Factorisation of Polynomials
Factorisation is one of the most important skills in algebra. Just as a number like 12 can be broken down into its prime factors 2 x 2 x 3, a polynomial can be broken down into simpler polynomial factors. In Class 9, you will learn to factorise quadratic and cubic polynomials using several methods: extracting common factors, splitting the middle term, using algebraic identities, and applying the Factor Theorem. Factorisation is the reverse process of expansion — while expansion takes (x + 2)(x + 3) and gives x^2 + 5x + 6, factorisation takes x^2 + 5x + 6 and breaks it back into (x + 2)(x + 3). This skill is essential for solving polynomial equations (finding zeroes), simplifying algebraic fractions, and working with algebraic identities. In the NCERT Class 9 curriculum, you are expected to factorise quadratic trinomials by splitting the middle term, recognise and apply standard identities (difference of squares, perfect square trinomials, sum and difference of cubes), and use the Factor Theorem to factorise cubic polynomials. This topic ties together everything you have learned about polynomials — degree, zeroes, the Remainder Theorem, and the Factor Theorem — into a powerful problem-solving toolkit.
What is Factorisation of Polynomials?
Factorisation of a polynomial is the process of expressing the polynomial as a product of two or more polynomials of lower degree.
If p(x) = f(x) x g(x) x h(x) x ..., then f(x), g(x), h(x), ... are called factors of p(x).
Methods of factorisation taught in Class 9:
Method 1: Taking out the common factor (HCF method)
If every term of the polynomial has a common factor, extract it.
Example: 6x^3 + 9x^2 = 3x^2(2x + 3).
Method 2: Grouping
Rearrange and group terms that share a common factor, then factor each group.
Example: x^3 + x^2 + x + 1 = x^2(x + 1) + 1(x + 1) = (x + 1)(x^2 + 1).
Method 3: Splitting the middle term (for quadratics)
For ax^2 + bx + c, find two numbers whose product is ac and whose sum is b.
Example: x^2 + 5x + 6. Find two numbers with product 6 and sum 5: 2 and 3.
x^2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3).
Method 4: Using algebraic identities
Recognise patterns that match standard identities:
a^2 - b^2 = (a + b)(a - b)
(a + b)^2 = a^2 + 2ab + b^2
(a - b)^2 = a^2 - 2ab + b^2
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Method 5: Factor Theorem
If p(a) = 0, then (x - a) is a factor of p(x). Use trial to find one zero, then divide to reduce the degree.
Example: For p(x) = x^3 - 6x^2 + 11x - 6, try x = 1: p(1) = 1 - 6 + 11 - 6 = 0. So (x - 1) is a factor. Divide to get x^2 - 5x + 6, which factors as (x - 2)(x - 3). Final: p(x) = (x - 1)(x - 2)(x - 3).
Factorisation of Polynomials Formula
Key identities used for factorisation:
1. Difference of squares:
a^2 - b^2 = (a + b)(a - b)
2. Perfect square trinomial (sum):
a^2 + 2ab + b^2 = (a + b)^2
3. Perfect square trinomial (difference):
a^2 - 2ab + b^2 = (a - b)^2
4. Sum of cubes:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
5. Difference of cubes:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
6. Three-variable identity:
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
Special case: If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc.
7. Splitting the middle term (quadratic ax^2 + bx + c):
Find two numbers p and q such that p + q = b and p x q = ac.
Then ax^2 + bx + c = ax^2 + px + qx + c, and factor by grouping.
8. Factor Theorem:
For polynomial p(x), if p(a) = 0, then (x - a) is a factor of p(x).
9. Remainder Theorem (used to test potential factors):
When p(x) is divided by (x - a), the remainder is p(a). If p(a) = 0, remainder is 0, confirming (x - a) is a factor.
Derivation and Proof
Why the splitting-the-middle-term method works:
Consider the quadratic ax^2 + bx + c. We want to factorise it as (dx + e)(fx + g) where d, e, f, g are chosen appropriately.
Expanding (dx + e)(fx + g) = dfx^2 + (dg + ef)x + eg.
Comparing with ax^2 + bx + c: df = a, dg + ef = b, eg = c.
Note that (dg)(ef) = (df)(eg) = ac. Also dg + ef = b.
So we need two numbers (dg and ef) whose product is ac and whose sum is b. This is exactly the splitting-the-middle-term condition.
Once we find such numbers, say p and q (where p + q = b and pq = ac), we write:
ax^2 + bx + c = ax^2 + px + qx + c
Then we group and factor: take common factors from the first two terms and the last two terms.
Why the Factor Theorem leads to factorisation:
The Factor Theorem states: if p(a) = 0, then (x - a) divides p(x) with zero remainder.
For a cubic p(x) = ax^3 + bx^2 + cx + d:
Step 1: Find a value a such that p(a) = 0. Usually try a = ±1, ±2, ±3 (factors of the constant term d, divided by factors of the leading coefficient a — this is the Rational Root Theorem).
Step 2: Once you find p(a) = 0, (x - a) is a factor. Divide p(x) by (x - a) using long division or synthetic division to get a quadratic q(x).
Step 3: Factorise the quadratic q(x) using splitting the middle term or identities.
Step 4: The complete factorisation is p(x) = (x - a) x q(x) = (x - a)(x - b)(x - c) if q(x) also has real factors.
Example: Factorise x^3 - 3x^2 - 4x + 12.
Try x = 2: p(2) = 8 - 12 - 8 + 12 = 0. So (x - 2) is a factor.
Divide: x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6) = (x - 2)(x - 3)(x + 2).
Types and Properties
Factorisation problems are classified by the method required:
Type 1: Common Factor Extraction
Used when all terms share a common factor (variable or coefficient).
Examples:
5x^2 + 10x = 5x(x + 2)
3a^2b - 6ab^2 + 9ab = 3ab(a - 2b + 3)
x^4 - x^3 + x^2 = x^2(x^2 - x + 1)
Type 2: Factorisation by Grouping
Used for 4-term polynomials where grouping reveals common factors.
Examples:
ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b)
x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x - 1)(x^2 + 1)
Type 3: Splitting the Middle Term (Quadratic Trinomials)
Used for ax^2 + bx + c. Find two numbers with product ac and sum b.
Examples:
x^2 + 7x + 12 = (x + 3)(x + 4) [3 x 4 = 12, 3 + 4 = 7]
2x^2 + 7x + 3 = (2x + 1)(x + 3) [split 7x as x + 6x; 1 x 6 = 6 = 2 x 3]
6x^2 - 5x - 6 = (2x - 3)(3x + 2) [split -5x as -9x + 4x; -9 x 4 = -36 = 6 x (-6)]
Type 4: Using Algebraic Identities
Used when the polynomial matches a standard pattern.
a^2 - b^2: x^2 - 49 = (x + 7)(x - 7)
a^2 + 2ab + b^2: x^2 + 6x + 9 = (x + 3)^2
a^3 + b^3: 8x^3 + 27 = (2x + 3)(4x^2 - 6x + 9)
a^3 - b^3: x^3 - 64 = (x - 4)(x^2 + 4x + 16)
Type 5: Factor Theorem (for cubics and higher)
Used when other methods do not immediately apply. Find a zero by trial, extract the linear factor, then factorise the remaining quadratic.
Example: x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3).
Solved Examples
Example 1: Example 1: Common factor extraction
Problem: Factorise: 4x^3 - 12x^2 + 8x.
Solution:
Find the HCF of all terms: HCF of 4x^3, 12x^2, 8x is 4x.
4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2)
Now factorise x^2 - 3x + 2 by splitting the middle term: find two numbers with product 2 and sum -3. These are -1 and -2.
x^2 - 3x + 2 = (x - 1)(x - 2)
Answer: 4x^3 - 12x^2 + 8x = 4x(x - 1)(x - 2).
Example 2: Example 2: Factorisation by grouping
Problem: Factorise: x^3 + 2x^2 + 3x + 6.
Solution:
Group the terms: (x^3 + 2x^2) + (3x + 6)
Factor each group: x^2(x + 2) + 3(x + 2)
Take out the common factor (x + 2): (x + 2)(x^2 + 3)
Answer: x^3 + 2x^2 + 3x + 6 = (x + 2)(x^2 + 3).
Note: x^2 + 3 cannot be factored further over real numbers (it has no real zeroes).
Example 3: Example 3: Splitting the middle term (simple quadratic)
Problem: Factorise: x^2 - 7x + 10.
Solution:
We need two numbers whose product is 10 (= 1 x 10 = a x c) and whose sum is -7 (= b).
The numbers are -5 and -2 (since (-5) x (-2) = 10 and (-5) + (-2) = -7).
Split: x^2 - 5x - 2x + 10
Group: x(x - 5) - 2(x - 5)
Factor: (x - 5)(x - 2)
Answer: x^2 - 7x + 10 = (x - 5)(x - 2).
Verification: (x - 5)(x - 2) = x^2 - 2x - 5x + 10 = x^2 - 7x + 10. Correct.
Example 4: Example 4: Splitting the middle term (leading coefficient not 1)
Problem: Factorise: 6x^2 + 11x + 3.
Solution:
Here a = 6, b = 11, c = 3. Product ac = 18. Sum = 11.
Find two numbers with product 18 and sum 11: 9 and 2.
Split: 6x^2 + 9x + 2x + 3
Group: 3x(2x + 3) + 1(2x + 3)
Factor: (2x + 3)(3x + 1)
Answer: 6x^2 + 11x + 3 = (2x + 3)(3x + 1).
Verification: (2x + 3)(3x + 1) = 6x^2 + 2x + 9x + 3 = 6x^2 + 11x + 3. Correct.
Example 5: Example 5: Using the difference of squares identity
Problem: Factorise: (a) 25x^2 - 36 (b) x^4 - 81
Solution:
(a) 25x^2 - 36 = (5x)^2 - 6^2 = (5x + 6)(5x - 6).
(b) x^4 - 81 = (x^2)^2 - 9^2 = (x^2 + 9)(x^2 - 9) = (x^2 + 9)(x + 3)(x - 3).
We applied the identity twice: first to x^4 - 81, then to x^2 - 9.
Note: x^2 + 9 cannot be factored further over real numbers.
Answer: x^4 - 81 = (x^2 + 9)(x + 3)(x - 3).
Example 6: Example 6: Using sum and difference of cubes
Problem: Factorise: (a) 27x^3 + 8 (b) x^3 - 125
Solution:
(a) 27x^3 + 8 = (3x)^3 + 2^3.
Using a^3 + b^3 = (a + b)(a^2 - ab + b^2), with a = 3x, b = 2:
= (3x + 2)((3x)^2 - (3x)(2) + 2^2) = (3x + 2)(9x^2 - 6x + 4).
(b) x^3 - 125 = x^3 - 5^3.
Using a^3 - b^3 = (a - b)(a^2 + ab + b^2), with a = x, b = 5:
= (x - 5)(x^2 + 5x + 25).
Example 7: Example 7: Using the Factor Theorem to factorise a cubic
Problem: Factorise: x^3 - 4x^2 + x + 6.
Solution:
Step 1: Try x = 1: p(1) = 1 - 4 + 1 + 6 = 4. Not zero.
Try x = -1: p(-1) = -1 - 4 - 1 + 6 = 0. Zero! So (x + 1) is a factor.
Step 2: Divide x^3 - 4x^2 + x + 6 by (x + 1):
Using long division or comparison: x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6).
Step 3: Factorise x^2 - 5x + 6: find two numbers with product 6 and sum -5: -2 and -3.
x^2 - 5x + 6 = (x - 2)(x - 3).
Answer: x^3 - 4x^2 + x + 6 = (x + 1)(x - 2)(x - 3).
Verification: The zeroes are x = -1, 2, 3. Check: (-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0. Correct.
Example 8: Example 8: Perfect square trinomial
Problem: Factorise: 4x^2 + 12x + 9.
Solution:
Check if this is a perfect square trinomial:
4x^2 = (2x)^2 and 9 = 3^2. Middle term = 2 x (2x) x 3 = 12x. Yes, it matches!
Using a^2 + 2ab + b^2 = (a + b)^2, with a = 2x, b = 3:
4x^2 + 12x + 9 = (2x + 3)^2.
Example 9: Example 9: Using the identity a^3 + b^3 + c^3 - 3abc
Problem: Factorise: x^3 + y^3 + z^3 - 3xyz.
Solution:
This directly matches the identity:
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
With a = x, b = y, c = z:
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
Special case: If x + y + z = 0, then x^3 + y^3 + z^3 = 3xyz.
Example 10: Example 10: Multi-step factorisation
Problem: Factorise completely: 2x^3 + 2x^2 - 12x.
Solution:
Step 1 (Common factor): 2x^3 + 2x^2 - 12x = 2x(x^2 + x - 6).
Step 2 (Split middle term): x^2 + x - 6. Need two numbers with product -6 and sum 1: 3 and -2.
x^2 + x - 6 = x^2 + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x + 3)(x - 2).
Step 3 (Combine): 2x^3 + 2x^2 - 12x = 2x(x + 3)(x - 2).
Verification: 2x(x + 3)(x - 2) = 2x(x^2 + x - 6) = 2x^3 + 2x^2 - 12x. Correct.
Real-World Applications
Factorisation of polynomials is one of the most widely used skills in mathematics:
Solving Equations: To solve x^2 - 5x + 6 = 0, factorise the left side as (x - 2)(x - 3) = 0, then set each factor to zero: x = 2 or x = 3. Factorisation converts a polynomial equation into simpler linear equations.
Simplifying Algebraic Fractions: To simplify (x^2 - 9)/(x + 3), factorise the numerator: (x + 3)(x - 3)/(x + 3) = x - 3 (for x not equal to -3). This is essential in calculus when finding limits.
Finding LCM and HCF of Polynomials: Similar to finding LCM and HCF of numbers using prime factorisation, you factorise polynomials to find their algebraic LCM and HCF.
Quadratic Formula Derivation: The quadratic formula is derived by factorising (completing the square) the general quadratic ax^2 + bx + c.
Physics and Engineering: Many physics problems lead to polynomial equations. The trajectory of a projectile, the vibration modes of a string, and the stability of a system are all analysed by factorising characteristic polynomials.
Number Theory: Factorisation of special polynomials like x^n - 1 is used in number theory and cryptography.
Key Points to Remember
- Factorisation is the reverse of expansion — expressing a polynomial as a product of simpler factors.
- Always look for a common factor first before trying other methods.
- For quadratic trinomials ax^2 + bx + c: find two numbers with product ac and sum b (splitting the middle term).
- Recognise algebraic identities: a^2 - b^2, a^2 ± 2ab + b^2, a^3 ± b^3.
- The Factor Theorem states: if p(a) = 0, then (x - a) is a factor of p(x).
- To factorise a cubic: use the Factor Theorem to find one linear factor, then factorise the resulting quadratic.
- Try x = ±1, ±2, ±3 (factors of the constant term) when looking for zeroes of a cubic.
- If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc — a useful shortcut.
- Always verify your factorisation by expanding the factors back.
- Complete factorisation means all factors are irreducible (cannot be factored further over real numbers).
Practice Problems
- Factorise by extracting the common factor: 15x^2y - 25xy^2 + 10xy.
- Factorise by grouping: 2xy + 3y + 2x + 3.
- Factorise by splitting the middle term: x^2 + 9x + 14.
- Factorise: 3x^2 - 14x + 8. (Hint: product = 24, sum = -14.)
- Factorise using identities: (a) 49x^2 - 64, (b) x^2 - 10x + 25, (c) 8x^3 - 125.
- Use the Factor Theorem to factorise: x^3 - 7x + 6.
- Factorise completely: x^4 - 16. (Hint: apply difference of squares twice.)
- If a + b + c = 0, show that 2a^3 + 2b^3 + 2c^3 = 6abc.
Frequently Asked Questions
Q1. What is factorisation of polynomials?
Factorisation is the process of breaking down a polynomial into a product of simpler polynomials. For example, x^2 - 5x + 6 = (x - 2)(x - 3). The factors (x - 2) and (x - 3) are simpler expressions that multiply together to give the original polynomial. It is the reverse of expansion (multiplying out brackets).
Q2. How do you factorise a quadratic by splitting the middle term?
For ax^2 + bx + c: (1) Find two numbers p and q such that p x q = ac and p + q = b. (2) Rewrite bx as px + qx. (3) Group the four terms into two pairs. (4) Factor each pair. (5) Extract the common binomial factor. For example, x^2 + 5x + 6: p x q = 6, p + q = 5, so p = 2 and q = 3. Then x^2 + 2x + 3x + 6 = x(x+2) + 3(x+2) = (x+2)(x+3).
Q3. What is the Factor Theorem?
The Factor Theorem states: if p(x) is a polynomial and p(a) = 0 for some value a, then (x - a) is a factor of p(x). The converse is also true: if (x - a) is a factor, then p(a) = 0. This theorem is used to find factors of cubic (and higher) polynomials by testing values. If p(2) = 0, then (x - 2) is a factor.
Q4. How do you factorise a cubic polynomial?
Step 1: Use the Factor Theorem — try x = 1, -1, 2, -2, 3, -3, etc. (factors of the constant term) to find a value a where p(a) = 0. Step 2: (x - a) is a factor. Divide p(x) by (x - a) to get a quadratic. Step 3: Factorise the quadratic by splitting the middle term or using identities. This gives the complete factorisation.
Q5. What algebraic identities are used in factorisation?
The main identities are: (1) a^2 - b^2 = (a+b)(a-b) — difference of squares. (2) a^2 + 2ab + b^2 = (a+b)^2 — perfect square. (3) a^2 - 2ab + b^2 = (a-b)^2 — perfect square. (4) a^3 + b^3 = (a+b)(a^2-ab+b^2) — sum of cubes. (5) a^3 - b^3 = (a-b)(a^2+ab+b^2) — difference of cubes. (6) a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).
Q6. What values should you try first when using the Factor Theorem?
Try the factors of the constant term (the term without x) divided by the factors of the leading coefficient. For p(x) = x^3 - 6x^2 + 11x - 6, the constant term is -6 and the leading coefficient is 1. Try x = ±1, ±2, ±3, ±6. Usually, x = 1 or x = -1 is tried first because the arithmetic is simplest.
Q7. Can every polynomial be factored?
Every polynomial can be factored into linear factors over the complex numbers (Fundamental Theorem of Algebra). However, over real numbers, some polynomials like x^2 + 1 or x^2 + x + 1 cannot be factored into linear factors because they have no real zeroes. These are called irreducible polynomials over the reals. In Class 9, you factor over real numbers, so some quadratics may remain as irreducible factors.
Q8. What is the difference between factorisation and finding zeroes?
They are closely related but not the same. Factorisation expresses the polynomial as a product: x^2 - 5x + 6 = (x-2)(x-3). Finding zeroes means solving the equation p(x) = 0: x = 2 or x = 3. Factorisation leads to zeroes (set each factor to zero), and knowing zeroes leads to factors (by the Factor Theorem). They are two sides of the same coin.
Q9. How do you verify a factorisation?
Multiply the factors back together and check that you get the original polynomial. For example, if you factored x^2 - 5x + 6 as (x-2)(x-3), verify by expanding: (x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6. Since it matches, the factorisation is correct. Always verify, especially in exams.
Q10. What if the splitting-the-middle-term method does not work?
If you cannot find two numbers with the required product and sum, the quadratic may not factorise over rational numbers. In that case, check if it fits an algebraic identity. If not, it may be irreducible over rationals. You can still find zeroes using the quadratic formula x = (-b ± sqrt(b^2-4ac))/(2a). If the discriminant b^2 - 4ac is negative, the polynomial has no real zeroes and cannot be factored over real numbers.
Related Topics
- Factor Theorem
- Algebraic Identities (Extended)
- Factorisation of Quadratic Expressions
- Factorisation Using Identities
- Polynomials in One Variable
- Degree of a Polynomial
- Types of Polynomials
- Value of a Polynomial
- Zeroes of a Polynomial
- Remainder Theorem
- (a + b)³ and (a - b)³ Identities
- a³ + b³ and a³ - b³ Identities
- Zeroes of Quadratic Polynomial
- Relationship Between Zeroes and Coefficients
