Polynomial Long Division & the Remainder Theorem
Divide polynomials with long division and synthetic division, and use the Remainder Theorem to find remainders fast — with worked SAT and ACT problems.
The Short Version
- Polynomial long division mirrors numerical long division: divide, multiply, subtract, bring down.
- Synthetic division is a fast shortcut when dividing by (x − a).
- Remainder Theorem: the remainder of P(x) ÷ (x − a) is just P(a).
- Factor Theorem: if P(a) = 0, then (x − a) is a factor. An SAT/ACT and Algebra II topic, beyond the SSAT.
You already know how to divide one number by another with long division. Dividing polynomials uses the exact same rhythm — divide, multiply, subtract, bring down — just with terms instead of digits. And there's a beautiful shortcut hiding here: to find the remainder when you divide by (x − a), you don't have to divide at all. You just evaluate the polynomial at a. That's the Remainder Theorem, and it turns a long problem into one substitution.
This guide covers long division, synthetic division, and the Remainder and Factor Theorems, with worked and practice problems matched to real test difficulty at Northside Tutoring.
Why Polynomial Division Matters
Polynomial division underpins factoring higher-degree polynomials, simplifying rational expressions, and finding roots. On the SAT and ACT it usually shows up as a Remainder-Theorem question — which is good news, because the theorem makes those questions fast. It's Algebra II content the SSAT doesn't reach.
Long Division, Step by Step
To divide, say, x² + 5x + 6 by x + 2, repeat the cycle: divide the leading terms (x² ÷ x = x), multiply back (x(x+2) = x²+2x), subtract, bring down, and repeat.
A remainder of 0 means the divisor divides evenly — it's a factor.
Synthetic Division: The Fast Lane
When the divisor is (x − a), synthetic division skips the bookkeeping: write the coefficients, bring down the first, multiply by a, add, repeat. It's faster and less error-prone than long division — just remember it only works for linear divisors of the form (x − a).
The Remainder Theorem
Here's the shortcut that saves the most time:
To find the remainder when P(x) = x² + 3x − 5 is divided by (x − 2), just compute P(2) = 4 + 6 − 5 = 5. No division needed.
Mind the sign
For division by (x − a), you plug in +a. Dividing by (x + 3) means a = −3, so evaluate P(−3). The sign flip is the most common Remainder-Theorem error.
The Factor Theorem
The Factor Theorem is the Remainder Theorem's special case: if P(a) = 0, the remainder is zero, so (x − a) divides evenly — it's a factor. This is how you test whether a given binomial is a factor, or confirm a root.
Which Method to Use
If a question only asks for the remainder when dividing by (x − a), use the Remainder Theorem — it's a single evaluation. If you need the full quotient, use synthetic division for (x − a) divisors and long division for anything more complex.
Where You'll See This — Test by Test
No reference sheet covers this. The Remainder Theorem is the SAT's and ACT's favorite way to test polynomial division — a quick substitution. It is beyond the SSAT, which doesn't cover polynomial division.
Digital SAT
Tests the Remainder and Factor Theorems — usually "what is the remainder" or "is (x − a) a factor."
Explore SAT Tutoring → College AdmissionsACT
Tests polynomial division and the Remainder Theorem, sometimes with synthetic division of a cubic.
Explore ACT Tutoring → Independent School AdmissionsSSAT
Not on the SSAT — this is Algebra II material. Strengthen factoring fundamentals with SSAT-level prep first.
Explore SSAT Tutoring → K-12 CurriculumAlgebra II
A core Algebra II skill for finding roots and factoring higher-degree polynomials.
Explore Algebra Tutoring →Watch the Lesson
Sometimes a diagram needs a voice. In the short video below, one of our Northside tutors walks through the core idea and works through test-style problems in real time.
Polynomial Division — In Plain English
A live walkthrough from our tutoring team.
— Featuring a Northside Tutoring instructor
Worked Example Problems
These problems are calibrated to the difficulty you'll actually see on test day. Try each one before opening the solution.
What is the remainder when P(x) = x² + 3x − 5 is divided by (x − 2)?
Show solution
By the Remainder Theorem, the remainder is P(2).
P(2) = 4 + 6 − 5 = 5.
Divide: (x² + 5x + 6) ÷ (x + 3).
Show solution
x + 3 corresponds to a = −3; P(−3) = 9 − 15 + 6 = 0, so it divides evenly.
The quotient is x + 2 (since (x+3)(x+2) = x²+5x+6).
Is (x − 1) a factor of P(x) = x³ − 2x² + x?
Show solution
By the Factor Theorem, check P(1): 1 − 2 + 1 = 0.
Since P(1) = 0, yes — (x − 1) is a factor.
What is the remainder when P(x) = 2x³ − x + 4 is divided by (x + 1)?
Show solution
(x + 1) means a = −1. P(−1) = 2(−1) − (−1) + 4 = −2 + 1 + 4 = 3.
If P(x) divided by (x − 4) leaves remainder 0, what does that tell you?
Show solution
A remainder of 0 means (x − 4) is a factor and x = 4 is a root of P(x).
Common Mistakes to Avoid
Three traps that catch students every year
- Sign error in the Remainder Theorem. Dividing by (x + 3) means evaluating at −3, not +3.
- Using synthetic division with a non-linear divisor. Synthetic division only works for (x − a). Use long division otherwise.
- Dropping a missing term. Include zero coefficients for absent powers (e.g., x³ + 0x² + ...) before dividing.
Practice Problems — You Try
Three problems below. Work each before checking the solution.
Find the remainder when P(x) = x² − 4x + 1 is divided by (x − 3).
Show solution
P(3) = 9 − 12 + 1 = −2.
Is (x + 2) a factor of P(x) = x² + x − 2?
Show solution
a = −2: P(−2) = 4 − 2 − 2 = 0. Yes, it's a factor.
When P(x) = x³ + kx − 6 is divided by (x − 2), the remainder is 4. Find k.
Show solution
P(2) = 8 + 2k − 6 = 2 + 2k. Set equal to 4: 2 + 2k = 4 → k = 1.
The Northside Method — How We Teach This 1-on-1
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