NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.2

NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.2

NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.2

?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) 4×2–8x+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 identity: (x+y)2 = x2+2xy+y2

(ii) p2–10p+25
= p2-2×5×p+52
= (p-5)2
Using identity: (x-y)2 = x2-2xy+y2

(iii) 25m2+30m+9
= (5m)2+2×5m×3+32
= (5m+3)2
Using identity: (x+y)2 = x2+2xy+y2

(iv) 49y2+84yz+36z2
=(7y)2+2×7y×6z+(6z)2
= (7y+6z)2
Using identity: (x+y)2 = x2+2xy+y2

(v) 4×2–8x+4
= (2x)2-2×4x+22
= (2x-2)2
Using identity: (x-y)2 = x2-2xy+y2

(vi) 121b2-88bc+16c2
= (11b)2-2×11b×4c+(4c)2
= (11b-4c)2
Using identity: (x-y)2 = x2-2xy+y2

(vii) (l+m)2-4lm (Hint: Expand (l+m)2 first)
Expand (l+m)2 using identity: (x+y)2 = x2+2xy+y2
(l+m)2-4lm = l2+m2+2lm-4lm
= l2+m2-2lm
= (l-m)2
Using identity: (x-y)2 = x2-2xy+y2

(viii) a4+2a2b2+b4
= (a2)2+2×a2×b2+(b2)2
= (a2+b2)2
Using identity: (x+y)2 = x2+2xy+y2

2. Factorise.

(i) 4p2–9q2
(ii) 63a2–112b2
(iii) 49×2–36
(iv) 16×5–144×3 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 identity: x2-y2 = (x+y)(x-y)

(ii) 63a2–112b2
= 7(9a2 –16b2)
= 7((3a)2–(4b)2)
= 7(3a+4b)(3a-4b)
Using identity: x2-y2 = (x+y)(x-y)

(iii) 49×2–36
= (7x)2 -62
= (7x+6)(7x–6)
Using identity: x2-y2 = (x+y)(x-y)

(iv) 16×5–144×3
= 16×3(x2–9)
= 16×3(x2–9)
= 16×3(x–3)(x+3)
Using identity: x2-y2 = (x+y)(x-y)

(v) (l+m) 2-(l-m) 2
= {(l+m)-(l–m)}{(l +m)+(l–m)}
Using 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 Identity: x2-y2 = (x+y)(x-y)

(vii) (x2–2xy+y2)–z2
= (x–y)2–z2
Using Identity: (x-y)2 = x2-2xy+y2
= {(x–y)–z}{(x–y)+z}
= (x–y–z)(x–y+z)
Using 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 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) 2×3+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) 2×3+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)( 2×2-2xz+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 the 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)