Factorisation by Grouping
Factorisation is the process of writing an algebraic expression as a product of its factors. It is the reverse of expansion (multiplication).
When an expression has four or more terms and no single common factor across all terms, we can use the method of grouping. In this method, terms are arranged in groups such that each group has a common factor, and after factoring each group, a common binomial factor emerges.
Factorisation by grouping is an important technique in the Class 8 NCERT Factorisation chapter. It extends the common factor method and prepares students for more advanced factorisation techniques used in Classes 9 and 10.
What is Factorisation by Grouping?
Definition: Factorisation by grouping is a method where the terms of an algebraic expression are rearranged into groups, each group is factorised separately, and then a common factor is extracted from the groups.
Key Terms:
- Factor: An expression that divides another expression exactly. For example, (x + 2) is a factor of x² + 5x + 6.
- Common Factor: A factor that appears in every term of an expression.
- Grouping: Arranging terms in pairs (or groups) that share a common factor.
- Binomial Factor: A common factor with two terms, such as (a + b) or (x - 3).
When to use grouping:
- The expression has 4 (or more) terms.
- No single factor is common to ALL terms.
- Terms can be paired so that each pair has a common factor.
- After factoring each pair, a common binomial factor appears.
Factorisation by Grouping Formula
Steps for Factorisation by Grouping:
- Arrange the terms in pairs (groups of 2) such that each pair has a common factor.
- Factor out the common factor from each pair.
- Identify the common binomial factor that appears in both groups.
- Factor out the common binomial factor.
- Write the expression as a product of two factors.
General Pattern:
ac + ad + bc + bd = a(c + d) + b(c + d) = (a + b)(c + d)
Important: If the initial grouping does not produce a common binomial factor, try rearranging the terms differently. Not all expressions can be factorised by grouping.
Derivation and Proof
Why does grouping work?
Grouping is based on the distributive property applied in reverse:
- We know that (a + b)(c + d) = ac + ad + bc + bd (expansion).
- Factorisation is the reverse: ac + ad + bc + bd = (a + b)(c + d).
- To go from the expanded form back to the factored form, we group: (ac + ad) + (bc + bd).
- Factor each group: a(c + d) + b(c + d).
- Now (c + d) is common: (c + d)(a + b).
Verification by expansion:
To check if your factorisation is correct, expand the factors back:
- If (a + b)(c + d) = ac + ad + bc + bd matches the original expression, the factorisation is correct.
Choosing the right grouping:
Sometimes the terms need to be rearranged before grouping. Consider the expression ax + by + bx + ay:
- Group as (ax + bx) + (ay + by) = x(a + b) + y(a + b) = (a + b)(x + y). This works.
- If we had grouped as (ax + by) + (bx + ay), no common factor emerges. So the grouping matters.
Types and Properties
Different patterns in factorisation by grouping:
1. Standard four-term grouping (2 + 2):
- ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)
2. Grouping with negative signs:
- ax - ay + bx - by = a(x - y) + b(x - y) = (a + b)(x - y)
3. Grouping where second group needs a negative sign factored out:
- ax + ay - bx - by = a(x + y) - b(x + y) = (a - b)(x + y)
4. Rearrangement needed before grouping:
- ax + by + ay + bx = (ax + bx) + (ay + by) = x(a + b) + y(a + b) = (a + b)(x + y)
5. Grouping in expressions with 6 terms (3 + 3 or 2 + 2 + 2):
- Less common, but same principle: group terms, factor each group, find common factor.
6. Grouping combined with common factor:
- Sometimes a common numerical factor can be taken out first, then grouping is applied to the remaining expression.
Solved Examples
Example 1: Example 1: Basic grouping
Problem: Factorise: ax + ay + bx + by
Solution:
Step 1: Group the terms
- (ax + ay) + (bx + by)
Step 2: Factor each group
- a(x + y) + b(x + y)
Step 3: Take out the common binomial factor (x + y)
- (x + y)(a + b)
Answer: ax + ay + bx + by = (x + y)(a + b)
Example 2: Example 2: Grouping with subtraction
Problem: Factorise: 6xy - 4y + 6 - 9x
Solution:
Step 1: Rearrange terms
- 6xy - 9x - 4y + 6
Step 2: Group
- (6xy - 9x) + (-4y + 6)
Step 3: Factor each group
- 3x(2y - 3) + (-2)(2y - 3)
- = 3x(2y - 3) - 2(2y - 3)
Step 4: Common binomial factor
- (2y - 3)(3x - 2)
Answer: 6xy - 4y + 6 - 9x = (2y - 3)(3x - 2)
Example 3: Example 3: Factoring out negative from a group
Problem: Factorise: x² + xz - xy - yz
Solution:
Step 1: Group
- (x² + xz) + (-xy - yz)
Step 2: Factor each group
- x(x + z) + (-y)(x + z)
- = x(x + z) - y(x + z)
Step 3: Common binomial factor
- (x + z)(x - y)
Verification: (x + z)(x - y) = x² - xy + xz - yz. Matches the original.
Answer: x² + xz - xy - yz = (x + z)(x - y)
Example 4: Example 4: Grouping with numerical coefficients
Problem: Factorise: 2a² + 6ab + a + 3b
Solution:
Step 1: Group
- (2a² + 6ab) + (a + 3b)
Step 2: Factor each group
- 2a(a + 3b) + 1(a + 3b)
Step 3: Common binomial factor
- (a + 3b)(2a + 1)
Answer: 2a² + 6ab + a + 3b = (a + 3b)(2a + 1)
Example 5: Example 5: Rearrangement needed
Problem: Factorise: pm + qn + pn + qm
Solution:
Step 1: Rearrange to bring similar terms together
- pm + pn + qm + qn
Step 2: Group
- (pm + pn) + (qm + qn)
Step 3: Factor each group
- p(m + n) + q(m + n)
Step 4: Common binomial factor
- (m + n)(p + q)
Answer: pm + qn + pn + qm = (m + n)(p + q)
Example 6: Example 6: All terms with same variable
Problem: Factorise: x³ + x² + x + 1
Solution:
Step 1: Group
- (x³ + x²) + (x + 1)
Step 2: Factor each group
- x²(x + 1) + 1(x + 1)
Step 3: Common binomial factor
- (x + 1)(x² + 1)
Verification: (x + 1)(x² + 1) = x³ + x + x² + 1 = x³ + x² + x + 1. Correct.
Answer: x³ + x² + x + 1 = (x + 1)(x² + 1)
Example 7: Example 7: Grouping with squared terms
Problem: Factorise: a²b + ab² + a + b
Solution:
Step 1: Group
- (a²b + ab²) + (a + b)
Step 2: Factor each group
- ab(a + b) + 1(a + b)
Step 3: Common binomial factor
- (a + b)(ab + 1)
Answer: a²b + ab² + a + b = (a + b)(ab + 1)
Example 8: Example 8: Three-term grouping after rearrangement
Problem: Factorise: 3x² - x - 3x + 1
Solution:
Step 1: Group
- (3x² - x) + (-3x + 1)
Step 2: Factor each group
- x(3x - 1) + (-1)(3x - 1)
- = x(3x - 1) - 1(3x - 1)
Step 3: Common binomial factor
- (3x - 1)(x - 1)
Verification: (3x - 1)(x - 1) = 3x² - 3x - x + 1 = 3x² - 4x + 1. But our expression is 3x² - x - 3x + 1 = 3x² - 4x + 1. Matches.
Answer: 3x² - 4x + 1 = (3x - 1)(x - 1)
Example 9: Example 9: Expression requiring common factor first
Problem: Factorise: 2ax + 2ay + 2bx + 2by
Solution:
Step 1: Take out the common numerical factor
- 2(ax + ay + bx + by)
Step 2: Group the expression inside the bracket
- 2[(ax + ay) + (bx + by)]
Step 3: Factor each group
- 2[a(x + y) + b(x + y)]
Step 4: Common binomial factor
- 2(x + y)(a + b)
Answer: 2ax + 2ay + 2bx + 2by = 2(x + y)(a + b)
Example 10: Example 10: Checking if grouping is possible
Problem: Can x² + 2x + 3y + 6 be factorised by grouping?
Solution:
Attempt: Try rearranging as (x² + 2x) + (3y + 6)
- x(x + 2) + 3(y + 2)
The binomial factors are (x + 2) and (y + 2) — they are not the same.
Try another grouping: (x² + 3y) + (2x + 6)
- No common factor in x² + 3y. Does not work.
Try: (x² + 6) + (2x + 3y)
- No common factor in either group.
Answer: This expression cannot be factorised by grouping. Not all four-term expressions are factorisable by grouping.
Real-World Applications
Simplifying Algebraic Expressions: Factorisation by grouping simplifies complex expressions, making them easier to evaluate and work with.
Solving Equations: Factorising expressions helps in solving polynomial equations. If (a + b)(c + d) = 0, then either a + b = 0 or c + d = 0.
Finding HCF and LCM of Expressions: Factorised forms help identify common factors and multiples of algebraic expressions.
Division of Polynomials: Factorisation simplifies division of one polynomial by another.
Higher Mathematics: Grouping is used in polynomial factorisation, partial fractions (Class 12), and solving systems of equations.
Key Points to Remember
- Factorisation by grouping is used when an expression has 4 or more terms and no single common factor across all terms.
- The method involves grouping terms in pairs, factoring each pair, and extracting a common binomial factor.
- The general pattern is: ac + ad + bc + bd = (a + b)(c + d).
- Sometimes terms need to be rearranged before grouping to make the method work.
- When factoring out from a group with subtraction, be careful with the negative sign: -(x + y) = -x - y.
- Always verify your factorisation by expanding the factors back.
- If a numerical factor is common to ALL terms, take it out first before grouping.
- Not all four-term expressions can be factorised by grouping.
- Factorisation is the reverse of expansion: expanding (a + b)(c + d) gives ac + ad + bc + bd.
- This method is a stepping stone to factorising quadratic expressions in higher classes.
Practice Problems
- Factorise: xy + xz + wy + wz.
- Factorise: a² + ab + ac + bc.
- Factorise: 2mp + 2mq + 3np + 3nq.
- Factorise: x³ - x² + x - 1.
- Factorise: 6ab - b² + 12ac - 2bc.
- Factorise: pq - pr - 3q + 3r.
- Factorise: 4st + 8su - 3t - 6u.
- Can ab + cd + ac + be be factorised by grouping? Justify your answer.
Frequently Asked Questions
Q1. What is factorisation by grouping?
It is a method of factorising algebraic expressions by arranging terms in groups (usually pairs), factoring each group separately, and then extracting a common binomial factor from the groups.
Q2. When should I use factorisation by grouping?
Use it when the expression has 4 or more terms, no single factor is common to all terms, but terms can be paired such that each pair shares a common factor.
Q3. What if my grouping does not give a common binomial factor?
Try rearranging the terms and grouping them differently. If no arrangement produces a common binomial factor, the expression may not be factorisable by this method.
Q4. How do I handle negative signs in grouping?
When you factor a negative from a group, all signs inside the bracket change. For example: -3x - 6 = -3(x + 2). Be careful to change ALL signs when factoring out a negative.
Q5. How do I verify my factorisation?
Expand the factors by multiplying them out. If the result matches the original expression, your factorisation is correct.
Q6. Can I use grouping for expressions with 3 terms?
Grouping is typically used for 4+ terms. For 3-term expressions (trinomials), use other methods like splitting the middle term (which is itself a form of grouping after splitting).
Q7. What is the difference between common factor method and grouping?
The common factor method extracts a factor common to ALL terms at once. Grouping is used when no single factor is common to all terms, but pairs of terms share common factors.
Q8. Can every four-term expression be factorised by grouping?
No. Some four-term expressions cannot be factorised by grouping regardless of how you arrange the terms. For example, x² + 2x + 3y + 5 has no grouping that produces a common binomial factor.
Q9. What comes after factoring each group?
After factoring each group, look for a common binomial factor in both groups. Factor it out to get the final answer as a product of two factors.
Q10. Is factorisation the reverse of expansion?
Yes. Expansion is multiplying factors to get an expression: (a+b)(c+d) = ac+ad+bc+bd. Factorisation is the reverse: writing the expression as a product of factors.
