AP Calculus · AP Pre-Calculus · Algebra II · NYC High School
Synthetic Division in Calculus — What It Is, Why It Matters, and 5 Practice Questions with Hidden Answers
Synthetic division is a small piece of pre-calculus that shows up over and over again in AP Calculus — when you factor a polynomial to find zeros, simplify a 0/0 limit, find a horizontal asymptote, or decompose an improper rational function before integration. This is the SOMATH class on it: what synthetic division actually is, when you can use it (and when you can’t), the four-step algorithm, why the last number equals p(c) (the Remainder Theorem), the trick for dividing by (ax + b), and then 5 practice questions with click-to-reveal step-by-step answers.
What’s in this class
1. What is synthetic division (and when can you use it)?
Suppose you want to divide a polynomial like p(x) = 2x³ − 5x² + 3x − 4 by a linear factor like (x − 2). You could set up polynomial long division and grind through it. That works, but it’s slow and easy to make sign errors. Synthetic division is a much faster shortcut that gives you the same answer — the same quotient and the same remainder — using only the coefficients, not the variables.
The rule for when synthetic division works
Synthetic division works when the divisor is linear — that is, when it has the form (x − c) for some number c. If the divisor looks like (ax + b) with a ≠ 1, you can still do synthetic division with a small extra step (see section 4).
If the divisor has degree 2 or higher — something like x² − 4 or x² + x + 1 — you cannot use synthetic division. You must use polynomial long division instead.
Sign warning. When you write the divisor as (x − c), the number you put outside the synthetic-division bracket is c, not the number that appears after the minus sign inside the parenthesis. Dividing by (x − 2) uses c = 2. Dividing by (x + 3) means (x − (−3)), so c = −3.
2. The 4-step algorithm
Let’s divide p(x) = 2x³ − 5x² + 3x − 4 by (x − 2). So c = 2.
Step 1 — Write the coefficients of p(x) in order
Write the coefficients including any zeros for missing terms. Here p(x) is already in order:
Put c = 2 to the left of the bar.
Step 2 — Bring the first coefficient straight down
Step 3 — Multiply and add across the row
Multiply the number you just brought down by c, write the product under the next coefficient, then add the column. Repeat all the way across.
2 × 2 = 4. Write 4 under −5. Add: −5 + 4 = −1.
−1 × 2 = −2. Write −2 under 3. Add: 3 + (−2) = 1.
1 × 2 = 2. Write 2 under −4. Add: −4 + 2 = −2.
Step 4 — Read the quotient and the remainder
The last row is the answer. The last number is the remainder; the numbers before it are the coefficients of the quotient, in decreasing degree, and the quotient has degree one less than p(x).
So the quotient is 2x² − x + 1 and the remainder is −2. In equation form:
(2x³ − 5x² + 3x − 4) ÷ (x − 2) = 2x² − x + 1 + (−2)/(x − 2)
Or equivalently:
2x³ − 5x² + 3x − 4 = (x − 2)(2x² − x + 1) − 2
3. Why the last number is p(c) — the Remainder Theorem
The Remainder Theorem
When you divide a polynomial p(x) by (x − c), the remainder equals p(c).
Quick proof. Polynomial division says p(x) = (x − c) · q(x) + r where r is a constant (because the divisor has degree 1). Plug in x = c: p(c) = 0 · q(c) + r = r. Done.
Check the example: p(2) = 2(8) − 5(4) + 3(2) − 4 = 16 − 20 + 6 − 4 = −2. That matches the last row of our synthetic-division table.
This is a huge shortcut. If a Regents, SAT, or AP question asks you to evaluate a polynomial at a number, synthetic division is often faster than plugging in.
4. Dividing by (ax + b) when a ≠ 1
Suppose the divisor is (2x + 3) instead of a monic linear divisor. Rewrite:
2x + 3 = 2(x + 3/2) = 2(x − (−3/2))
So c = −3/2. Do synthetic division with c = −3/2. Whatever quotient you get, divide every coefficient of the quotient by a = 2. The remainder does not change.
Why the remainder is unchanged
p(x) = (x − c) q(x) + r = (1/a)(ax + b) q(x) + r, so p(x) = (ax + b) · (q(x)/a) + r. The quotient scales down by a, but the constant remainder is unchanged.
5. Where synthetic division shows up in Calculus
- Factoring polynomials to find zeros. Once you find one rational root (usually via the Rational Root Theorem: possible rational roots are ±(factor of constant) / (factor of leading coefficient)), synthetic division reduces the polynomial by one degree so you can factor what remains. This shows up in curve-sketching and in finding critical points of higher-degree polynomials.
- Simplifying 0/0 limits. If lim (x→a) f(x)/g(x) gives 0/0, then (x − a) divides both f and g. Synthetic-divide each polynomial by (x − a) and evaluate the reduced expression at a.
- Horizontal asymptotes and slant asymptotes of rational functions. When the degree of the numerator equals the degree of the denominator (or is one greater), polynomial division tells you the horizontal or slant asymptote. Synthetic division is the fastest form of that division when the denominator is linear.
- Setting up partial fractions. Partial-fraction decomposition requires a proper rational function (degree of numerator < degree of denominator). If the numerator has larger degree, do the division first — synthetic division if the denominator is linear.
- Testing rational roots on the AP exam. When a free-response question hands you a polynomial with a known integer root, synthetic division reduces it to a quadratic you can solve with the quadratic formula.
6. Five practice questions — click to reveal each answer
Question 1 — A clean quadratic reduction
Divide p(x) = x³ − 6x² + 11x − 6 by (x − 1). Write the quotient and remainder. Then use the Factor Theorem to fully factor p(x).
Show answer
Set up. c = 1. Coefficients: 1, −6, 11, −6.
Read the answer. Quotient x² − 5x + 6, remainder 0.
Since the remainder is 0, (x − 1) is a factor of p(x) (Factor Theorem). The quotient factors as (x − 2)(x − 3), so:
p(x) = (x − 1)(x − 2)(x − 3)
Sanity check. p(1) = 1 − 6 + 11 − 6 = 0. Matches the remainder.
Question 2 — Missing terms (careful with zeros)
Divide p(x) = 2x⁴ − 3x² + 5 by (x + 1). What is the quotient and remainder? What is p(−1)?
Show answer
Set up. (x + 1) = (x − (−1)), so c = −1. Write every coefficient, including zeros for missing degrees. The polynomial is degree 4, so we need coefficients for x⁴, x³, x², x, 1. Missing terms are x³ and x. Coefficients: 2, 0, −3, 0, 5.
Read the answer. Quotient 2x³ − 2x² − x + 1, remainder 4. So p(−1) = 4 by the Remainder Theorem.
Sanity check. Plug in: p(−1) = 2(1) − 3(1) + 5 = 2 − 3 + 5 = 4. ✓
Question 3 — A 0/0 limit
Evaluate lim (x→2) (x³ − 7x + 6) / (x − 2).
Show answer
Check the form. Numerator at x = 2: 8 − 14 + 6 = 0. Denominator at x = 2: 0. Good — 0/0 form. So (x − 2) divides the numerator, and we can simplify by synthetic division.
Synthetic-divide x³ + 0x² − 7x + 6 by (x − 2). c = 2:
Quotient: x² + 2x − 3, remainder 0 ✓. So for x ≠ 2:
(x³ − 7x + 6) / (x − 2) = x² + 2x − 3
The limit equals the value of x² + 2x − 3 at x = 2: 4 + 4 − 3 = 5.
Answer: lim (x→2) (x³ − 7x + 6) / (x − 2) = 5.
Question 4 — Rational Root Theorem + synthetic division
Fully factor p(x) = 2x³ − 3x² − 11x + 6 over the real numbers, then list all real zeros.
Show answer
Rational Root Theorem. Possible rational roots are ±(factor of 6) / (factor of 2) = ±1, ±2, ±3, ±6, ±1/2, ±3/2.
Test candidates. Try x = 3: 2(27) − 3(9) − 11(3) + 6 = 54 − 27 − 33 + 6 = 0. ✓ So (x − 3) is a factor.
Divide. Synthetic-divide by c = 3:
Quotient: 2x² + 3x − 2. Factor: 2x² + 3x − 2 = (2x − 1)(x + 2).
Fully factored: p(x) = (x − 3)(2x − 1)(x + 2)
Real zeros: x = 3, x = 1/2, x = −2.
Question 5 — Non-monic linear divisor
Divide p(x) = 6x³ + x² − 5x + 2 by (2x − 1). Give the quotient and the remainder.
Show answer
Rewrite the divisor. 2x − 1 = 2(x − 1/2), so use c = 1/2 for the synthetic step, then divide the quotient coefficients by a = 2. The remainder is unchanged.
Synthetic-divide by c = 1/2 with coefficients 6, 1, −5, 2:
Intermediate quotient (from the row before scaling): 6x² + 4x − 3 with remainder 1/2.
Scale the quotient by 1/a = 1/2. Divide each quotient coefficient by 2: 6/2 = 3, 4/2 = 2, −3/2.
Final quotient: 3x² + 2x − 3/2. Remainder: 1/2 (unchanged).
Check. Multiply: (2x − 1)(3x² + 2x − 3/2) + 1/2
= 6x³ + 4x² − 3x − 3x² − 2x + 3/2 + 1/2
= 6x³ + x² − 5x + 2 ✓
Also, by the Remainder Theorem applied at c = 1/2: p(1/2) = 6(1/8) + 1/4 − 5/2 + 2 = 3/4 + 1/4 − 5/2 + 2 = 1 − 5/2 + 2 = 1/2 ✓
FAQ
Can synthetic division give a fractional quotient?
Yes. Nothing in the algorithm requires integer coefficients. If c is a fraction, or if the polynomial has fractional coefficients, you’ll get fractional numbers in the quotient row. That’s fine.
What if the polynomial has a missing term?
You must include a zero as the coefficient for every missing degree. If you skip a missing term, the columns don’t line up and every subsequent multiply-and-add is wrong. Question 2 above shows this: 2x⁴ − 3x² + 5 becomes 2, 0, −3, 0, 5.
Is synthetic division on the AP Calculus exam?
Not as a stand-alone skill (no question asks “divide these two polynomials” on the AP), but as a tool it’s essential. Free-response questions regularly hand you a polynomial with a known root and expect you to factor it, or give you a 0/0 limit that has a polynomial numerator. Synthetic division is the fastest way through both.
When should I use polynomial long division instead?
Whenever the divisor is not linear. Anything of degree 2 or higher — like x² − 4, x² + 1, or x³ − 8 — requires polynomial long division. Synthetic division has no version for higher-degree divisors.
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