NCERT Solutions Class 8 Mathematics
Chapter – 14 (Factorisation)
The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 14 Factorisation Exercise 14.2 has been provided here to help the students in solving the questions from this exercise.
Chapter 14: Factorisation
- NCERT Solution Class 8 Maths Ex – 14.1
- NCERT Solution Class 8 Maths Ex – 14.3
- NCERT Solution Class 8 Maths Ex – 14.4
Exercise – 14.2
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
Solution –
(i) a2+8a+16
= a2+2×4×a+42
= (a+4)2
Using the identity (x+y)2 = x2+2xy+y2
(ii) p2–10p+25
= p2-2×5×p+52
= (p-5)2
Using the identity (x-y)2 = x2-2xy+y2
(iii) 25m2+30m+9
= (5m)2+2×5m×3+32
= (5m+3)2
Using the identity (x+y)2 = x2+2xy+y2
(iv) 49y2+84yz+36z2
= (7y)2+2×7y×6z+(6z)2
= (7y+6z)2
Using the identity (x+y)2 = x2+2xy+y2
(v) 4x2–8x+4
= (2x)2-2×4x+22
= (2x-2)2
Using the identity (x-y)2 = x2-2xy+y2
(vi) 121b2-88bc+16c2
= (11b)2-2×11b×4c+(4c)2
= (11b-4c)2
Using the identity (x-y)2 = x2-2xy+y2
(vii) (l+m)2-4lm
Expand (l+m)2 using the identity (x+y)2 = x2+2xy+y2
(l+m)2-4lm = l2+m2+2lm-4lm
= l2+m2-2lm
= (l-m)2
Using the identity (x-y)2 = x2-2xy+y2
(viii) a4+2a2b2+b4
= (a2)2+2×a2×b2+(b2)2
= (a2+b2)2
Using the identity (x+y)2 = x2+2xy+y2
2. Factorise.
(i) 4p2–9q2
(ii) 63a2–112b2
(iii) 49x2–36
(iv) 16x5–144x3 differ
(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)
Using the identity x2-y2 = (x+y)(x-y)
(ii) 63a2–112b2
= 7(9a2 –16b2)
= 7((3a)2–(4b)2)
= 7(3a+4b)(3a-4b)
Using the identity x2-y2 = (x+y)(x-y)
(iii) 49x2–36
= (7x)2 -62
= (7x+6)(7x–6)
Using the identity x2-y2 = (x+y)(x-y)
(iv) 16x5–144x3
= 16x3(x2–9)
= 16x3(x2–9)
= 16x3(x–3)(x+3)
Using the identity x2-y2 = (x+y)(x-y)
(v) (l+m) 2-(l-m) 2
= {(l+m)-(l–m)}{(l +m)+(l–m)}
Using the identity x2-y2 = (x+y)(x-y)
= (l+m–l+m)(l+m+l–m)
= (2m)(2l)
= 4ml
(vi) 9x2y2–16
= (3xy)2-42
= (3xy–4)(3xy+4)
Using the identity x2-y2 = (x+y)(x-y)
(vii) (x2–2xy+y2)–z2
= (x–y)2–z2
Using the identity (x-y)2 = x2-2xy+y2
= {(x–y)–z}{(x–y)+z}
= (x–y–z)(x–y+z)
Using the identity x2-y2 = (x+y)(x-y)
(viii) 25a2–4b2+28bc–49c2
= 25a2–(4b2-28bc+49c2 )
= (5a)2-{(2b)2-2(2b)(7c)+(7c)2}
= (5a)2-(2b-7c)2
Using the identity x2-y2 = (x+y)(x-y) , we have
= (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
Solution –
(i) ax2+bx
= x(ax+b)
(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
= lm+m+l+1
= m(l+1)+(l+1)
= (m+1)(l+1)
(vi) y(y+z)+9(y+z)
= (y+9)(y+z)
(vii) 5y2–20y–8z+2yz
= 5y(y–4)+2z(y–4)
= (y–4)(5y+2z)
(viii) 10ab+4a+5b+2
= 5b(2a+1)+2(2a+1)
= (2a+1)(5b+2)
(ix) 6xy–4y+6–9x
= 6xy–9x–4y+6
= 3x(2y–3)–2(2y–3)
= (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
Solution –
(i) a4–b4
= (a2)2-(b2)2
= (a2-b2) (a2+b2)
= (a – b)(a + b)(a2+b2)
(ii) p4–81
= (p2)2-(9)2
= (p2-9)(p2+9)
= (p2-32)(p2+9)
= (p-3)(p+3)(p2+9)
(iii) x4–(y+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}
= (x–y–z)(x+y+z) {x2+(y+z)2}
(iv) x4–(x–z) 4
= (x2)2-{(x-z)2}2
= {x2-(x-z)2}{x2+(x-z)2}
= { x-(x-z)}{x+(x-z)} {x2+(x-z)2}
= z(2x-z)( x2+x2-2xz+z2)
= z(2x-z)( 2x2-2xz+z2)
(v) a4–2a2b2+b4
= (a2)2-2a2b2+(b2)2
= (a2-b2)2
= ((a–b)(a+b))2
= (a – b)2 (a + b)2
5. Factorise the following expressions.
(i) p2+6p+8
(ii) q2–10q+21
(iii) p2+6p–16
Solution –
(i) p2+6p+8
We observed that 8 = 4×2 and 4+2 = 6
p2+6p+8
= p2+2p+4p+8
= p(p+2)+4(p+2)
= (p+2)(p+4)
This implies that p2+6p+8 = (p+2)(p+4)
(ii) q2–10q+21
We observed that 21 = -7×-3 and -7+(-3) = -10
q2–10q+21
= q2–3q-7q+21
= q(q–3)–7(q–3)
= (q–7)(q–3)
This implies that q2–10q+21 = (q–7)(q–3)
(iii) p2+6p–16
We observed that -16 = -2×8 and 8+(-2) = 6
p2+6p–16
= p2–2p+8p–16
= p(p–2)+8(p–2)
= (p+8)(p–2)
This implies that p2+6p–16 = (p+8)(p–2)
