Mathematics › 🌌 Grade 11 › Polynomial Division: the Remainder, Factor, and Rational Root Theorems

Polynomial Division: the Remainder, Factor, and Rational Root Theorems

Dividing a polynomial by a linear factor is arithmetic in disguise — and the leftover remainder tells you, for free, whether you have found a root.

Grade 11Algebra 2 / Pre-Calculus

Polynomial Division: the Remainder, Factor, and Rational Root Theorems — illustration

💡

The big idea: Just as 17 ÷ 5 leaves a quotient and a remainder, a polynomial f(x) divided by (x − c) leaves a polynomial quotient and a number remainder. Synthetic division computes that split with nothing but multiply-and-add on the coefficients. The payoff is the Remainder Theorem: the leftover number equals f(c). Push that one step further and the Factor Theorem turns a division problem into a root-finding tool, while the Rational Root Theorem tells you which values of c are even worth trying.

🎯 By the end, you'll be able to

Perform synthetic division of a polynomial by a linear factor (x − c)
Apply the Remainder Theorem to evaluate f(c) without direct substitution
Use the Factor Theorem to test whether (x − c) is a factor of f(x)
Generate candidate rational roots with the Rational Root Theorem and factor a cubic completely

📎 You should already know

Multiplying and adding polynomials
Long division of numbers

Division leaves a quotient and a remainder

Dividing 17 by 5 gives quotient 3 and remainder 2, because 17 = 5·3 + 2. Polynomials divide the same way. Dividing f(x) by a divisor d(x) produces a quotient q(x) and a remainder r(x) that is “smaller” than d(x) — of lower degree — so that f(x) = d(x)·q(x) + r(x).

When the divisor is linear, (x − c), the remainder must have degree less than 1 — that is, it is just a number. That single fact is what makes everything in this lesson work.

\[ f(x) = (x-c)\,q(x) + r,\qquad r \in \mathbb{R} \]

Dividing by a linear factor always leaves a constant remainder r.

Synthetic division: the shortcut

Synthetic division is long division of a polynomial by (x − c), stripped down to just its coefficients. Write the coefficients of f(x) in a row, bring down the first one, then repeatedly multiply the last number written by c and add it into the next column. The numbers you generate, except the last, are the coefficients of the quotient q(x); the very last number is the remainder.

⚠️ Missing powers still need a coefficient

Every degree from the highest down to 0 must appear in the coefficient row, even the ones that are absent from f(x) — write a 0 for them. For f(x) = x³ − 1, the x² and x terms are missing, so the row is [1, 0, 0, −1], not [1, −1]. Skipping a zero shifts every coefficient into the wrong column and corrupts the whole division.

🎮 Synthetic Division Machine LIVE

Step through multiply-and-add synthetic division and watch the quotient and remainder build up column by column.

✨ Multiply-then-ADD, and the remainder self-checks

The synthetic division algorithm is multiply by c, then add — never subtract, unlike ordinary long division. And because the last number in the bottom row is the remainder, the Remainder Theorem below means that number is also f(c). That gives you a free check: compute f(c) a second way and see if the two answers agree.

The Remainder & Factor Theorems

Substitute x = c into f(x) = (x − c)q(x) + r: the first term vanishes because (c − c) = 0, leaving f(c) = r. So the remainder from dividing f(x) by (x − c) is exactly the value of the function at c — no substitution arithmetic required once the division is done. That is the Remainder Theorem.

Push one step further: if the remainder is 0, then f(c) = 0 — c is a root — and f(x) = (x − c)q(x) exactly, with no leftover. So (x − c) is a factor of f(x) exactly when c is a root exactly when synthetic division by c leaves remainder 0. This is the Factor Theorem — three statements that are all the same fact seen three ways.

\[ \text{Remainder Theorem: } f(x) \div (x-c) \text{ leaves remainder } f(c) \]

The remainder of synthetic division by (x − c) equals f(c).

🔑 Same divisor, opposite sign

Synthetic division always divides by the root c, not by the factor itself. To divide by (x + 2), write it as (x − (−2)) and use c = −2. Using +2 instead of −2 is the single most common synthetic-division mistake.

The Rational Root Theorem: where to even look

A cubic could have a root anywhere on the number line — testing values blindly is hopeless. The Rational Root Theorem narrows the search: if a polynomial with integer coefficients has a rational root p⁄q (in lowest terms), then p must divide the constant term and q must divide the leading coefficient. That turns an infinite search into a short, finite list of candidates to test by synthetic division — find one rational root, divide it out, and repeat on the smaller quotient until the polynomial is factored completely. It is the same division that later clears a shared factor from a rational function's numerator and denominator.

\[ \text{root } = \dfrac{p}{q}: \quad p \mid a_0 \ (\text{constant term}), \quad q \mid a_n \ (\text{leading coefficient}) \]

Candidate rational roots come from factors of the constant term over factors of the leading coefficient.

📝 Worked example: Use synthetic division to divide f(x) = x³ + 2x² − 5x − 6 by (x − 2), and state f(2).

Coefficients of f(x), including every power: \( [1,\ 2,\ -5,\ -6] \). Since the divisor is (x − 2), use \( c = 2 \).
Bring down the 1. Multiply \( 1 \times 2 = 2 \), add to the next coefficient: \( 2 + 2 = 4 \). Multiply \( 4 \times 2 = 8 \), add: \( -5 + 8 = 3 \). Multiply \( 3 \times 2 = 6 \), add: \( -6 + 6 = 0 \).
The bottom row is \( [1,\ 4,\ 3,\ 0] \): quotient \( q(x) = x^2+4x+3 \), remainder \( 0 \).

✓ f(2) = <strong>0</strong> (the remainder), so x = 2 is a root and (x − 2) is a factor: f(x) = (x − 2)(x² + 4x + 3).

📝 Worked example: Factor f(x) = x³ − 4x² + x + 6 completely.

Constant term 6 has factors ±1, ±2, ±3, ±6; leading coefficient 1 has factor ±1, so every candidate is one of these integers.
Try \( c = -1 \) by synthetic division on \( [1,\ -4,\ 1,\ 6] \): bring down 1; \( 1(-1)=-1,\ -4+(-1)=-5 \); \( -5(-1)=5,\ 1+5=6 \); \( 6(-1)=-6,\ 6+(-6)=0 \). Remainder 0, so \( x=-1 \) is a root and \( q(x)=x^2-5x+6 \).
The quotient \( x^2 - 5x + 6 \) factors by inspection into \( (x-2)(x-3) \).

✓ f(x) = <strong>(x + 1)(x − 2)(x − 3)</strong>, with roots −1, 2, and 3.

Check your understanding

1. To divide a polynomial by (x + 3) using synthetic division, what value of c should you use?

(x + 3) is (x − (−3)), so the root used in synthetic division is c = −3, not +3.

2. The coefficient row for f(x) = x⁴ − 3x + 2 is:

Degrees 4 down to 0 must all appear; x³ and x² are missing so they get 0s: [1, 0, 0, −3, 2].

3. Dividing f(x) by (x − c) leaves remainder 5. What is f(c)?

The Remainder Theorem says the remainder of division by (x − c) equals f(c) exactly, so f(c) = 5.

4. Synthetic division of f(x) by (x − 4) gives remainder 0. Which statement is FALSE?

A zero remainder confirms x = 4 is a root and (x − 4) a factor, but says nothing about how many other roots f(x) may have.

5. For f(x) = 2x³ + x − 6, which of these is a valid candidate from the Rational Root Theorem?

Candidates are p/q with p dividing the constant term 6 and q dividing the leading coefficient 2; 3/2 fits (p = 3, q = 2), while 4, 1/4, and 5 do not.

✅ Key takeaways

Dividing f(x) by (x − c) leaves a constant remainder: f(x) = (x − c)q(x) + r.
Synthetic division computes q(x) and r from coefficients alone by repeated multiply-then-add — insert zeros for any missing powers.
Remainder Theorem: the remainder equals f(c), giving a free way to evaluate the function.
Factor Theorem: remainder 0 ⇔ (x − c) is a factor ⇔ c is a root — and synthetic division always uses c, the opposite sign of (x + c).
Rational Root Theorem narrows candidate roots p/q to factors of the constant term over factors of the leading coefficient, enabling a cubic to be factored completely.