The Division Algorithm
When you divide polynomial p(x) by divisor d(x), the result is:
p(x) = d(x) · q(x) + r(x)
dividend = divisor × quotient + remainder
dividend p(x)
The polynomial being divided
divisor d(x)
The polynomial you divide by
quotient q(x)
The result of the division
remainder r(x)
What's left over; degree < degree of d(x)
Polynomial Long Division — 5-Step Method
Use this when the divisor has degree ≥ 1. Works for any divisor.
Divide
Divide leading term of dividend by leading term of divisor
Multiply
Multiply that result by the entire divisor
Subtract
Subtract from the current dividend (change all signs)
Bring down
Bring down the next term
Repeat
Repeat until degree of remainder < degree of divisor
Worked Example 1
Divide (2x³ − 3x² + x − 5) ÷ (x − 2)
Step 1: 2x³ ÷ x = 2x²
Multiply: 2x²(x − 2) = 2x³ − 4x²
Subtract: (2x³ − 3x²) − (2x³ − 4x²) = x²
Step 2: x² ÷ x = x
Multiply: x(x − 2) = x² − 2x
Subtract: (x² + x) − (x² − 2x) = 3x
Step 3: 3x ÷ x = 3
Multiply: 3(x − 2) = 3x − 6
Subtract: (3x − 5) − (3x − 6) = 1
Result: 2x² + x + 3 + 1/(x − 2)
quotient = 2x² + x + 3, remainder = 1
Synthetic Division
A shortcut for dividing by (x − c). Only works with linear divisors of this exact form.
Write c (not −c) in the box, then write the coefficients of p(x) — include zeros for missing terms
Bring down the first coefficient
Multiply it by c and write the result under the next coefficient
Add down the column
Repeat: multiply, add, until done. Last number = remainder; others = quotient coefficients
Worked Example 2
Divide (3x³ + 2x² − 5x + 1) ÷ (x + 1) [c = −1]
Coefficients: 3 | 2 | −5 | 1 c = −1
3 | 2 | −5 | 1
↓ | −3 | 1 | 4
3 | −1 | −4 | 5
Bring 3 down → multiply by −1 → −3 → add to 2 → −1 → multiply by −1 → 1 → add to −5 → −4 → multiply by −1 → 4 → add to 1 → 5
Result: 3x² − x − 4 + 5/(x + 1)
Remainder Theorem
If p(x) is divided by (x − c), the remainder = p(c)
Worked Example 3
Find the remainder when p(x) = x⁴ − 2x² + 3 is divided by (x − 2)
By Remainder Theorem: remainder = p(2)
p(2) = (2)⁴ − 2(2)² + 3 = 16 − 8 + 3 = 11
No long division needed — just evaluate p at c.
Factor Theorem
(x − c) is a factor of p(x) if and only if p(c) = 0
Equivalently: c is a zero of p(x) ↔ (x − c) is a factor of p(x)
Worked Example 4
Show (x − 3) is a factor of p(x) = x³ − 6x² + 11x − 6 and fully factor
p(3) = 27 − 54 + 33 − 6 = 0 ✓ → (x − 3) is a factor
Synthetic division with c = 3, coefficients 1 | −6 | 11 | −6:
1 | −6 | 11 | −6
↓ | 3 | −9 | 6
1 | −3 | 2 | 0
Quotient: x² − 3x + 2 = (x − 1)(x − 2)
p(x) = (x − 1)(x − 2)(x − 3)
Zeros: x = 1, 2, 3
Rational Root Theorem
For a polynomial with integer coefficients, all rational roots have the form:
p/q
p = factor of the constant term | q = factor of the leading coefficient
Worked Example 5
Find all rational roots of p(x) = 2x³ − x² − 7x + 6
Constant term = 6, factors: ±1, ±2, ±3, ±6
Leading coefficient = 2, factors: ±1, ±2
Possible rational roots: ±1, ±2, ±3, ±6, ±1/2, ±3/2
Test x = 1: p(1) = 2 − 1 − 7 + 6 = 0 ✓
Synthetic division with c = 1:
2 | −1 | −7 | 6
↓ | 2 | 1 | −6
2 | 1 | −6 | 0
Quotient: 2x² + x − 6 = (2x − 3)(x + 2)
Roots: x = 1, x = 3/2, x = −2
When to Use Each Method
| Situation | Method | Why |
|---|---|---|
| Divisor is (x − c) | Synthetic division | Fastest — coefficients only |
| Divisor is quadratic | Long division | Synthetic requires linear divisor |
| Divisor has coefficient ≠ 1 (e.g., 2x − 3) | Long division | Synthetic restricted to monic linear |
| Need to evaluate p(c) quickly | Remainder Theorem + synthetic | Plug in c instead of dividing |
| Testing possible roots | Rational Root Theorem → synthetic | Generate candidates, test with synthetic |
| Factor a polynomial completely | RRT → Factor Theorem → synthetic | Find one root, factor out, repeat |
Exam Strategy
Always check degree
The quotient degree = dividend degree − divisor degree. If your answer doesn't match, you made an arithmetic error.
Don't skip zero placeholders
If p(x) = x³ + 1, the coefficients are 1, 0, 0, 1. Skipping the zeros will destroy your synthetic division.
Verify with multiplication
Always check: q(x)·d(x) + r(x) = p(x). Takes 30 seconds and catches sign errors before they cost you.
Quick Reference
Frequently Asked Questions
When can I use synthetic division instead of long division?
Synthetic division only works when dividing by a linear binomial of the form (x − c), where c is a constant. If the divisor is quadratic or has a leading coefficient other than 1 (like 2x − 3), you must use polynomial long division.
What does the Remainder Theorem say?
The Remainder Theorem states that when a polynomial p(x) is divided by (x − c), the remainder equals p(c). This means you can evaluate a polynomial at a specific value by doing synthetic division — much faster than direct substitution for high-degree polynomials.
How do I use the Rational Root Theorem?
The Rational Root Theorem says all rational roots of a polynomial with integer coefficients have the form p/q, where p divides the constant term and q divides the leading coefficient. List all possibilities, test them with synthetic division, and use the resulting depressed polynomial to find remaining roots.
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