor zero. For x − 3 evaluate P(3); for x + 3 evaluate P(−3). Getting this backwards flips the sign of every odd-power term and is by far the most common mistake.

Can I check my answer with division?+

Yes, and it is worth doing once or twice. Run the same division synthetically — the last cell of the bottom row is the remainder, and it should match the value you got by substituting.

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

No. Sets are produced on demand, free, and need no account. Reload for new problems as often as you like.

P(a)Remainder and Factor Theorem Practice Problems