Before we go into solving the question for Class 8 Maths Exercise 12.2 Factorisation, lets have a look at brief notes for the chapter. This will help you solve the questions. You should go through it properly before attempting the questions –
- Factoring Quadratic Expressions:
- Identify coefficients and constants in the expression ax² + bx + c.
- Find two numbers that multiply to ac and add to b.
- Rewrite the middle term using these two numbers.
- Factor by grouping.
- Special Factoring Formulas:
- Difference of squares: a² – b² = (a – b)(a + b)
- Perfect square trinomial: a² ± 2ab + b² = (a ± b)²
Example 1: Factorize x² + 5x + 6.
Step 1: Identify coefficients: a = 1, b = 5, c = 6.
Step 2: Find two numbers that multiply to 6 and add to 5 (2 and 3).
Step 3: Rewrite: x² + 2x + 3x + 6.
Step 4: Factor by grouping: x(x + 2) + 3(x + 2).
Step 5: Final factors: (x + 2)(x + 3).
Example 2: Factorize 4x² – 9.
Step 1: Recognize as a difference of squares.
Step 2: Rewrite as (2x)² – 3².
Step 3: Apply the formula: (2x – 3)(2x + 3).
Example 3: Factorize x² – 6x + 9.
Step 1: Recognize as a perfect square trinomial.
Step 2: Rewrite as (x – 3)².
Step 3: Final factors: (x – 3)(x – 3).
Example 4: Factorize x² – 4.
Step 1: Recognize as a difference of squares.
Step 2: Rewrite as (x)² – (2)².
Step 3: Apply the formula: (x – 2)(x + 2).
NCERT Solutions for Class 8 Maths Exercise 12.2 Chapter 12 Factorisation
1. Factorise the following expressions.
(i) a² + 8a + 16 (ii) p² – 10p + 25 (iii) 25m² + 30m + 9
(iv) 49y² + 84yz + 36z² (v) 4x² – 8x + 4
(vi) 121b² – 88bc + 16c²
(vii) (l + m)² – 4lm (Hint: Expand ( l + m)² first)
(viii) a⁴ + 2a²b² + b⁴
(i) a² + 8a + 16
= a² + 4a + 4a + 16
= a(a + 4) + 4(a + 4)
= (a + 4)(a + 4)
= (a + 4)²
Method 2:
a² + 8a + 16
= (a)² + 8a + (4)²
The 1st and the 3rd terms are perfect squares
The expression can be written as a² + 2ab + b²
a² + 2ab + b² = (a + b)²
(a)² + 8a + (4)² = (a)² + 2 × 4 × a + (4)² is in the form of above expression
We can write above as (a + 4)² or (a + 4)(a + 4)
(ii) p² – 10p + 25
= p² – 5p – 5p + 25
= p(p – 5) – 5(p – 5)
= (p – 5)(p – 5)
= (p – 5)²
Method 2:
p² – 10p + 25
= (p)² – 2 × 5 × p + (5)²
The 1st and the 3rd terms are perfect squares
The expression is similar to a² – 2ab + b² = (a – b)²
(p)² – 2 × 5 × p + (5)² = (p – 5)² = (p – 5)(p – 5)
(iii) 25m² + 30m + 9
= 25m² + 15m + 15m + 9
= 5m(5m + 3) + 3(5m + 3)
= (5m + 3)(5m + 3)
= (5m + 3)²
Method 2:
25m² + 30m + 9
= (5m)² + 2 × 3 × 5m + (3)²
The 1st and the 3rd terms are perfect squares
The expression is similar to a² + 2ab + b² = (a + b)²
(5m)² + 2 × 3 × 5m + (3)² = (5m + 3)² = (5m + 3)(5m + 3)
(iv) 49y² + 84yz + 36z²
= 49y² + 42yz + 42yz + 36z²
= 7y(7y + 6z) + 6z(7y + 6z)
= (7y + 6z)(7y + 6z)
= (7y + 6z)²
Method 2:
49y² + 84yz + 36z²
= (7y)² + 2 × 7y × 6z + (6z)²
The 1st and the 3rd terms are perfect squares
The expression is similar to a² + 2ab + b² = (a + b)²
(7y)² + 2 × 7y × 6z + (6z)² = (7y + 6z)² = (7y + 6z)(7y + 6z)
(v) 4x² – 8x + 4
= 4x² – 4x – 4x + 4
= 4x(x – 1) – 4(x – 1)
= 4(x – 1)(x – 1)
= 4(x – 1)²
(vi) 121b² – 88bc + 16c²
= 121b² – 44bc – 44bc + 16c²
= 11b(11b – 4c) – 4c(11b – 4c)
= (11b – 4c)(11b – 4c)
= (11b – 4c)²
Method 2:
Given expression: 121b² – 88bc + 16c²
Recognizing the perfect square form:
= (11b)² – 2 × 11b × 4c + (4c)²
Comparing with the perfect square formula a² – 2ab + b² = (a – b)²: = (11b – 4c)²
= (11b – 4c)(11b – 4c)
(vii) (l + m)² – 4lm
= l² + 2lm + m² – 4lm
= l² – 2lm + m²
= (l – m)(l – m)
= (l – m)²
Method 2:
(l + m)² – 4lm
Recognizing the perfect square form:
= l² + 2lm + m² – 4lm
Arranging to match the perfect square formula a² – 2ab + b² = (a – b)²: = l² – 2lm + m²
= (l – m)² = (l – m)(l – m)
(viii) a⁴ + 2a²b² + b⁴
= a⁴ + a²b² + a²b² + b⁴ (as 2a²b² = a²b² + a²b²)
= a²(a² + b²) + b²(a² + b²)
= (a² + b²)(a² + b²)
= (a² + b²)²
Method 2:
a⁴ + 2a²b² + b⁴
Recognizing the perfect square form: = (a²)⁴ + 2 × a² × b² + (b²)⁴
Comparing with the perfect square formula a² + 2ab + b² = (a + b)²: = (a² + b²)²
Final expression: = (a² + b²)(a² + b²)
2. Factorise.
(i) 4p² – 9q² (ii) 63a² – 112b² (iii) 49x² – 36
(iv) 16x⁵ – 144x² (v) (l + m)² – (l – m)²
(vi) 9x²y² – 16 (vii) (x² – 2xy + y²) – z²
(viii) 25a² – 4b² + 28bc – 49c²
(i) 4p² – 9q²
= (2p)² – (3q)² [Hint: (2p)² = 2p × 2p]
= (2p – 3q)(2p + 3q) [From Formula: a² – b² = (a – b) (a + b)]
(ii) 63a² – 112b²
= 7 × 9a² – 7 × 16b²
= 7(9a² – 16b²)
= 7((3a)² – (4b)²) [Here we can use the identity formula a² – b² = (a=b)(a-b)]
= 7(3a – 4b)(3a + 4b)
There are 3 factors of 63a² – 112b² which are 7, (3a – 4b), and (3a + 4b)
(iii) 49x² – 36
= (7x)² – 6²
= (7x – 6)(7x + 6) [From Formula: a² – b² = (a – b) (a + b)]
(iv) 16x⁵ – 144x²
= 16x²(x³ – 9)
= 16x²(x – 3)(x² + 3x + 9)
(v) (l + m)² – (l – m)²
= (l + m + l – m)(l + m – l + m)
= (2l)(2m)
= 4lm
(vi) 9x²y² – 16
= (3xy)² – 4²
= (3xy – 4)(3xy + 4)
(vii) (x² – 2xy + y²) – z²
= (x – y)² – z²
= (x – y – z)(x – y + z)
(viii) 25a² – 4b² + 28bc – 49c²
= (5a)² – (2b – 7c)²
= (5a – 2b + 7c)(5a + 2b – 7c)
3. Factorise the expressions.
(i) ax² + bx (ii) 7p² + 21q² (iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an² (v) (lm + l) + m + 1
(vi) y (y + z) + 9 (y + z) (vii) 5y² – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2 (ix) 6xy – 4y + 6 – 9x
(i) ax² + bx
Expanding – ax² = a × x × x and bx = b × x
ax² + bx = a × x × x + b × x
Taking x as common – x(ax + b)
Factors of the expression is x and (ax + b)
(ii) 7p² + 21q²
Expanding – 7p² = 7 × p × p and 21q² = 21 × q × q
7p² + 21q² = 7 × p × p + 21 × q × q
Factoring out the common factor of 7 – 7(p² + 3q²)
Factors of the expression are 7 and (p² + 3q²)
(iii) 2x³ + 2xy² + 2xz²
Expanding – 2x³ = 2 × x³, 2xy² = 2 × x × y², 2xz² = 2 × x × z²
2x³ + 2xy² + 2xz² = 2 × x³ + 2 × x × y² + 2 × x × z²
Taking 2x as common – 2x(x² + y² + z²)
Factors of the expression are 2x and (x² + y² + z²)
(iv) am² + bm² + bn² + an²
Given Expression: am² + bm² + bn² + an²
Step 1: Group and factorize.
Group as m²(a + b) + n²(b + a).
Step 2: Factor out the common binomial.
Factorize to (a + b)(m² + n²).
(v) (lm + l) + m + 1
Given Expression: (lm + l) + m + 1
Step 1: Factor out the common term.
Factorize l out of the first part and rewrite to get l(m + 1) + (m + 1).
Step 2: Factor out the common binomial.
Factorize to (m + 1)(l + 1).
(vi) y(y + z) + 9(y + z)
Expanding – y(y + z) = y × (y + z), 9(y + z) = 9 × (y + z)
y(y + z) + 9(y + z) = y × (y + z) + 9 × (y + z)
Taking (y + z) as common – (y + z)(y + 9)
Factors of the expression are (y + z) and (y + 9)
(vii) 5y² – 20y – 8z + 2yz
Given Expression: 5y² – 20y – 8z + 2yz
Step 1: Group and factorize.
Group as 5y(y – 4) – 2z(4 – y).
Step 2: Factor out the common binomial.
Factorize to (5y – 2z)(y – 4).
(viii) 10ab + 4a + 5b + 2
Given Expression: 10ab + 4a + 5b + 2
Step 1: Group and factorize.
Group as 2(5ab + 2a) + 1(5b + 2).
Step 2: Factor out the common binomial.
Factorize to (2a + 1)(5b + 2).
(ix) 6xy – 4y + 6 – 9x
Given Expression: 6xy – 4y + 6 – 9x
Step 1: Group and factorize.
Group as 2y(3x – 2) + 3(2 – 3x).
Step 2: Factor out the common binomial.
Factorize to (2y + 3)(3x – 2).
4. Factorise.
(i) a⁴ – b⁴ (ii) p⁴ – 81 (iii) x⁴ – (y + z)⁴
(iv) x⁴ – (x – z)⁴ (v) a⁴ – 2a²b² + b⁴
(i) a⁴ – b⁴
Given Expression: a⁴ – b⁴
Step 1: Recognize as a difference of squares.
Rewrite as (a² + b²)(a² – b²).
Step 2: Factorize the second term.
Using the formula a² – b² = (a – b)(a + b), factorize to (a² + b²)(a + b)(a – b).
(ii) p⁴ – 81
Given Expression: p⁴ – 81
Step 1: Recognize as a difference of squares.
Rewrite as (p² + 9)(p² – 9).
Step 2: Factorize the second term.
Note that 9 is 3². Rewrite as (p² + 9)(p² – 3²).
Step 3: Apply the formula again.
Factorize to (p² + 9)(p + 3)(p – 3).
(iii) x⁴ – (y + z)⁴
Given Expression: x⁴ – (y + z)⁴
Step 1: Recognize as a difference of squares.
Rewrite as (x² + (y + z)²)(x² – (y + z)²).
Step 2: Expand and factorize.
Expand the second term and factorize to (x² + y² + 2yz + z²)(x + y + z)(x – y – z).
(iv) x⁴ – (x – z)⁴
Given Expression: x⁴ – (x – z)⁴
Step 1: Recognize as a difference of squares.
Rewrite as (x² + (x – z)²)(x² – (x – z)²).
Step 2: Expand and factorize.
Expand the second term and factorize to (x² + x² – 2xz + z²)(x + x – z)(x – x + z).
(v) a⁴ – 2a²b² + b⁴
Given Expression: a⁴ – 2a²b² + b⁴
Step 1: Recognize as a perfect square trinomial.
Rewrite as (a² – b²)².
Step 2: Apply the difference of squares formula.
Factorize to (a – b)²(a + b)².
5. Factorise the following expressions.
(i) p² + 6p + 8 (ii) q² – 10q + 21 (iii) p² + 6p – 16
(i) p² + 6p + 8
Given Expression: p² + 6p + 8
Step 1: Split the middle term.
Rewrite as p² + 4p + 2p + 8.
Step 2: Group and factorize.
Group as p(p + 4) + 2(p + 4).
Step 3: Factor out the common binomial.
Factorize to (p + 2)(p + 4).
(ii) q² – 10q + 21
Given Expression: q² – 10q + 21
Step 1: Split the middle term.
Rewrite as q² – 7q – 3q + 21.
Step 2: Group and factorize.
Group as q(q – 7) – 3(q – 7).
Step 3: Factor out the common binomial.
Factorize to (q – 3)(q – 7).
(iii) p² + 6p – 16
Given Expression: p² + 6p – 16
Step 1: Split the middle term.
Rewrite as p² + 8p – 2p – 16.
Step 2: Group and factorize.
Group as p(p + 8) – 2(p + 8).
Step 3: Factor out the common binomial.
Factorize to (p – 2)(p + 8).
