In Chapter 14 of Class 8 Maths, Factorisation, students learn how to break down algebraic expressions into their simplest factors. Exercise 14.2 focuses on applying the distributive property and common factors to factorise algebraic expressions.
This set of problems helps strengthen the understanding of taking out common terms and rewriting expressions as products of factors. The NCERT solutions provided here offer clear, step-by-step explanations to help students easily understand and solve each question in Exercise 14.2. These solutions are especially useful for self-study and exam preparation.
Class 8 Maths Ch14 Factorisation Ex 14.2-NCERT Solutions
Question 1.Factorise the following expressions.
(ii) p2 – 10p + 25
Here, 5 + 5 = 10 and 5 × 5 = 25
p2 – 10p + 25
= p2 – 5p – 5p + 5 × 5
= (p2 – 5p) + (-5p + 25)
= p(p – 5) – 5(p – 5)
= (p – 5) (p – 5)
= (p – 5)2
(iii) 25m2 + 30m + 9
Here, 15 + 15 = 30 and 15 × 15 = 25 × 9 = 225
25m2 + 30m + 9
= 25m2 + 15m + 15m + 9
= (25m2 + 15m) + (15m + 9)
= 5m(5m + 3) + 3(5m + 3)
= (5m + 3) (5m + 3)
= (5m + 3)2
(iv) 49y2 + 84yz + 36z2
Here, 42 + 42 = 84 and 42 × 42 = 49 × 36 = 1764
49y2 + 84yz + 36z2
= 49y2 + 42yz + 42yz + 36z2
= 7y(7y + 6z) +6z(7y + 6z)
= (7y + 6z) (7y + 6z)
= (7y + 6z)2
(v) 4x2 – 8x + 4
= 4(x2 – 2x + 1) [Taking 4 common]
= 4(x2 – x – x + 1)
= 4[x(x – 1) -1(x – 1)]
= 4(x – 1)(x – 1)
= 4(x – 1)2
(vi) 121b2 – 88bc + 16c2
Here, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936
121b2 – 88bc + 16c2
= 121b2 – 44bc – 44bc + 16c2
= 11b(11b – 4c) – 4c(11b – 4c)
= (11b – 4c) (11b – 4c)
= (11b – 4c)2
(vii) (l + m)2 – 4lm
Expanding (l + m)2, we get
l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
= l2 – Im – lm + m2
= l(l – m) – m(l – m)
= (l – m) (l – m)
= (l – m)2
(viii) a4 + 2a2b2 + b4
= a4 + a2b2 + a2b2 + b4
= a2(a2 + b2) + b2(a2 + b2)
= (a2 + b2)(a2 + b2)
= (a2 + b2)2
Ex 14.2 Class 8 Maths Question 2.
Factorise.
(i) 4p2 – 9q2
(ii) 63a2 – 112b2
(iii) 49x2 – 36
(iv) 16x5 – 144x3
(v) (l + m)2 – (l – m)2
(vi) 9x2y2 – 16
(vii) (x2 – 2xy + y2) – z2
(viii) 25a2 – 4b2 + 28bc – 49c2
Solution:
(i) 4p2 – 9q2
= (2p)2 – (3q)2
= (2p – 3q) (2p + 3q)
[∵ a2 – b2 = (a + b)(a – b)]
(ii) 63a2 – 112b2
= 7(9a2 – 16b2)
= 7 [(3a)2 – (4b)2]
= 7(3a – 4b)(3a + 4b)
[∵ a2 – b2 = (a + b)(a – b)]
(iii) 49x2 – 36 = (7x)2 – (6)2
= (7x – 6) (7x + 6)
[∵ a2 – b2 = (a + b)(a – b)]
(iv) 16x5 – 144x3 = 16x3 (x2 – 9)
= 16x3 [(x)2 – (3)2]
= 16x3(x – 3)(x + 3)
[∵ a2 – b2 = (a + b)(a – b)]
(v) (l + m)2 – (l – m)2
= (l + m) – (l – m)] [(l + m) + (l – m)]
[∵ a2 – b2 = (a + b)(a – b)]
= (l + m – l + m)(l + m + l – m)
= (2m) (2l)
= 4ml
(vi) 9x2y2 – 16 = (3xy)2 – (4)2
= (3xy – 4)(3xy + 4)
[∵ a2 – b2 = (a + b)(a – b)]
(vii) (x2 – 2xy + y2) – z2
= (x – y)2 – z2
= (x – y – z) (x – y + z)
[∵ a2 – b2 = (a + b)(a – b)]
(viii) 25a2 – 4b2 + 28bc – 49c2
= 25a2 – (4b2 – 28bc + 49c2)
= (5a)2 – (2b – 7c)2
= [5a – (2b – 7c)] [5a + (2b – 7c)]
= (5a – 2b + 7c)(5a + 2b – 7c)
Ex 14.2 Class 8 Maths Question 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
Solution:
(i) ax2 + bx = x(ax + 5)
(ii) 7p2 + 21q2 = 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)
(iv) am2 + bm2 + bn2 + an2
= m2 (a + b) + n2(a + b)
= (a + b)(m2 + n2)
(v) (lm + l) + m + 1
= l(m + 1) + (m + 1)
= (m + 1) (l + 1)
(vi) y(y + z) + 9(y + z) = (y + z)(y + 9)
(vii) 5y2 – 20y – 8z + 2yz
= 5y2 – 20y + 2yz – 8z
= 5y(y – 4) + 2z(y – 4)
= (y – 4) (5y + 2z)
(viii) 10ab + 4a + 5b + 2
= 2a(5b + 2) + 1(5b + 2)
= (5b + 2)(2a + 1)
(ix) 6xy – 4y + 6 – 9x
= 6xy – 4y – 9x + 6
= 2y(3x – 2) – 3(3x – 2)
= (3x – 2) (2y – 3)
Ex 14.2 Class 8 Maths Question 4.
Factorise.
(i) a4 – b4
(ii) p4 – 81
(iii) x4 – (y + z)4
(iv) x4 – (x – z)4
(v) a4 – 2a2b2 + b4
Solution:
(i) a4 – b4 – (a2)2 – (b2)2
[∵ a2 – b2 = (a – b)(a + b)]
= (a2 – b2) (a2 + b2)
= (a – b) (a + b) (a2 + b2)
(ii) p4 – 81 = (p2)2 – (9)2
= (p2 – 9) (p2 + 9)
[∵ a2 – b2 = (a – b)(a + b)]
= (p – 3)(p + 3) (p2 + 9)
(iii) x4 – (y + z)4 = (x2)2 – [(y + z)2]2
[∵ a2 – b2 = (a – b)(a + b)]
= [x2 – (y + z)2] [x2 + (y + z)2]
= [x – (y + z)] [x + (y + z)] [x2 + (y + z)2]
= (x – y – z) (x + y + z) [x2 + (y + z)2]
(iv) x4 – (x – z)4 = (x2)2 – [(y – z)2]2
= [x2 – (y – z)2] [x2 + (y – z)2]
= (x – y + z) (x + y – z) (x2 + (y – z)2]
(v) a4 – 2a2b2 + b4
= a4 – a2b2 – a2b2 + b4
= a2(a2 – b2) – b2(a2 – b2)
= (a2 – b2)(a2 – b2)
= (a2 – b2)2
= [(a – b) (a + b)]2
= (a – b)2 (a + b)2
Ex 14.2 Class 8 Maths Question 5.
Factorise the following expressions.
(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16
Solution:
(i) p2 + 6p + 8
Here, 2 + 4 = 6 and 2 × 4 = 8
p2 + 6p + 8
= p2 + 2p + 4p + 8
= p (p + 2) + 4(p + 2)
= (p + 2) (p + 4)
(ii) q2 – 10q + 21
Here, 3 + 7 = 10 and 3 × 7 = 21
q2 – 10q + 21
= q2 – 3q – 7q + 21
= q(q – 3) – 7(q – 3)
= (q – 3) (q – 7)
(iii) p2 + 6p – 16
Here, 8 – 2 = 6 and 8 × 2 = 16
p2 + 6p – 16
= p2 + 8p – 2p – 16
= p(p + 8) – 2(p + 8)
= (p + 8) (p – 2)
Q1: 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
(viii) a⁴ + 2a²b² + b⁴
Solutions
(i) a² + 8a + 16
= a² + 2×4×a + 4²
= (a + 4)²
(ii) p² – 10p + 25
= p² – 2×5×p + 5²
= (p – 5)²
(iii) 25m² + 30m + 9
= (5m)² + 2×5m×3 + 3²
= (5m + 3)²
(iv) 49y² + 84yz + 36z²
= (7y)² + 2×7y×6z + (6z)²
= (7y + 6z)²
(v) 4x² – 8x + 4
= (2x)² – 2×2x×2 + 2²
= (2x – 2)²
(vi) 121b² – 88bc + 16c²
= (11b)² – 2×11b×4c + (4c)²
= (11b – 4c)²
(vii) (l + m)² – 4lm
= l² + 2lm + m² – 4lm
= l² – 2lm + m²
= (l – m)²
(viii) a⁴ + 2a²b² + b⁴
= (a²)² + 2a²b² + (b²)²
= (a² + b²)²
Q2: Factorise
(i) 4p² − 9q²
(ii) 63a² − 112b²
(iii) 49x² − 36
(iv) 16x⁵ − 144x³
(v) (l + m)² − (l − m)²
(vi) 9x²y² − 16
(vii) (x² − 2xy + y²) − z²
(viii) 25a² − 4b² + 28bc − 49c²
Solutions
(i) 4p² − 9q²
= (2p)² − (3q)²
= (2p − 3q)(2p + 3q)
(ii) 63a² − 112b²
= Common factor is 7
= 7(9a² − 16b²)
= 7[(3a)² − (4b)²]
= 7(3a − 4b)(3a + 4b)
(iii) 49x² − 36
= (7x)² − 6²
= (7x − 6)(7x + 6)
(iv) 16x⁵ − 144x³
= Common factor is 16x³
= 16x³(x² − 9)
= 16x³[(x − 3)(x + 3)]
(v) (l + m)² − (l − m)²
Use identity: a² − b² = (a − b)(a + b)
Let a = (l + m), b = (l − m)
= [(l + m) − (l − m)][(l + m) + (l − m)]
= [l + m − l + m][l + m + l − m]
= [2m][2l] = 4lm
(vi) 9x²y² − 16
= (3xy)² − 4²
= (3xy − 4)(3xy + 4)
(vii) (x² − 2xy + y²) − z²
= (x − y)² − z²
Use identity: a² − b² = (a − b)(a + b)
= [(x − y) − z][(x − y) + z]
= (x − y − z)(x − y + z)
(viii) 25a² − 4b² + 28bc − 49c²
Group the middle terms:
= 25a² − (4b² − 28bc + 49c²)
= 25a² − [(2b − 7c)²]
Now apply: a² − b² = (a − b)(a + b)
= [5a − (2b − 7c)][5a + (2b − 7c)]
= (5a − 2b + 7c)(5a + 2b − 7c)
Q3: Factorise the expressions
(i) ax² + bx
(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2x²
(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
Solutions
(i) ax² + bx
= x(ax + b)
(ii) 7p² + 21q²
= 7(p² + 3q²)
(iii) 2x³ + 2xy² + 2x²
= 2x(x² + y² + x)
(iv) am² + bm² + bn² + an²
Group terms:
= (am² + bm²) + (bn² + an²)
= m²(a + b) + n²(b + a)
= m²(a + b) + n²(a + b)
= (a + b)(m² + n²)
(v) (lm + l) + m + 1
Group and factor:
= l(m + 1) + (m + 1)
= (m + 1)(l + 1)
(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)
(vii) 5y² − 20y − 8z + 2yz
Group:
= (5y² − 20y) + (−8z + 2yz)
= 5y(y − 4) − 2z(4 − y)
= 5y(y − 4) + 2z(y − 4)
= (y − 4)(5y + 2z)
(viii) 10ab + 4a + 5b + 2
Group:
= (10ab + 4a) + (5b + 2)
= 2a(5b + 2) + 1(5b + 2)
= (5b + 2)(2a + 1)
(ix) 6xy − 4y + 6 − 9x
Group:
= (6xy − 4y) + (6 − 9x)
= 2y(3x − 2) − 3(3x − 2)
= (3x − 2)(2y − 3)
Q4: Factorise
(i) a⁴ − b⁴
(ii) p⁴ − 81
(iii) x⁴ − (y + z)⁴
(iv) x⁴ − (x − z)⁴
(v) a⁴ − 2a²b² + b⁴
Solutions
(i) a⁴ − b⁴
= (a²)² − (b²)²
= (a² − b²)(a² + b²)
Now factor a² − b² again:
= (a − b)(a + b)(a² + b²)
(ii) p⁴ − 81
= p⁴ − 3⁴
= (p²)² − 9²
= (p² − 9)(p² + 9)
Now factor p² − 9:
= (p − 3)(p + 3)(p² + 9)
(iii) x⁴ − (y + z)⁴
= (x²)² − [(y + z)²]²
Use identity: a² − b² = (a − b)(a + b)
= [x² − (y + z)²][x² + (y + z)²]
Now factor x² − (y + z)²:
= [(x − (y + z))(x + (y + z))][x² + (y + z)²]
= (x − y − z)(x + y + z)(x² + (y + z)²)
(iv) x⁴ − (x − z)⁴
Let A = x², B = (x − z)²
So: x⁴ − (x − z)⁴ = A² − B²
= (A − B)(A + B)
= [x² − (x − z)²][x² + (x − z)²]
Now expand and simplify:
First part:
x² − (x − z)²
= x² − [x² − 2xz + z²] = 2xz − z²
Second part:
x² + (x − z)² = x² + x² − 2xz + z² = 2x² − 2xz + z²
Final answer:
= (2xz − z²)(2x² − 2xz + z²)
(v) a⁴ − 2a²b² + b⁴
This is a perfect square trinomial:
= (a² − b²)²
Now factor further:
= [(a − b)(a + b)]²
= (a − b)²(a + b)²
Q5: Factorise using mid-term method
(i) p² + 6p + 8
Find two numbers whose sum = 6 and product = 8 → 2 and 4
= p² + 2p + 4p + 8
= (p² + 2p) + (4p + 8)
= p(p + 2) + 4(p + 2)
= (p + 2)(p + 4)
(ii) q² − 10q + 21
Find two numbers whose sum = -10 and product = 21 → -3 and -7
= q² − 3q − 7q + 21
= (q² − 3q) − (7q − 21)
= q(q − 3) − 7(q − 3)
= (q − 3)(q − 7)
(iii) p² + 6p − 16
Find two numbers whose sum = 6 and product = -16 → 8 and -2
= p² + 8p − 2p − 16
= (p² + 8p) − (2p + 16)
= p(p + 8) − 2(p + 8)
= (p + 8)(p − 2)
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