Factorisation - REET Maths Notes
Factorisation in REET Maths tests whether you can apply algebraic identities correctly - especially the subtle difference between a3+b3 and a3-b3. This page gives every identity, method, and REET 2024's questions with full solutions.
- 2 questions in REET 2024
- Class 8-9 level
- 6 min read
How REET asks this topic
Both REET 2024 questions were based directly on applying the cube identities (a3-b3 and a3+b3) - factorising x3-8 and expanding x3+y3 were both asked. In other words, the biggest challenge here is remembering the correct sign (+/-) in each identity, because the two identities look almost the same.
Key algebraic identities
| Identity | Example |
|---|---|
| (a+b)^2 = a^2+2ab+b^2 | (x+3)^2 = x^2+6x+9 |
| (a-b)^2 = a^2-2ab+b^2 | (x-3)^2 = x^2-6x+9 |
| a^2-b^2 = (a+b)(a-b) | x^2-9 = (x+3)(x-3) |
| (x+a)(x+b) = x^2+(a+b)x+ab | (x+2)(x+3) = x^2+5x+6 |
| a^3+b^3 = (a+b)(a^2-ab+b^2) | x^3+27 = (x+3)(x^2-3x+9) |
| a^3-b^3 = (a-b)(a^2+ab+b^2) | x^3-8 = (x-2)(x^2+2x+4) |
| (a-b)^3 = a^3-3a^2b+3ab^2-b^3 | (x-1)^3 = x^3-3x^2+3x-1 |
In the a3+b3 trinomial, the middle sign is negative (-ab), while in the a3-b3 trinomial it's positive (+ab) - this is the point students mix up most.
Worked examples (based on REET 2024)
x^3-8 = x^3-2^3. Using the identity a^3-b^3=(a-b)(a^2+ab+b^2) with a=x, b=2: x^3-8 = (x-2)(x^2+2x+4). So (x^2+2x+4) is the correct factor - whereas (x^2-2x+4) would be a factor of x^3+8, not x^3-8.
For x^3+y^3, using the identity a^3+b^3=(a+b)(a^2-ab+b^2) with a=x, b=y: x^3+y^3=(x+y)(x^2-xy+y^2). To verify, expand it: (x+y)(x^2-xy+y^2) = x^3-x^2y+xy^2+x^2y-xy^2+y^3 = x^3+y^3, which checks out.
Methods of factorisation
Common factor method: for example, in 3x^2+6x, taking out the common factor 3x gives 3x(x+2).
Regrouping method: the terms are split into pairs and a common factor is pulled from each pair, e.g. ax+ay+bx+by = a(x+y)+b(x+y) = (a+b)(x+y).
Identity method: factors are written directly once you recognise a match with one of the identities above.
Splitting the middle term: in an expression of the form x^2+bx+c, you find two factors of c whose sum is b, e.g. in x^2+7x+12, 3x4=12 and 3+4=7, so (x+3)(x+4).
Another useful identity
- Square of three terms
- (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca - useful for factorising expressions with three variables.
- Check
- With a=1,b=2,c=3, left side: (1+2+3)^2=36; right side: 1+4+9+2x2+2x6+2x3=14+4+12+6=36 - both equal.
- Common mistake
- Students often write (a+b+c)^2 as just a^2+b^2+c^2 and forget the three cross-product terms (2ab, 2bc, 2ca) - give this special attention.
Why factorisation matters
Factorisation is the process of writing an algebraic expression as the product of two or more simpler expressions. It's the foundation for solving quadratic equations, simplifying fractions, and further algebraic operations, which is why it stays important in REET's Maths section.
When an expression looks hard to factorise, first check whether any common factor can be pulled out, then see whether it matches a standard identity (like a^2-b^2 or a^3+/-b^3) - this systematic order solves most factorisation questions.
Splitting the middle term when the coefficient isn't 1
When the coefficient a in a trinomial ax^2+bx+c isn't 1, first find the product a x c, then find two factors of it that add up to b, and finally split the middle term into those two parts and apply the regrouping method.
Example: factorise 2x^2+7x+3. Here a x c = 2x3=6; you need two factors that add up to 7 - 6 and 1 (6x1=6, 6+1=7). So 2x^2+6x+x+3 = 2x(x+3)+1(x+3) = (2x+1)(x+3). Check by expanding: (2x+1)(x+3)=2x^2+6x+x+3=2x^2+7x+3, which checks out.
Another example: factorise 3x^2-11x+6. Here a x c = 3x6=18; you need two factors that add up to -11 - -9 and -2 (-9x-2=18, -9+(-2)=-11). So 3x^2-9x-2x+6 = 3x(x-3)-2(x-3) = (3x-2)(x-3). Check: (3x-2)(x-3)=3x^2-9x-2x+6=3x^2-11x+6, which checks out.
How to teach this to children
Classroom activity
In Classes 8-9, show an identity like (a+b)^2 using algebra tiles or a coloured-paper area model - an a x a square, a b x b square, and two a x b rectangles combine to form the area of (a+b)^2, making the identity visible.
Teaching aids
Make cards for all the key identities and run a 'matching activity' - expressions on one side, their factorised form on the other.
Common misconception
Students often write a^3-b^3's factors as (a-b)(a^2-ab+b^2) incorrectly (the sign flips), or mistake a^3+b^3 for (a+b)^3. Clear this up by checking with numbers (like a=2, b=1).
Assessment idea
Have students solve an expression two ways - by factorising, then by substituting values to verify both sides match.
Cross-curricular link
Connect factorisation to area problems (like given the area of a rectangular field, finding its sides), so the relationship between algebra and geometry becomes clear.
Trick to remember / exam tip
An easy way to remember the correct sign in a3+/-b3: the first bracket (a+/-b) keeps the same sign as in the original question; the middle sign of the trinomial in the second bracket always flips; and the sign of b^2 at the end is always positive. So for a3-b3: first sign '-', middle sign flips to '+', and b^2 positive - (a-b)(a^2+ab+b^2).
Questions asked in REET 2024 (with solutions)
Question 107
A factor of (x³ – 8) is :
- A(x + 2)
- B(x² + 2x + 4)
- C(x – 4)
- D(x² – 2x + 4)
Explanation
सर्वसमिका a³ − b³ = (a − b)(a² + ab + b²) से x³ − 8 = x³ − 2³ = (x − 2)(x² + 2x + 4)। अतः (x² + 2x + 4) इसका एक गुणनखण्ड है। (x² − 2x + 4) तो x³ + 8 का गुणनखण्ड होता है।
Question 109
x³ + y³ is equal to
- A(x + y)(x² + xy + y²)
- B(x − y)(x² + xy + y²)
- C(x + y)(x² − xy + y²)
- D(x + y)(x² + xy − y²)
Explanation
सर्वसमिका: x³ + y³ = (x + y)(x² − xy + y²)। जाँच: (x + y)(x² − xy + y²) = x³ − x²y + xy² + x²y − xy² + y³ = x³ + y³। विकल्प (B) का गुणनफल x³ − y³ होता है।
Practice Questions
Question 1
a² − b² is equal to:
- A(a+b)²
- B(a−b)(a+b)
- C(a−b)²
- Da²+b²
Explanation
The identity is a^2-b^2 = (a+b)(a-b).
Question 2
The expansion of (x+5)(x+5) is:
- Ax²+25
- Bx²+10x+25
- Cx²+5x+25
- Dx²−10x+25
Explanation
From (a+b)^2=a^2+2ab+b^2, (x+5)^2=x^2+10x+25.
Question 3
The factorisation of x²+7x+12 is:
- A(x+2)(x+6)
- B(x+3)(x+4)
- C(x+1)(x+12)
- D(x−3)(x−4)
Explanation
We need factors of 12 that add up to 7: 3 and 4 (3x4=12, 3+4=7), giving (x+3)(x+4).
Question 4
Which is the correct factorisation of a³ − b³?
- A(a−b)(a²+ab+b²)
- B(a−b)(a²−ab+b²)
- C(a+b)(a²−ab+b²)
- D(a−b)(a²+2ab+b²)
Explanation
The identity is a^3-b^3 = (a-b)(a^2+ab+b^2), where the middle sign of the trinomial is positive.
Question 5
The factorisation of 4x² − 9 is:
- A(2x−3)(2x+3)
- B(4x−3)(x+3)
- C(2x−9)(2x+1)
- D(x−3)(4x+3)
Explanation
4x^2-9 = (2x)^2-3^2 = (2x-3)(2x+3), from the a^2-b^2 identity.
Question 6
One factor of x³ + 27 is:
- A(x+3)
- B(x−3)
- C(x²+3x+9)
- D(x+9)
Explanation
27=3^3, so by the a^3+b^3 identity, x^3+27=(x+3)(x^2-3x+9). (x+3) is a correct factor.
Question 7
In the expansion of (a−b)², the middle term is:
- A+2ab
- B−2ab
- Cab
- D−ab
Explanation
(a-b)^2=a^2-2ab+b^2, so the middle term is -2ab.
Question 8
If x + 1/x = 5, what is the value of x² + 1/x²?
- A21
- B23
- C25
- D27
Explanation
x^2+1/x^2 = (x+1/x)^2-2 = 5^2-2 = 25-2 = 23.
Question 9
Using the identity a²−b², what is the value of 999² − 1²?
- A998000
- B999000
- C998999
- D997999
Explanation
999^2-1^2=(999-1)(999+1)=998x1000=998000.
Question 10
Which is the most effective teaching method for teaching identities like a²−b², a³+b³ to Class 8-9 students?
- Aबीजगणितीय टाइल्स/क्षेत्रफल मॉडल से सर्वसमिका को चित्रात्मक रूप से दिखाना
- Bकेवल सूत्र रटवाना
- Cबिना समझाए सीधे उत्तर बताना
- Dकोई गतिविधि न कराना
Explanation
Showing an identity visually through algebra tiles/area models helps children understand its geometric meaning, building genuine understanding rather than rote memorisation of a formula.
Frequently Asked Questions
How many questions come from Factorisation in REET?+
REET 2024 had 1-2 questions from this topic in the Maths section, both based on the cube identities (a3+b3, a3-b3).
What is the formula for a3-b3?+
a^3-b^3 = (a-b)(a^2+ab+b^2), where the middle sign of the trinomial is positive.
What is the formula for a3+b3?+
a^3+b^3 = (a+b)(a^2-ab+b^2), where the middle sign of the trinomial is negative.
Where is the splitting-the-middle-term method used?+
For factorising trinomials of the type x^2+bx+c, where you look for factors of c whose sum equals b.
What's the difference between factorisation and expansion?+
In expansion, brackets are opened up to turn an expression into a sum of terms, while in factorisation, a sum of terms is written back as a product (in brackets) - the two are opposite processes.
Is every binomial factorisable?+
No, not every binomial has a simple factorisation - it depends on whether it matches a standard identity (like a^2-b^2).
What should you check first when factorising?+
First check whether all the terms share a common factor - once that's pulled out, the remaining expression often matches a standard identity, making the solution easier.
Can a factorisation answer be written in more than one form?+
Writing the factors in a different order (like (x+3)(x+4) or (x+4)(x+3)) is valid since multiplication is commutative, but the factors themselves should remain the same.
How do you check whether a factorisation is correct?+
Multiply the resulting factors together and see whether they match the original expression - this check should be done at the end of every factorisation question, so even a small sign error gets caught in time.
What is the expansion of (a-b)^3?+
(a-b)^3 = a^3-3a^2b+3ab^2-b^3 - this is the cube expansion identity, which differs in sign from (a+b)^3=a^3+3a^2b+3ab^2+b^3.
When does the special result a3+b3+c3 = 3abc hold true?+
When a+b+c=0. Example: a=1, b=2, c=-3 (sum=0). Check: a^3+b^3+c^3=1+8+(-27)=-18; 3abc=3x1x2x(-3)=-18 - both equal, proving the result.
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Change Log
- 17 September 2026 — English page published — mirrors the fact-checked Hindi page, independently bilingual-accuracy checked