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## Download PDF of NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

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## Exercise 14.1 Page No: 208

**1. Find the common factors of the given terms.**

** (i) 12x, 36 **

**(ii) 2y, 22xy**

**(iii) 14 pq, 28p ^{2}q^{2}**

**(iv) 2x, 3x ^{2}, 4**

**(v) 6 abc, 24ab ^{2}, 12a^{2}b **

**(vi) 16 x ^{3}, – 4x^{2} , 32 x**

**(vii) 10 pq, 20qr, 30 rp**

**(viii) 3x ^{2}y^{3} , 10x^{3}y^{2} , 6x^{2}y^{2}z**

**Solution: **

(i) Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

Common factors of 12x and 36 are 2, 2, 3

and , 2×2×3 = 12

(ii) Factors of 2y and 22xy

2y = 2×y

22xy = 2×11×x×y

Common factors of 2y and 22xy are 2, y

and ,2×y = 2y

(iii) Factors of 14pq and 28p^{2}q

14pq = 2x7xpxq

28p^{2}q = 2x2x7xpxpxq

Common factors of 14 pq and 28 p^{2}q are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x^{2}and 4

2x = 2×x

3x^{2}= 3×x×x

4 = 2×2

Common factors of 2x, 3x^{2 }and 4 is 1.

(v) Factors of 6abc, 24ab^{2} and 12a^{2}b

6abc = 2×3×a×b×c

24ab^{2} = 2×2×2×3×a×b×b

12 a^{2 }b = 2×2×3×a×a×b

Common factors of 6 abc, 24ab^{2} and 12a^{2}b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) Factors of 16x^{3 }, -4x^{2}and 32x

16 x^{3 }= 2×2×2×2×x×x×x

– 4x^{2} = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x^{3 }, – 4x^{2 }and 32x are 2,2, x

and, 2×2×x = 4x

(vii) Factors of 10 pq, 20qr and 30rp

10 pq = 2×5×p×q

20qr = 2×2×5×q×r

30rp= 2×3×5×r×p

Common factors of 10 pq, 20qr and 30rp are 2, 5

and, 2×5 = 10

(viii) Factors of 3x^{2}y^{3} , 10x^{3}y^{2} and 6x^{2}y^{2}z

3x^{2}y^{3} = 3×x×x×y×y×y

10x^{3 }y^{2} = 2×5×x×x×x×y×y

6x^{2}y^{2}z = 3×2×x×x×y×y×z

Common factors of 3x^{2}y^{3}, 10x^{3}y^{2} and 6x^{2}y^{2}z are x^{2}, y^{2}

and, x^{2}×y^{2} = x^{2}y^{2}

** 2.Factorise the following expressions**

**(i) 7x–42 **

**(ii) 6p–12q **

**(iii) 7a ^{2}+ 14a**

**(iv) -16z+20 z ^{3}**

**(v) 20l ^{2}m+30alm**

**(vi) 5x ^{2}y-15xy^{2}**

**(vii) 10a ^{2}-15b^{2}+20c^{2}**

**(viii) -4a ^{2}+4ab–4 ca**

**(ix) x ^{2}yz+xy^{2}z +xyz^{2}**

**(x) ax ^{2}y+bxy^{2}+cxyz**

**Solution: **

(vii) 10a^{2}-15b^{2}+20c^{2}

10a^{2 }= 2×5×a×a

– 15b^{2 }= -1×3×5×b×b

20c^{2} = 2×2×5×c×c

Common factor of 10 a^{2} , 15b^{2} and 20c^{2} is 5

10a^{2}-15b^{2}+20c^{2} = 5(2a^{2}-3b^{2}+4c^{2} )

(viii) – 4a^{2}+4ab-4ca

– 4a^{2} = -1×2×2×a×a

4ab = 2×2×a×b

– 4ca = -1×2×2×c×a

Common factor of – 4a^{2} , 4ab , – 4ca are 2, 2, a i.e. 4a

So,

– 4a^{2}+4 ab-4 ca = 4a(-a+b-c)

(ix) x^{2}yz+xy^{2}z+xyz^{2}

x^{2}yz = x×x×y×z

xy^{2}z = x×y×y×z

xyz^{2} = x×y×z×z

Common factor of x^{2}yz , xy^{2}z and xyz^{2} are x, y, z i.e. xyz

Now, x^{2}yz+xy^{2}z+xyz^{2 }= xyz(x+y+z)

(x) ax^{2}y+bxy^{2}+cxyz

ax^{2}y = a×x×x×y

bxy^{2 }= b×x×y×y

cxyz = c×x×y×z

Common factors of a x^{2}y ,bxy^{2 }and cxyz are xy

Now, ax^{2}y+bxy^{2}+cxyz = xy(ax+by+cz)

**3. Factorise.**

**(i) x ^{2}+xy+8x+8y**

**(ii) 15xy–6x+5y–2**

**(iii) ax+bx–ay–by**

**(iv) 15pq+15+9q+25p**

**(v) z–7+7xy–xyz**

**Solution**:

## Exercise 14.2 Page No: 223

**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

= a^{2}+2×4×a+4^{2}

= (a+4)^{2}

Using identity: (x+y)^{2} = x^{2}+2xy+y^{2}

(ii) p^{2}–10p+25

= p^{2}-2×5×p+5^{2}

= (p-5)^{2}

Using identity: (x-y)^{2} = x^{2}-2xy+y^{2}

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

= (5m)^{2}-2×5m×3+3^{2}

= (5m+3)^{2}

Using identity: (x+y)^{2} = x^{2}+2xy+y^{2}

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

=(7y)^{2}+2×7y×6z+(6z)^{2}

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

Using identity: (x+y)^{2} = x^{2}+2xy+y^{2}

(v) 4x^{2}–8x+4

= (2x)^{2}-2×4x+2^{2}

= (2x-2)^{2}

Using identity: (x-y)^{2} = x^{2}-2xy+y^{2}

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

= (11b)^{2}-2×11b×4c+(4c)^{2}

= (11b-4c)^{2}

Using identity: (x-y)^{2} = x^{2}-2xy+y^{2}

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

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

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

= l^{2}+m^{2}-2lm

= (l-m)^{2}

Using identity: (x-y)^{2} = x^{2}-2xy+y^{2}

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

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

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

Using identity: (x+y)^{2} = x^{2}+2xy+y^{2}

**2. Factorise.**

**(i) 4p ^{2}–9q^{2}**

**(ii) 63a ^{2}–112b^{2}**

**(iii) 49x ^{2}–36**

**(iv) 16x ^{5}–144x^{3} differ**

**(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}

= (2p)^{2}-(3q)^{2}

= (2p-3q)(2p+3q)

Using identity: x^{2}-y^{2} = (x+y)(x-y)

(ii) 63a^{2}–112b^{2}

= 7(9a^{2} –16b^{2})

= 7((3a)^{2}–(4b)^{2})

= 7(3a+4b)(3a-4b)

Using identity: x^{2}-y^{2} = (x+y)(x-y)

(iii) 49x^{2}–36

= (7a)^{2} -6^{2}

= (7a+6)(7a–6)

Using identity: x^{2}-y^{2} = (x+y)(x-y)

(iv) 16x^{5}–144x^{3}

= 16x^{3}(x^{2}–9)

= 16x^{3}(x^{2}–9)

= 16x^{3}(x–3)(x+3)

Using identity: x^{2}-y^{2} = (x+y)(x-y)

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

= {(l+m)-(l–m)}{(l +m)+(l–m)}

Using Identity: x^{2}-y^{2} = (x+y)(x-y)

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

= 4 ml

(vi) 9x^{2}y^{2}–16

= (3xy)^{2}-4^{2}

= (3xy–4)(3xy+4)

Using Identity: x^{2}-y^{2} = (x+y)(x-y)

(vii) (x^{2}–2xy+y^{2})–z^{2 }

= (x–y)^{2}–z^{2}

Using Identity: (x-y)^{2} = x^{2}-2xy+y^{2}

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using Identity: x^{2}-y^{2} = (x+y)(x-y)

(viii) 25a^{2}–4b^{2}+28bc–49c^{2}

= 25a^{2}–(4b^{2}-28bc+49c^{2} )

= (5a)^{2}-{(2b)^{2}-2(2b)(7c)+(7c)^{2}}

= (5a)^{2}-(2b-7c)^{2}

Using Identity: x^{2}-y^{2} = (x+y)(x-y) , we have

= (5a+2b-7c)(5a-2b-7c)

**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 = x(ax+b)

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

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

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

(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) 5y^{2}–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) 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}

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

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

= (a – b)(a + b)(a^{2}+b^{2})

(ii) p^{4}–81

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

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

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

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

(iii) x^{4}–(y+z)^{ 4} = (x^{2})^{2}-[(y+z)^{2}]^{2 }

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

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

= (x–y–z)(x+y+z) {x^{2}+(y+z)^{2}}

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

= {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})

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

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

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

= ((a–b)(a+b))^{2}

**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

We observed that, 8 = 4×2 and 4+2 = 6

p^{2}+6p+8 can be written as p^{2}+2p+4p+8

Taking Common terms, we get

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

Again p+2 is common in both the terms.

= (p+2)(p+4)

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

(ii) q^{2}–10q+21

Observed that, 21 = -7×-3 and -7+(-3) = -10

q^{2}–10q+21 = q^{2}–3q-7q+21

= q(q–3)–7(q–3)

= (q–7)(q–3)

This implies q^{2}–10q+21 = (q–7)(q–3)

(iii) p^{2}+6p–16

We observed that, -16 = -2×8 and 8+(-2) = 6

p^{2}+6p–16 = p^{2}–2p+8p–16

= p(p–2)+8(p–2)

= (p+8)(p–2)

So, p^{2}+6p–16 = (p+8)(p–2)

## Exercise 14.3 Page No: 227

**1. Carry out the following divisions.**

**(i) 28x ^{4} ÷ 56x**

**(ii) –36y ^{3} ÷ 9y^{2}**

**(iii) 66pq ^{2}r^{3} ÷ 11qr^{2}**

**(iv) 34x ^{3}y^{3}z^{3} ÷ 51xy^{2}z^{3}**

**(v) 12a ^{8}b^{8} ÷ (– 6a^{6}b^{4})**

**Solution: **

(i)28x^{4} = 2×2×7×x×x×x×x

56x = 2×2×2×7×x

**2. Divide the given polynomial by the given monomial.**

**(i)(5x ^{2}–6x) ÷ 3x**

**(ii)(3y ^{8}–4y^{6}+5y^{4}) ÷ y^{4}**

**(iii) 8(x ^{3}y^{2}z^{2}+x^{2}y^{3}z^{2}+x^{2}y^{2}z^{3})÷ 4x^{2 }y^{2 }z^{2}**

**(iv)(x ^{3}+2x^{2}+3x) ÷2x**

**(v) (p ^{3}q^{6}–p^{6}q^{3}) ÷ p^{3}q^{3}**

**Solution: **

**3. Work out the following divisions.**

**(i) (10x–25) ÷ 5**

**(ii) (10x–25) ÷ (2x–5)**

**(iii) 10y(6y+21) ÷ 5(2y+7)**

**(iv) 9x ^{2}y^{2}(3z–24) ÷ 27xy(z–8)**

**(v) 96abc(3a–12)(5b–30) ÷ 144(a–4)(b–6)**

**Solution: **

(i) (10x–25) ÷ 5 = 5(2x-5)/5 = 2x-5

(ii) (10x–25) ÷ (2x–5) = 5(2x-5)/( 2x-5) = 5

(iii) 10y(6y+21) ÷ 5(2y+7) = 10y×3(2y+7)/5(2y+7) = 6y

(iv) 9x^{2}y^{2}(3z–24) ÷ 27xy(z–8) = 9x^{2}y^{2}×3(z-8)/27xy(z-8) = xy

**4. Divide as directed.**

**(i) 5(2x+1)(3x+5)÷ (2x+1)**

**(ii) 26xy(x+5)(y–4)÷13x(y–4)**

**(iii) 52pqr(p+q)(q+r)(r+p) ÷ 104pq(q+r)(r+p)**

**(iv) 20(y+4) (y ^{2}+5y+3) ÷ 5(y+4)**

**(v) x(x+1) (x+2)(x+3) ÷ x(x+1)**

**Solution: **

**5. Factorise the expressions and divide them as directed.**

**(i) (y ^{2}+7y+10)÷(y+5)**

**(ii) (m ^{2}–14m–32)÷(m+2)**

**(iii) (5p ^{2}–25p+20)÷(p–1)**

**(iv) 4yz(z ^{2}+6z–16)÷2y(z+8)**

**(v) 5pq(p ^{2}–q^{2})÷2p(p+q)**

**(vi) 12xy(9x ^{2}–16y^{2})÷4xy(3x+4y)**

**(vii) 39y ^{3}(50y^{2}–98) ÷ 26y^{2}(5y+7)**

**Solution: **

**(i) (y ^{2}+7y+10)÷(y+5)**

First solve for equation, (y^{2}+7y+10)

(y^{2}+7y+10) = y^{2}+2y+5y+10 = y(y+2)+5(y+2) = (y+2)(y+5)

Now, (y^{2}+7y+10)÷(y+5) = (y+2)(y+5)/(y+5) = y+2

**(ii) (m ^{2}–14m–32)÷ (m+2)**

Solve for m^{2}–14m–32, we have

m^{2}–14m–32 = m^{2}+2m-16m–32 = m(m+2)–16(m+2) = (m–16)(m+2)

Now, (m^{2}–14m–32)÷(m+2) = (m–16)(m+2)/(m+2) = m-16

**(iii) (5p ^{2}–25p+20)÷(p–1)**

Step 1: Take 5 common from the equation, 5p^{2}–25p+20, we get

5p^{2}–25p+20 = 5(p^{2}–5p+4)

Step 2: Factorize p^{2}–5p+4

p^{2}–5p+4 = p^{2}–p-4p+4 = (p–1)(p–4)

Step 3: Solve original equation

(5p^{2}–25p+20)÷(p–1) = 5(p–1)(p–4)/(p-1) = 5(p–4)

**(iv) 4yz(z ^{2} + 6z–16)÷ 2y(z+8)**

Factorize z^{2}+6z–16,

z^{2}+6z–16 = z^{2}-2z+8z–16 = (z–2)(z+8)

Now, 4yz(z^{2}+6z–16) ÷ 2y(z+8) = 4yz(z–2)(z+8)/2y(z+8) = 2z(z-2)

**(v) 5pq(p ^{2}–q^{2}) ÷ 2p(p+q)**

p^{2}–q^{2} can be written as (p–q)(p+q) using identity.

5pq(p^{2}–q^{2}) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5/2q(p–q)

**(vi) 12xy(9x ^{2}–16y^{2}) ÷ 4xy(3x+4y)**

Factorize 9x^{2}–16y^{2} , we have

9x^{2}–16y^{2} = (3x)^{2}–(4y)^{2} = (3x+4y)(3x-4y) using identity: p^{2}–q^{2} = (p–q)(p+q)

Now, 12xy(9x^{2}–16y^{2}) ÷ 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)

**(vii) 39y ^{3}(50y^{2}–98) ÷ 26y^{2}(5y+7)**

st solve for 50y

^{2}–98, we have

50y^{2}–98 = 2(25y^{2}–49) = 2((5y)^{2}–7^{2}) = 2(5y–7)(5y+7)

Now, 39y^{3}(50y^{2}–98) ÷ 26y^{2}(5y+7) =

## Exercise 14.4 Page No: 228

**1. 4(x–5) = 4x–5**

**Solution: **

4(x- 5)= 4x – 20 ≠ 4x – 5 = RHS

The correct statement is 4(x-5) = 4x–20

**2. x(3x+2) = 3x ^{2}+2**

**Solution: **

LHS = x(3x+2) = 3x^{2}+2x ≠ 3x^{2}+2 = RHS

The correct solution is x(3x+2) = 3x^{2}+2x

**3. 2x+3y = 5xy**

**Solution: **

LHS= 2x+3y ≠ R. H. S

The correct statement is 2x+3y = 2x+3 y

**4. x+2x+3x = 5x**

**Solution:**

LHS = x+2x+3x = 6x ≠ RHS

The correct statement is x+2x+3x = 6x

**5. 5y+2y+y–7y = 0**

**Solution: **

LHS = 5y+2y+y–7y = y ≠ RHS

The correct statement is 5y+2y+y–7y = y

**6. 3x+2x = 5x ^{2}**

**Solution:**

LHS = 3x+2x = 5x ≠ RHS

The correct statement is 3x+2x = 5x

**7. (2x) ^{ 2}+4(2x)+7 = 2x^{2}+8x+7 **

**Solution: **

LHS = (2x)^{ 2}+4(2x)+7 = 4x^{2}+8x+7 ≠ RHS

The correct statement is (2x)^{ 2}+4(2x)+7 = 4x^{2}+8x+7

**8. (2x) ^{ 2}+5x = 4x+5x = 9x**

**Solution: **

LHS = (2x)^{ 2}+5x = 4x^{2}+5x ≠ 9x = RHS

The correct statement is(2x)^{ 2}+5x = 4x^{2}+5x

**9. (3x + 2) ^{ 2} = 3x^{2}+6x+4 **

**Solution:**

LHS = (3x+2)^{ 2} = (3x)^{2}+2^{2}+2x2x3x = 9x^{2}+4+12x ≠ RHS

The correct statement is (3x + 2)^{ 2} = 9x^{2}+4+12x

**10. Substituting x = – 3 in**

**(a) x ^{2} + 5x + 4 gives (– 3)^{ 2}+5(– 3)+4 = 9+2+4 = 15**

**(b) x ^{2} – 5x + 4 gives (– 3)^{ 2}– 5( – 3)+4 = 9–15+4 = – 2**

**(c) x ^{2} + 5x gives (– 3)^{ 2}+5(–3) = – 9–15 = – 24**

**Solution: **

(a) Substituting x = – 3 in x^{2}+5x+4, we have

x^{2}+5x+4 = (– 3)^{ 2}+5(– 3)+4 = 9–15+4 = – 2. This is the correct answer.

(b) Substituting x = – 3 in x^{2}–5x+4

x^{2}–5x+4 = (–3)^{ 2}–5(– 3)+4 = 9+15+4 = 28. This is the correct answer

(c)Substituting x = – 3 in x^{2}+5x

x^{2}+5x = (– 3)^{ 2}+5(–3) = 9–15 = -6. This is the correct answer

**11.(y–3) ^{2} = y^{2}–9**

**Solution:**

LHS = (y–3)^{2} , which is similar to (a–b)^{2} identity, where (a–b)^{ 2} = a^{2}+b^{2}-2ab.

(y – 3)^{2} = y^{2}+(3)^{ 2}–2y×3 = y^{2}+9 –6y ≠ y^{2} – 9 = RHS

The correct statement is (y–3)^{2} = y^{2} + 9 – 6y

**12. (z+5) ^{ 2} = z^{2}+25**

**Solution: **

LHS = (z+5)^{2} , which is similar to (a +b)^{2} identity, where (a+b)^{ 2} = a^{2}+b^{2}+2ab.

(z+5)^{ 2} = z^{2}+5^{2}+2×5×z = z^{2}+25+10z ≠ z^{2}+25 = RHS

The correct statement is (z+5)^{ 2} = z^{2}+25+10z

**13. (2a+3b)(a–b) = 2a ^{2}–3b^{2}**

**Solution: **

LHS = (2a+3b)(a–b) = 2a(a–b)+3b(a–b)

= 2a^{2}–2ab+3ab–3b^{2}

= 2a^{2}+ab–3b^{2}

≠ 2a^{2}–3b^{2} = RHS

The correct statement is (2a +3b)(a –b) = 2a^{2}+ab–3b^{2}

**14. (a+4)(a+2) = a ^{2}+8**

**Solution: **

LHS = (a+4)(a+2) = a(a+2)+4(a+2)

= a^{2}+2a+4a+8

= a^{2}+6a+8

≠ a^{2}+8 = RHS

The correct statement is (a+4)(a+2) = a^{2}+6a+8

**15. (a–4)(a–2) = a ^{2}–8**

**Solution: **

LHS = (a–4)(a–2) = a(a–2)–4(a–2)

= a^{2}–2a–4a+8

= a^{2}–6a+8

≠ a^{2}-8 = RHS

The correct statement is (a–4)(a–2) = a^{2}–6a+8

**16. 3x ^{2}/3x^{2} = 0**

**Solution: **

LHS = 3x^{2}/3x^{2} = 1 ≠ 0 = RHS

The correct statement is 3x^{2}/3x^{2} = 1

**17.** **(3x ^{2}+1)/3x^{2} = 1 + 1 = 2**

**Solution: **

LHS = (3x^{2}+1)/3x^{2} = (3x^{2}/3x^{2})+(1/3x^{2}) = 1+(1/3x^{2}) ≠ 2 = RHS

The correct statement is (3x^{2}+1)/3x^{2} = 1+(1/3x^{2})

**18. 3x/(3x+2) = ½ **

**Solution: **

LHS = 3x/(3x+2) ≠ 1/2 = RHS

The correct statement is 3x/(3x+2)** = **3x/(3x+2)

**19. 3/(4x+3) = 1/4x**

**Solution: **

LHS = 3/(4x+3) ≠ 1/4x

The correct statement is 3/(4x+3) = 3/(4x+3)

**20. (4x+5)/4x = 5**

**Solution:**

LHS = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠ 5 = RHS

The correct statement is (4x+5)/4x = 1 + (5/4x)

Solution:

LHS = (7x+5)/5 = (7x/5)+ 5/5 = (7x/5)+1 ≠ 7x = RHS

The correct statement is (7x+5)/5 = (7x/5) +1

Also Access |

CBSE Notes for Class 8 Maths Chapter 14 |

**NCERT Solutions for Class 8 Maths Chapter 14 Factorisation**

Class 8 NCERT exercise wise questions and answers will help students frame a perfect solution in the Maths Exam. These exercise questions can provide students with a topic wise preparation strategy. Some of the important topics introduced in Class 8 NCERT Solutions Maths are Factorisation, Factors of natural numbers, Factors of algebraic expressions and Division of Algebraic Expressions.

### NCERT Solutions For Class 8 Maths Chapter 14 Exercises:

Get detailed solutions for all the questions listed under the below exercises:

Exercise 14.1 Solutions: 3 Questions (Short answer type)

Exercise 14.2 Solutions: 5 Questions (Short answer type)

Exercise 14.3 Solutions: 5 Questions (Short answer type)

Exercise 14.4 Solutions: 21 Questions (Short answer type)

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

NCERT Class 8 Maths Chapter 14, deals primarily with the factorisation of the numbers and algebraic expressions using algebraic identities. Students will also learn about, Division of Algebraic Expressions, Division of a monomial by another monomial, Division of a polynomial by a monomial and Division of Polynomial by Polynomial.

**The main topics covered in this chapter include:**

Exercise | Topic |

14.1 | Introduction |

14.2 | What is Factorisation? |

14.3 | Division of Algebraic Expressions |

14.4 | Division of Algebraic Expressions |

14.5 | Can you Find the Error? |

Key Features of NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

- These NCERT solutions help students in understanding the concepts clearly.
- Simple and precise language is used to explain the topics.
- All concepts have been explained in detail.
- Subject experts have consolidated all exercise questions at one place for practice.
- NCERT Solutions are helpful for the preparation of competitive exams.

## Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 14

### Are NCERT Solutions for Class 8 Maths Chapter 14 important from the exam point of view?

### How many exercises are there in NCERT Solutions for Class 8 Maths Chapter 14?

### What are the main topics covered in the NCERT Solutions for Class 8 Maths Chapter 14?

14.1 – Introduction

14.2 – Definition of Factorisation

14.3 – Division of Algebraic Expressions

14.4 – Division of Algebraic Expressions

14.5 – Can you Find the Error?