1st Year Math Chapter 9 Exercise 9.2 Notes: Synthetic Division (Punjab Board 2026-27)
Complete concept notes, formulas, and solved examples for ICS/FSc Part 1 Mathematics, Punjab Board (PECTAA), Single National Curriculum 2026-27 session.
Exercise 9.2 introduces synthetic division, a faster alternative to the long division method from Exercise 9.1, used specifically for dividing a polynomial by a linear factor (x − c).
What Does This Exercise Cover?
This exercise teaches the compact synthetic division process using only coefficients, and shows how it connects directly to the Remainder and Factor Theorems from Exercise 9.1.
Key Concepts
1. What Is Synthetic Division?
Synthetic division is a shortcut method for dividing a polynomial by a linear factor of the form (x − c), using only the polynomial’s coefficients instead of writing out every term.
2. Steps of Synthetic Division
Write the value of c and the polynomial’s coefficients (in descending order, using 0 for any missing term) in a row. Bring down the leading coefficient, then repeatedly multiply by c and add to the next coefficient.
3. Reading the Result
The final row of numbers gives the coefficients of the quotient, which has one degree less than the original polynomial, with the very last number being the remainder.
4. Connecting Synthetic Division to the Remainder and Factor Theorems
The remainder from synthetic division is exactly P(c), so synthetic division doubles as a fast way to evaluate a polynomial or test whether (x − c) is a factor.
Solved Examples
Example 1: Use synthetic division to divide x³ − 6x² + 11x − 6 by x − 2.
Coefficients: 1, −6, 11, −6; c = 2.
Bring down 1. Multiply 1×2 = 2, add to −6: −4.
Multiply −4×2 = −8, add to 11: 3.
Multiply 3×2 = 6, add to −6: 0.
Answer: Quotient: x² − 4x + 3, Remainder: 0
Example 2: Use synthetic division to find P(−1) for P(x) = 2x³ + 3x² − 4x + 1.
Coefficients: 2, 3, −4, 1; c = −1.
Bring down 2. Multiply 2×(−1) = −2, add to 3: 1.
Multiply 1×(−1) = −1, add to −4: −5.
Multiply −5×(−1) = 5, add to 1: 6.
Answer: P(−1) = 6 (the final remainder)
Example 3: Use synthetic division to determine whether (x + 3) is a factor of P(x) = x³ + 2x² − 5x − 6.
c = −3, coefficients: 1, 2, −5, −6.
Bring down 1. Multiply 1×(−3) = −3, add to 2: −1.
Multiply −1×(−3) = 3, add to −5: −2.
Multiply −2×(−3) = 6, add to −6: 0.
Answer: Remainder = 0, so (x + 3) is a factor
Sample MCQs
1. Synthetic division is a shortcut method used to divide a polynomial by:
a) Any polynomial b) A linear factor of the form (x − c) c) A quadratic factor d) A constant only
Answer: b) A linear factor of the form (x − c)
2. In synthetic division, the first step is to:
a) Multiply all coefficients by c b) Bring down the leading coefficient c) Add all the coefficients together d) Find the remainder first
Answer: b) Bring down the leading coefficient
3. The last number obtained in synthetic division represents:
a) The leading coefficient of the quotient b) The remainder c) The value of c d) The degree of the polynomial
Answer: b) The remainder
4. The degree of the quotient obtained via synthetic division is:
a) The same as the original polynomial b) One less than the original polynomial c) One more than the original polynomial d) Always 1
Answer: b) One less than the original polynomial
5. Synthetic division can be used to quickly find:
a) The sum of a series b) The value of P(c) c) The derivative of a function d) The inverse of a matrix
Answer: b) The value of P(c)
Important Short Questions
- Explain what synthetic division is used for.
- List the basic steps of synthetic division.
- What does the last number in a synthetic division row represent?
- Use synthetic division to find the remainder when P(x) = x² − 5x + 6 is divided by x − 2.
- Why is the degree of the quotient always one less than the degree of the original polynomial?
Important Long Questions
- Use synthetic division to divide x³ − 6x² + 11x − 6 by x − 2, identifying the quotient and remainder.
- Use synthetic division to find P(−2) for P(x) = x⁴ + 3x³ − x + 5.
- Use synthetic division to determine whether (x − 4) is a factor of P(x) = x³ − 6x² + 11x − 6.
- Compare synthetic division and long division as two methods for dividing a polynomial by (x − c), noting the advantages of each.
How to Approach This Exercise Effectively
- Always write a 0 for any missing power of x in the coefficient row — skipping this shifts every following calculation.
- Keep the multiply-then-add pattern consistent all the way across the row; a single arithmetic slip early on throws off every later value.
- Remember that the sign of c in synthetic division matches the root, not the divisor — for (x + 3), use c = −3.
- Use synthetic division as your default method for evaluating P(c) instead of direct substitution when the polynomial has many terms — it’s often faster and less error-prone.
- Cross-check a synthetic division result against long division occasionally, especially while still building confidence with the method.
FAQs
Q: Why is synthetic division faster than long division?
A: It works only with coefficients, skipping the need to write out full polynomial terms at every step, which makes the calculation more compact and quicker to perform.
Q: Can synthetic division be used for divisors that aren’t linear?
A: No — synthetic division only works for linear divisors of the form (x − c); other divisors require standard long division.
Q: What does it mean if the remainder from synthetic division is zero?
A: It means (x − c) divides the polynomial exactly, so (x − c) is a factor and c is a root of the polynomial.
Q: How does synthetic division relate to the Remainder Theorem?
A: The final number produced by synthetic division is exactly P(c), which is precisely what the Remainder Theorem predicts — synthetic division is simply an efficient way to compute it.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 9: Division of Polynomials, Exercise 9.2.