Factorisation of Polynomials
Introduction
Factorisation of polynomials is a critical skill in ICSE Class 10 algebra. It involves expressing a polynomial as a product of its factors using the Remainder Theorem and Factor Theorem. These theorems help you find roots and factors of polynomials without long division.
Key Terms
- Polynomial — An expression of the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀.
- Degree — The highest power of the variable in the polynomial.
- Zero/Root — A value α such that p(α) = 0. Corresponds to factor (x − α).
- Remainder Theorem — When p(x) is divided by (x − a), the remainder is p(a).
- Factor Theorem — (x − a) is a factor of p(x) if and only if p(a) = 0.
Remainder Theorem
If a polynomial p(x) is divided by (x − a), the remainder is p(a).
This means you can find the remainder by simply substituting x = a, without performing long division.
Example: Find the remainder when x³ − 3x² + 2x + 1 is divided by (x − 2).
p(2) = (2)³ − 3(2)² + 2(2) + 1 = 8 − 12 + 4 + 1 = 1
Remainder = 1
If p(a) = 0, then (x − a) is a factor.
Factor Theorem
(x − a) is a factor of p(x) if and only if p(a) = 0.
This is a special case of the Remainder Theorem where the remainder is zero.
Proof
If (x − a) is a factor, then p(x) = (x − a) × q(x), so p(a) = 0 × q(a) = 0. Conversely, if p(a) = 0, then by Remainder Theorem, remainder is 0, so (x − a) is a factor.
Splitting the Middle Term (Quadratic Polynomials)
For a quadratic polynomial ax² + bx + c:
Step 1: Find two numbers whose product = ac and sum = b. Step 2: Split the middle term. Step 3: Factor by grouping.
Example: Factorise 6x² + 11x − 10
- a = 6, b = 11, c = −10
- ac = −60, b = 11
- Numbers: 15 and −4 (product = −60, sum = 11)
- 6x² + 15x − 4x − 10
- 3x(2x + 5) − 2(2x + 5)
- (2x + 5)(3x − 2)
Factorising Cubic Polynomials
For cubic expressions like ax³ + bx² + cx + d:
Step 1: Use trial and error to find a root α such that p(α) = 0. (Check factors of d/a as possible roots.) Step 2: Factor out (x − α) using synthetic division or long division. Step 3: Factorise the resulting quadratic.
Example: Factorise x³ − 6x² + 11x − 6
p(1) = 1 − 6 + 11 − 6 = 0 → (x − 1) is a factor.
Divide: (x³ − 6x² + 11x − 6) ÷ (x − 1)
Using synthetic division:
1 | 1 -6 11 -6
| 1 -5 6
----------------
1 -5 6 0
Quotient: x² − 5x + 6 = (x − 2)(x − 3)
Therefore: x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3)
Worked Examples
Example 1: Using Factor Theorem to Find Unknown Coefficient
If (x − 2) is a factor of 2x³ + kx² − 3x + 10, find k.
Solution: p(2) = 0 (by Factor Theorem) 2(2)³ + k(2)² − 3(2) + 10 = 0 16 + 4k − 6 + 10 = 0 4k + 20 = 0 k = −5
Example 2: Factorise Using Factor Theorem
Factorise 6x³ + x² − 5x − 2.
Solution: Check p(1) = 6 + 1 − 5 − 2 = 0 → (x − 1) is a factor.
Divide: (6x³ + x² − 5x − 2) ÷ (x − 1)
Synthetic division:
1 | 6 1 -5 -2
| 6 7 2
----------------
6 7 2 0
Quotient: 6x² + 7x + 2 = 6x² + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2) = (3x + 2)(2x + 1)
Therefore: 6x³ + x² − 5x − 2 = (x − 1)(3x + 2)(2x + 1)
Comparison: Remainder Theorem vs Factor Theorem
| Aspect | Remainder Theorem | Factor Theorem |
|---|---|---|
| Statement | p(a) = remainder when divided by (x − a) | (x − a) is a factor if p(a) = 0 |
| Remainder | Any real number | Always zero |
| Application | Find remainder without division | Find factors and roots |
| Relationship | General case | Special case of Remainder Theorem |
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Sign error in (x − a): treating (x + 2) as x − (−2) | For (x + 2), substitute x = −2, not x = 2 |
| Forgetting to check all possible factors for cubic | Try ±1, ±2, ±3, ± factors of constant term |
| Incorrect synthetic division setup | Bring down leading coefficient, multiply, add |
| Quotient degree error | If dividing cubic by linear, quotient is quadratic |
ICSE Exam Focus
This topic carries 8–12 marks in ICSE exams. Questions include:
- Using Remainder Theorem to find remainders.
- Using Factor Theorem to find unknown coefficients.
- Factorising cubic expressions.
- Solving cubic equations by factorisation.
Marks Blueprint:
| Topic | Marks |
|---|---|
| Applying Remainder Theorem | 3 |
| Finding unknown coefficient using Factor Theorem | 3 |
| Factorising cubic polynomial | 4–6 |
| Solving cubic equation | 3 |
Self-Test Questions
-
Find the remainder when 2x³ − 5x² + 3x + 7 is divided by (x − 3).
-
Show that (x − 1) is a factor of x³ − 7x² + 14x − 8. Factorise the expression completely.
-
If (x + 2) is a factor of x³ − kx² + 5x + 2, find k.
-
Factorise 2x³ − 3x² − 11x + 6 using the Factor Theorem.
-
Solve the equation: x³ − 2x² − 5x + 6 = 0.
-
State the Remainder Theorem and explain how it leads to the Factor Theorem.
In ICSE, factorisation questions often appear as part of larger algebra problems. Master the Factor Theorem — it is your most efficient tool for handling cubic polynomials.
