NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.2 include all questions and answers listed under the exercise. NCERT Solutions cover each concept thoroughly using simple language. It is essential that Class 8 students practise the concepts learned as much as possible. Without practising the CBSE Class 8 NCERT solutions, students may not be exam ready, which is essential. Therefore, the NCERT solutions act as a great resource for scoring well in the exam.
NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.2
Access Other Exercise Solutions of Class 8 Maths Chapter 14 Factorisation
Exercise 14.1 Solutions: 3 Questions (Short answer type)
Exercise 14.3 Solutions: 5 Questions (Short answer type)
Exercise 14.4 Solutions: 21 Questions (Short answer type)
Access Answers to NCERT Class 8 Maths Chapter 14 Factorisation 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 (Hint: Expand (l+m)2 first)
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)
= 4 ml
(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 can be written as p2+2p+4p+8
Taking Common terms, we get
p2+6p+8 = p2+2p+4p+8 = p(p+2)+4(p+2)
Again p+2 is common in both terms.
= (p+2)(p+4)
This implies, p2+6p+8 = (p+2)(p+4)
(ii) q2–10q+21
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 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)
So, p2+6p–16 = (p+8)(p–2)
Also, explore –Â
NCERT Solutions for Class 8 Maths Chapter 14
i like byju’s very much when i am in trouble so i open byju’s an start learning from it
good
It’s more useful for me thnk u to give knowledge to me
Very good
It’s more useful for me
if i am struggling to do maths it is helpful
It is very helpful as i do not go any kind of tutions or classes and byjus really helps me a lot i suggest my friends too to join byjus and i score good marks too by your all subs recorded session ans your solutions really it’s very helpful for me and i understand that how much effort you take to make students champions