Class 8 | NCERT Solution Maths Chapter 14 | Factorisation | Exercise 14.2

Exercise 14.2 Page No: 223

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 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) 4x2–8x+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) 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 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) 49x2–36

= (7a)2 -62

= (7a+6)(7a–6)

Using identity: x2-y2 = (x+y)(x-y)

(iv) 16x5–144x3

= 16x3(x2–9)

= 16x3(x2–9)

= 16x3(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) 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

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)