# Class 8 NCERT Solutions – Chapter 14 Factorisation – Exercise 14.2

**Question 1: **Factorise the following expressions.

### (i) a^{2}+8a+16

### (ii) p^{2}–10p+25

### (iii) 25m^{2}+30m+9

### (iv) 49y^{2}+84yz+36z^{2}

### (v) 4x^{2}–8x+4

### (vi) 121b^{2}–88bc+16c^{2}

### (vii) (l+m)^{2}–4lm

### (Hint: Expand (l+m)^{2} first)

### (viii) a^{4}+2a^{2}b^{2}+b^{4}

**Solution:**

(i) a^{2}+8a+16

Ans:Given: a

^{2}+8a+16Since, 8a and 16 can be substituted by 2×4×a and 4

^{2}respectively we get,= a

^{2}+2×4×a+4^{2}Therefore, by using the identity :

(x+y)^{2}= x^{2}+2xy+y^{2}a

^{2}+8a+16 = (a+4)^{2}

(ii) p^{2}–10p+25

Ans:Given: p

^{2}–10p+25Since, 10p and 25 can be substituted by 2×5×p and 5

^{2}respectively we get,= p

^{2}-2×5×p+5^{2}Therefore, by using the identity : (x-y)

^{2}= x^{2}-2xy+y^{2}p

^{2}–10p+25 = (p-5)^{2}

(iii)25m^{2}+30m+9

Ans:Given: 25m

^{2}+30m+9Since, 25m

^{2 }, 30m and 9 can be substituted by (5m)^{2}, 2×5m×3 and 3^{2}respectively we get,= (5m)

^{2 }+ 2×5m×3 + 3^{2}Therefore, by using the identity: (x+y)

^{2 }= x^{2}+2xy+y^{2}25m

^{2}+30m+9 = (5m+3)^{2}

(iv)49y^{2}+84yz+36z^{2}

Ans:Given: 49y

^{2}+84yz+36z^{2 }Since, 49y

^{2}, 84yz and 36z^{2}can be substituted by (7y)^{2}, 2×7y×6z and (6z)^{2}respectively we get,=(7y)

^{2}+2×7y×6z+(6z)^{2}Therefore, by using the identity: (x+y)

^{2}= x^{2}+2xy+y^{2}49y

^{2}+84yz+36z^{2}= (7y+6z)^{2}

(v)4x^{2}–8x+4

Ans:Given: 4x

^{2}–8x+4Since, 4x

^{2}, 8x and 4 can be substituted by (2x)^{2}, 2×4x and 2^{2}respectively we get,= (2x)

^{2}-2×4x+2^{2}Therefore, by using the identity: (x-y)

^{2}= x^{2}-2xy+y^{2}4x

^{2}–8x+4 = (2x-2)^{2}

(vi)121b^{2}-88bc+16c^{2}

Ans:Given: 121b

^{2}-88bc+16c^{2}Since, 121b

^{2}, 88bc and 16c^{2}can be substituted by (11b)^{2}, 2×11b×4c and (4c)^{2}respectively we get,= (11b)

^{2}-2×11b×4c+(4c)^{2}Therefore, by using the identity: (x-y)

^{2}= x^{2}-2xy+y^{2}121b

^{2}-88bc+16c^{2}= (11b-4c)^{2}

(vii)(l+m)^{2}-4lm (Hint: Expand (l+m)^{2}_{ }first)

Ans:Given: (l+m)

^{2}-4lmBy expanding (l+m)

^{2}using identity: (x+y)^{2}= x^{2}+2xy+y^{2}, we get,(l+m)

^{2}-4lm = l^{2}+m^{2}+2lm-4lm(l+m)

^{2}-4lm = l^{2}+m^{2}-2lm = l^{2}-2lm+m^{2}Therefore, by using the identity: (x-y)

^{2}= x^{2}-2xy+y^{2}(l+m)

^{2}-4lm = (l-m)^{2}

(viii)a^{4}+2a^{2}b^{2}+b^{4}

Ans:Given: a

^{4}+2a^{2}b^{2}+b^{4}Since, a

^{4 }and b^{4}can be substituted by (a^{2})^{2}and (b^{2})^{2}respectively we get,= (a

^{2})^{2}+2×a^{2}×b^{2}+(b^{2})^{2}Therefore, by using the identity: (x+y)

^{2}= x^{2}+2xy+y^{2}a

^{4}+2a^{2}b^{2}+b^{4}= (a^{2}+b^{2})^{2}

**Question 2: **Factorise.

### (i) 4p^{2}–9q^{2}

### (ii) 63a^{2}–112b^{2}

### (iii) 49x^{2}–36

### (iv) 16x^{5}–144x^{3}

### (v) (l+m)^{2}-(l-m)^{2}

### (vi) 9x^{2}y^{2}–16

### (vii) (x^{2}–2xy+y^{2})–z^{2}

### (viii) 25a^{2}–4b^{2}+28bc–49c^{2}

**Solution:**

(i) 4p^{2}–9q^{2}

Ans:Given: 4p

^{2}–9q^{2}Since, 4p

^{2}and 9q^{2}can be substituted by (2p)^{2}and (3q)^{2}respectively we get,= (2p)

^{2}-(3q)^{2}Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y)4p

^{2}–9q^{2}= (2p-3q)(2p+3q)

(ii) 63a^{2}–112b^{2}

Ans:Given: 63a

^{2}–112b^{2}63a

^{2}–112b^{2}= 7(9a^{2}–16b^{2})Since, 9a

^{2}and 16b^{2}can be substituted by (3a)^{2}and (4b)^{2}respectively we get,= 7((3a)

^{2}–(4b)^{2})Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y)= 7(3a+4b)(3a-4b)

(iii)49x^{2}–36

Ans:Given: 49x

^{2}–36Since, 49x

^{2}and 36 can be substituted by (7x)^{2}and 6^{2}respectively we get,= (7x)

^{2}– 6^{2}Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y)49x

^{2}–36 = (7x+6)(7x–6)

(iv)16x^{5}–144x^{3}

Ans:Given: 16x

^{5}– 144x^{3}16x

^{5 }– 144x^{3}= 16x^{3}(x^{2}–9)Since, 9 can be substituted by 3

^{2}respectively we get,= 16x

^{3}(x^{2}–3^{2})Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y)16x

^{5}–144x^{3}= 16x^{3}(x–3)(x+3)

(v) (l+m)^{2 }– (l-m)^{2}

Ans:Given: (l+m)

^{2}– (l-m)^{2 }By expanding (l+m)

^{2}– (l-m)^{2}using identity: x^{2}-y^{2}= (x+y)(x-y) , we get,= {(l+m)-(l–m)}{(l +m)+(l–m)}

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

(l+m)

^{2}– (l-m)^{2}= 4 ml

(vi)9x^{2}y^{2}–16

Ans:Given: 9x

^{2}y^{2}–16Since, 9x

^{2}y^{2}and 16 can be substituted by (3xy)^{2}and 4^{2}respectively we get,= (3xy)

^{2}-4^{2}Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y)9x

^{2}y^{2}–16 = (3xy–4)(3xy+4)

(vii)(x^{2}–2xy+y^{2})–z^{2}

Ans:Given: (x

^{2}–2xy+y^{2})–z^{2}By compressing x

^{2}–2xy+y^{2}using identity: (x-y)^{2}= x^{2}-2xy+y^{2}, we get,(x

^{2}–2xy+y^{2})–z^{2}= (x–y)^{2}–z^{2}Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y)= {(x–y)–z}{(x–y)+z}

(x

^{2}–2xy+y^{2})–z^{2}= (x–y–z)(x–y+z)

(viii) 25a^{2}–4b^{2}+28bc–49c^{2}

Ans:Given: 25a

^{2}–4b^{2}+28bc–49c^{2}25a

^{2}–4b^{2}+28bc–49c^{2}= 25a^{2}–(4b^{2}-28bc+49c^{2})Since, 25a

^{2}, 4b^{2}, 28bc and 49c^{2}can be substituted by (5a)^{2}, (2b)^{2}, 2(2b)(7c) and (7c)^{2}respectively we get,= (5a)

^{2}-{(2b)^{2}-2(2b)(7c)+(7c)^{2}}Therefore, by using the identity: (x-y)

^{2}= x^{2}-2xy+y^{2}= (5a)

^{2}-(2b-7c)^{2}and by using Identity: x

^{2}-y^{2}= (x+y)(x-y) , we get25a

^{2}–4b^{2}+28bc–49c^{2}= (5a+2b-7c)(5a-2b+7c)

**Question 3: **Factorise the expressions.

### (i) ax^{2}+bx

### (ii) 7p^{2}+21q^{2}

### (iii) 2x^{3}+2xy^{2}+2xz^{2}

### (iv) am^{2}+bm^{2}+bn^{2}+an^{2}

### (v) (lm+l)+m+1

### (vi) y(y+z)+9(y+z)

### (vii) 5y^{2}–20y–8z+2yz

### (viii) 10ab+4a+5b+2

### (ix)6xy–4y+6–9x

**Solution:**

(i) ax^{2}+bx

Ans:Given: ax

^{2}+bxTaking x as common, we get

ax

^{2}+bx = x(ax+b)

(ii) 7p^{2}+21q^{2 }

Ans:Given: 7p

^{2}+21q^{2 }Taking 7 as common, we get,

7p

^{2}+21q^{2}= 7(p^{2}+3q^{2})

(iii) 2x^{3}+2xy^{2}+2xz^{2}

Ans:Given: 2x

^{3}+2xy^{2}+2xz^{2}Taking 2x as common, we get,

2x

^{3}+2xy^{2}+2xz^{2}= 2x(x^{2}+y^{2}+z^{2})

(iv)am^{2}+bm^{2}+bn^{2}+an^{2 }

Ans:Given: am

^{2}+bm^{2}+bn^{2}+an^{2}Taking m2 and n2 as common, we get,

= m

^{2}(a+b)+n^{2}(a+b)Taking (a+b) as common, we get,

am

^{2}+bm^{2}+bn^{2}+an^{2}= (a+b)(m^{2}+n^{2})

(v)(lm+l)+m+1

Ans:Given: (lm+l)+m+1

= lm+m+l+1

Taking m as common, we get,

= m(l+1)+(l+1)

(lm+l)+m+1 = (m+1)(l+1)

(vi)y(y+z)+9(y+z)

Ans:Given: y(y+z)+9(y+z)

Taking (y+z) as common, we get,

y(y+z)+9(y+z) = (y+9)(y+z)

(vii)5y^{2}–20y–8z+2yz

Ans:Given: 5y

^{2}–20y–8z+2yzTaking 5y and 2z as common, we get,

= 5y(y–4)+2z(y–4)

Taking (y-4) as common, we get,

5y

^{2}–20y–8z+2yz = (y–4)(5y+2z)

(viii)10ab+4a+5b+2

Ans:Given: 10ab+4a+5b+2

Taking 5b and 2 as common, we get,

= 5b(2a+1)+2(2a+1)

Taking (2a+1) as common, we get,

10ab+4a+5b+2 = (2a+1)(5b+2)

(ix) 6xy–4y+6–9x

Ans:Given: 6xy–4y+6–9x

= 6xy–9x–4y+6

Taking 3x and 2 as common, we get,

= 3x(2y–3)–2(2y–3)

Taking (2y-3) as common, we get,

6xy–4y+6–9x = (2y–3)(3x–2)

**Question 4: **Factorise.

### (i) a^{4}–b^{4}

### (ii) p^{4}–81

### (iii) x^{4}–(y+z)^{4}

### (iv) x^{4}–(x–z)^{4}

### (v) a^{4}–2a^{2}b^{2}+b^{4}

**Solution:**

(i)a^{4}–b^{4}

Ans:Given: a

^{4}–b^{4}Since, a

^{4}and b^{4}can be substituted by (a^{2})^{2}and (b^{2})^{2}respectively we get,= (a

^{2})^{2}-(b^{2})^{2}Therefore, by using Identity: x

^{2}-y^{2}= (x+y)(x-y) , we get= (a

^{2}-b^{2}) (a^{2}+b^{2})a

^{4}–b^{4}= (a – b)(a + b)(a^{2}+b^{2})

(ii) p^{4}–81

Ans:Given: p

^{4}–81Since, p

^{4}and 81 can be substituted by (p^{2})^{2}and (9)^{2}respectively we get,= (p

^{2})^{2}-(9)^{2}Therefore, by using Identity: x

^{2}-y^{2}= (x+y)(x-y) , we get= (p

^{2}-9)(p^{2}+9)= (p

^{2}-3^{2})(p^{2}+9)p

^{4}–81 =(p-3)(p+3)(p^{2}+9)

(iii)x^{4}–(y+z)^{4}

Ans:Given: x

^{4}–(y+z)^{4}Since, x

^{4 }and (y+z)^{4}can be substituted by (x^{2})^{2}and [(y+z)^{2}]^{2}respectively we get,= (x

^{2})^{2}-[(y+z)^{2}]^{2}Therefore, by using the identity: x

^{2}-y^{2}= (x+y)(x-y) , we get= {x

^{2}-(y+z)^{2}}{ x^{2}+(y+z)^{2}}= {(x –(y+z)(x+(y+z)}{x

^{2}+(y+z)^{2}}x

^{4}–(y+z)^{4}= (x–y–z)(x+y+z) {x^{2}+(y+z)^{2}}

(iv) x^{4}–(x–z)^{4}

Ans:Given: x

^{4}–(x–z)^{4}Since, x

^{4}and (x-z)^{4}can be substituted by (x^{2})^{2 }and [(x-z)^{2}]^{2}respectively we get,= (x

^{2})^{2}-{(x-z)^{2}}^{2}By using Identity: x

^{2}-y^{2}= (x+y)(x-y) , we get= {x

^{2}-(x-z)^{2}}{x^{2}+(x-z)^{2}}= { x-(x-z)}{x+(x-z)} {x

^{2}+(x-z)^{2}}= z(2x-z)( x

^{2}+x^{2}-2xz+z^{2})x

^{4}–(x–z)^{4}= z(2x-z)( 2x^{2}-2xz+z^{2})

(v)a^{4}–2a^{2}b^{2}+b^{4}

Ans:Given: a

^{4}–2a^{2}b^{2}+b^{4}Since, a

^{4}and b^{4}can be substituted by (a^{2})^{2}and (b^{2})^{2}respectively we get,= (a

^{2})^{2}-2a^{2}b^{2}+(b^{2})^{2}Therefore, by using the identity: (x-y)

^{2}= x^{2}-2xy+y^{2}= (a

^{2}-b^{2})^{2}And by using Identity: x

^{2}-y^{2}= (x+y)(x-y) , we geta

^{4}–2a^{2}b^{2}+b^{4}= ((a–b)(a+b))^{2}

**Question 5.** Factorise the following expressions.

### (i) p^{2}+6p+8

### (ii) q^{2}–10q+21

### (iii) p^{2}+6p–16

**Solution:**

(i) p^{2}+6p+8

Ans:Given: p

^{2}+6p+8Since, 6p and 8 can be substituted by 2p+4p and 4×2 respectively we get,

= p

^{2}+2p+4p+8Taking p and 4 common terms, we get

= p(p+2)+4(p+2)

Again taking (p+2) as common, we get

= (p+2)(p+4)

p

^{2}+6p+8 = (p+2)(p+4)

(ii)q^{2}–10q+21

Ans:Given: q

^{2}–10q+21Since, 10q and 21 can be substituted by (-3q)+(-7q) and (-3)×(-7) respectively we get,

= q

^{2}–3q-7q+21Taking q and 7 common terms, we get

= q(q–3)–7(q–3)

Again taking (q–3) as common, we get

= (q–7)(q–3)

q

^{2}–10q+21 = (q–7)(q–3)

(iii) p^{2}+6p–16

Ans:Given: p

^{2}+6p+16Since, 6p and 16 can be substituted by 8p +(-2p) and (-2)×8 respectively we get,

= p

^{2}–2p+8p–16Taking p and 8 common terms, we get

= p(p–2)+8(p–2)

Again taking (p-2) as common, we get

= (p+8)(p–2)

p

^{2}+6p–16 = (p+8)(p–2)

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