NCERT Solutions for Class 8 Maths Chapter 14 Factorization

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Written by Team Trustudies
Updated at 2021-02-11


NCERT solutions for class 8 Maths Chapter 14 Factorization Exercise 14.1

1.Find the common factors of the given terms.
(i) \(12x, 36\)

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

(iii)\( 14pq, 28p^2q^2\)

(iv)\( 2x, 3x^2, 4\)

(v) \(6abc, 24ab^2, 12a^2b\)

(vi) \(16x^3, -4x^2, 32x\)

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

(viii) \(3x^2y^3, 10x^3y^2, 6x^2y^2z\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i) Given:12x, 36

\(=(2 \times 2 \times 3 \times x) and (2 \times 2 \times3 \times 3)\)

So we have the common factors are \(2 \times 2 \times 3 = 12\)

Hence, the common factor = 12

(ii) Given that: 2y, 22xy

\(= (2 \times y) and (2 \times 11 \times x \times y)\)

We have the common factors are \(2 \times y = 2y\)

Hence, the common factor = 2y

(iii) Given that: \(14pq, 28p^2q^2\)

\(= (2 \times 7 \times p \times q) and (2 \times 2 \times 7 \times p \times p \times q \times q)\)

So, we have the common factors are \(2 \times 7 \times p \times q = 14pq\)

Hence, the common factor = 14pq

(iv) Given that:\(2x, 3x^2, 4\)

\(= (2 \times x), (3 \times x \times x) \) and \((2 \times 2)\)

We have the common factor as 1

Hence, the common factor = 1 [∵ 1 is a factor of every number]

(v)Given that: \(abc, 24ab^2, 12a^2b\)

\(= (2 \times 3 \times a \times b \times c), (2 \times 2 \times 2 \times 3 \times a \times b \times b)\) and \((2 \times 2 \times 3 \times a \times a \times b)\)

So, we have the common factors are \(2 \times 3 \times a \times b = 6ab\)

Hence, the common factor = 6ab

(vi) Given that:\(16x^3, -4x^2, 32x\)

\(= (2 \times 2 \times 2 \times 2 \times x \times x \times x), -(2 \times 2 \times x \times x), (2 \times 2 \times 2 \times 2 \times 2 \times x)\)

We have the common factors are \(2 \times 2 \times x = 4x\)

Hence, the common factor = 4x

(vii)Given that: 10pq, 20qr, 30rp

\(= (2 \times 5 \times p \times q), (2 \times 2 \times 5 \times q \times r), (2 \times 3 \times 5 \times r \times p)\)

So,we have common factors are \(2 \times 5 = 10\)

Hence, the common factor = 10

(viii) Given that:\(3x^2y^2, 10x^3y^2, 6x^2y^2z\)

\(= (3 \times x \times x \times y \times y), (2 \times 5 \times x \times x \times x \times y \times y), (2 \times 3 \times x \times x \times y \times y \times z)\)

So, the common factors are \(x \times x \times y \times y = x^2y^2\)

Hence, the common factor = \(x^2y^2\).

2. Factorise the following expressions.

(i) \(7x – 42\)

(ii) \(6p – 12q\)

(iii) \(7a^2 + 14a\)

(iv) \(-16z + 20z^3\)

(v) \(20l^2m + 30alm\)

(vi) \(5x^2y – 15xy^2\)

(vii) \(10a^2 – 15b^2 + 20c^2\)

(viii) \(-4a^2 + 4ab – 4ca\)

(ix) \(x^2yz + xy^2z + xyz^2\)

(x) \(ax^2y + bxy^2 + cxyz\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)We have: \(7x – 42 = 7(x – 6)\)

(ii)We have: \(6p – 12q = 6(p – 2q)\)

(iii) We have: \(7a^2 + 14a = 7a(a + 2)\)

(iv)We have: \(-16z + 20z^3 = 4z(-4 + 5z^2)\)

(v) We have: \(20l^2m + 30alm = 10lm(2l + 3a)\)

(vi) We have: \(5x^2y – 15xy^2 = 5xy(x – 3y)\)

(vii) We have: \(10a^2 – 15b^2 + 20c^2 = 5(2a^2 – 3b^2 + 4c^2)\)

(viii)We have: \(-4a^2 + 4ab – 4ca = 4a(-a + b – c)\)

(ix)We have: \(x^2yz + xy^2z + xyz^2 = xyz(x + y + z)\)

(x) We have: \(ax^2y + bxy^2 + cxyz = xy(ax + by + cz)\)

3.Factorise:
(i) \(x2 + xy + 8x + 8y\)

(ii) \(15xy – 6x + 5y – 2\)

(iii) \(ax + bx – ay – by\)

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

(v) \(z – 7 + 7xy – xyz\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)We have: \(x2 + xy + 8x + 8y\)

now, grouping the terms, we have \(x2 + xy + 8x + 8y\)

\(= x(x + y) + 8(x + y)\)

\(= (x + y)(x + 8)\)

Thus, the required factors =\((x + y)(x + 8)\)

(ii)We have: \(15xy – 6x + 5y – 2\)

Now, grouping the terms, we have \((15xy – 6x) + (5y – 2)\)

\(= 3x(5y – 2) + (5y – 2)\)

\(= (5y – 2)(3x + 1)\)

Thus, the required factors =\( (5y – 2)(3x + 1)\)

(iii) We have: \(ax + bx – ay – by\)

Now, grouping the terms, we have \(= (ax – ay) + (bx – by)\)

\(= a(x – y) + b(x – y)\)

\(= (x – y)(a + b)\)

Thus, the required factors = \((x – y)(a + b)\)

(iv) We have: \(15pq + 15 + 9q + 25p\)

Now, grouping the terms, we have \(= (15pq + 25p) + (9q + 15)\)

\(= 5p(3q + 5) + 3(3q + 5)\)

\(= (3q + 5) (5p + 3)\)

Thus, the required factors = \((3q + 5) (5p + 3)\)

(v) We have:\(z – 7 + 7xy – xyz\)

Now,grouping the terms, we have \(= (-xyz + 7xy) + (z – 7)\)

\(= -xy(z – 7) + 1 (z – 7)\)

\(= (-xy + 1) (z – 1)\)

Thus, the required factor =\( -(1 – xy) (z – 7)\)

NCERT solutions for class 8 Maths Chapter 14 Factorization Exercise 14.2

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) \(a^4 + 2a^2b^2 + b^4\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i) Given:\(a^2 + 8a + 16\)

Here, 4 + 4 = 8 and \(4 \times 4 = 16\)

\(a^2 + 8a +16\)

\(= a^2 + 4a + 4a + 4 \times 4\)

\(= (a^2 + 4a) + (4a + 16)\)

\(= a(a + 4) + 4(a + 4)\)

\(= (a + 4) (a + 4)\)

\(= (a + 4)^2\)

(ii) Given:\(p^2 – 10p + 25\)

Here, 5 + 5 = 10 and \(5 \times 5 = 25\)

\(=p^2 – 10p + 25\)

\(= p^2 – 5p – 5p + 5 \times 5\)

\(= (p^2 – 5p) + (-5p + 25)\)

\(= p(p – 5) – 5(p – 5)\)

\(= (p – 5) (p – 5)\)

\(= (p – 5)^2\)

(iii) Given:\(25m^2 + 30m + 9\)

Here, \(15 + 15 = 30\) and \(15 \times 15 = 25 \times 9 = 225\)

\(25m^2 + 30m + 9\)

\(= 25m^2 + 15m + 15m + 9\)

\(= (25m^2 + 15m) + (15m + 9)\)

\(= 5m(5m + 3) + 3(5m + 3)\)

\(= (5m + 3) (5m + 3)\)

\(= (5m + 3)^2\)

(iv) Given:\(49y^2 + 84yz + 36z^2\)

Here, \(42 + 42 = 84\) and \(42 \times 42 = 49 \times 36 = 1764\)

\(49y^2 + 84yz + 36z^2\)

\(= 49y^2 + 42yz + 42yz + 36z^2\)

\(= 7y(7y + 6z) +6z(7y + 6z)\)

\(= (7y + 6z) (7y + 6z)\)

\(= (7y + 6z)^2\)

(v) Given:\(4x^2 – 8x + 4\)

\(= 4(x^2 – 2x + 1) [Taking \;4 \;common]\)

\(= 4(x^2 – x – x + 1)\)

\(= 4[x(x – 1) -1(x – 1)]\)

\(= 4(x – 1)(x – 1)\)

\(= 4(x – 1)^2\)

(vi) Given: \(121b^2 – 88bc + 16c^2\)

Here, \(44 + 44 = 88\) and \(44 \times 44 = 121 \times 16 = 1936\)

\(121b^2 – 88bc + 16c^2\)

\(= 121b^2 – 44bc – 44bc + 16c^2\)

\(= 11b(11b – 4c) – 4c(11b – 4c)\)

\(= (11b – 4c) (11b – 4c)\)

\(= (11b – 4c)^2\)

(vii) Given:\((l + m)^2 – 4lm\)

Expanding the expression, \( (l + m)^2\), we have

\(l^2 + 2lm + m^2 – 4lm\)

\(= l^2 – 2lm + m^2\)

\(= l^2 – Im – lm + m^2\)

\(= l(l – m) – m(l – m)\)

\(= (l – m) (l – m)\)

\(= (l – m)^2\)

(viii) Given:\(a^4 + 2a^2b^2 + b^4\).So we have,

\(= a^4 + a^2b^2 + a^2b^2 + b^4\)

\(= a^2(a^2 + b^2) + b^2(a^2 + b^2)\)

\(= (a^2 + b^2)(a^2 + b^2)\)

\(= (a^2 + b^2)^2\)

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) \(9x^2y^2 – 16\)

(vii) \((x^2 – 2xy + y^2) – z^2\)

(viii) \(25a^2 – 4b^2 + 28bc – 49c^2\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given: \(4p^2 – 9q^2\)

\(= (2p)^2 – (3q)^2\)

\(= (2p – 3q) (2p + 3q)\quad\)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(ii)Given: \(63a^2 – 112b^2\)

\(= 7(9a^2 – 16b^2)\)

\(= 7 [(3a)^2 – (4b)^2]\)

\(= 7(3a – 4b)(3a + 4b)\quad\)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(iii) Given:\(49x^2 – 36\)

\(49x^2 – 36 = (7x)^2 – (6)^2\)

\(= (7x – 6) (7x + 6)\quad\)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(iv)Given: \(16x^5 – 144x^3 = 16x^3 (x^2 – 9)\)

\(= 16x^3 [(x)^2 – (3)^2]\)

\(= 16x^3(x – 3)(x + 3)\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(v) Given:\((l + m)^2 – (l – m)^2\)

\(= (l + m) – (l – m)] [(l + m) + (l – m)]\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

\(= (l + m – l + m)(l + m + l – m)\)

\(= (2m) (2l)\)

\(= 4ml\)

(vi)Given \(9x^2y^2 – 16 = (3xy)^2 – (4)^2\)

\(= (3xy – 4)(3xy + 4)\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)]\)

(vii)Given: \((x^2 – 2xy + y^2) – z^2\)

\(= (x – y)^2 – z^2\)

\(= (x – y – z) (x – y + z)\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)]\)

(viii) Given:\(25a^2 – 4b^2 + 28bc – 49c^2\)

\(= 25a^2 – (4b^2 – 28bc + 49c^2)\)

\(= (5a)^2 – (2b – 7c)^2\)

\(= [5a – (2b – 7c)] [5a + (2b – 7c)]\)

\(= (5a – 2b + 7c)(5a + 2b – 7c)\)

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) \(6xy – 4y + 6 – 9x\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given: \(ax^2 + bx \)

\(= x(ax + 5)\)

(ii) Given:\(7p^2 + 21q^2\)

\( = 7(p^2 + 3q^2)\)

(iii) Given:\(2x^3 + 2xy^2 + 2xz^2\)

\( = 2x(x^2 + y^2 + z^2)\)

(iv)Given: \(am^2 + bm^2 + bn^2 + an^2\)

\(= m^2 (a + b) + n^2(a + b)\)

\(= (a + b)(m^2 + n^2)\)

(v)Given: \((lm + l) + m + 1\)

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

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

(vi) Given:\(y(y + z) + 9(y + z) \)

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

(vii)Given: \(5y^2 – 20y – 8z + 2yz\)

\(= 5y^2 – 20y + 2yz – 8z\)

\(= 5y(y – 4) + 2z(y – 4)\)

\(= (y – 4) (5y + 2z)\)

(viii)Given: \(10ab + 4a + 5b + 2\)

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

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

(ix)Given:\(6xy – 4y + 6 – 9x\)

\(= 6xy – 4y – 9x + 6\)

\(= 2y(3x – 2) – 3(3x – 2)\)

\(= (3x – 2) (2y – 3)\)

4. Factorise.
(i) \(a^4 – b^4\)

(ii) \(p^4 – 81\)

(iii) \(x^4 – (y + z)^4\)

(iv) \(x^4 – (x – z)^4\)

(v) \(a^4 – 2a^2b^2 + b^4\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i) Given:\(a^4 – b^4 \)

\(=(a^2)^2 – (b^2)^2\quad\)[∵ \(a^2 – b^2 = (a – b)(a + b)\)]

\(= (a^2 – b^2) (a^2 + b^2)\)

\(= (a – b) (a + b) (a^2 + b^2)\)

(ii) Given:\(p^4 – 81 = (p^2)^2 – (9)^2\)

\(= (p^2 – 9) (p^2 + 9)\quad\)[∵\( a^2 – b^2 = (a – b)(a + b)]\)

\(= (p – 3)(p + 3) (p^2 + 9)\)

(iii) Given:\(x^4 – (y + z)4 \) \(= (x^2)^2 – [(y + z)^2]^2\quad\)[∵ \(a^2 – b^2 = (a – b)(a + b)\)]

\(= [x^2 – (y + z)^2] [x^2 + (y + z)^2]\)

\(= [x – (y + z)] [x + (y + z)] [x^2 + (y + z)^2]\)

\(= (x – y – z) (x + y + z) [x^2 + (y + z)^2]\)

(iv) Given:\(x^4 – (x – z)^4 = (x^2)^2 – [(y – z)^2]^2\)

\(= [x^2 – (y – z)^2] [x^2 + (y – z)^2]\)

\(= (x – y + z) (x + y – z) (x^2 + (y – z)^2]\)

(v)Given: \(a^4 – 2a^2b^2 + b^4\)

\(= a^4 – a^2b^2 – a^2b^2 + b^4\)

\(= a^2(a^2 – b^2) – b^2(a^2 – b^2)\)

\(= (a^2 – b^2)(a^2 – b^2)\)

\(= (a^2 – b^2)^2\)

\(= [(a – b) (a + b)]^2\)

\(= (a – b)^2 (a + b)^2\)

5. Factorise the following expressions.

(i) \(p^2 + 6p + 8\)

(ii) \(q^2 – 10q + 21\)

(iii) \(p^2 + 6p – 16\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given:\( p^2 + 6p + 8\)

Here, 2 + 4 = 6 and \(2 \times 4 = 8\)

\(p^2 + 6p + 8\)

\(= p^2 + 2p + 4p + 8\)

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

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

(ii)Given: \(q^2 – 10q + 21\)

Here, 3 + 7 = 10 and \(3 \times 7 = 21\)

\(q^2 – 10q + 21\)

\(= q^2 – 3q – 7q + 21\)

\(= q(q – 3) – 7(q – 3)\)

\(= (q – 3) (q – 7)\)

(iii) Given:\(p^2 + 6p – 16\)

Here, 8 – 2 = 6 and \(8 \times 2 = 16\)

\(p^2 + 6p – 16\)

\(= p^2 + 8p – 2p – 16\)

\(= p(p + 8) – 2(p + 8)\)

\(= (p + 8) (p – 2)\)

NCERT solutions for class 8 Maths Chapter 14 Factorization Exercise 14.3

1. Carry out the following divisions.
(i) \(28x^4 ÷ 56x\)

(ii) \(-36y^3 ÷ 9y^2\)

(iii) \(66pq^2r^3 ÷ 11qr^2\)

(iv) \(34x^3y^3z^3 ÷ 51xy^2z^3\)

(v) \(12a^8b^8 ÷ (-6a^6b^4)\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given:\(28x^4 ÷ 56x\)

\(=\frac{28\times x \times x \times x \times x}{28\times 2 \times x}\)

\(=\frac{x \times x \times x}2=\frac{x^3}2\)

(ii)Given:\(-36y^3 ÷ 9y^2\)

\(=\frac{-1\times 4 \times 9 \times y \times y \times y}{9\times y \times y}\)

\(=\frac{-1 \times 4 \times y}1=-4y\)

(iii)Given: \(66pq^2r^3 ÷ 11qr^2\)

\(=\frac{6\times 11 \times p \times q \times q \times r \times r \times r}{11\times q \times r \times r}\)

\(=\frac{6 \times p \times q \times r}1=6pqr\)

(iv)Given: \(34x^3y^3z^3 ÷ 51xy^2z^3\)

\(=\frac{6\times 11 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{17\times 3 \times x \times y\times z \times z \times z}\)

\(=\frac{2 \times x \times x \times y}3=\frac{2x^2y}{3}\)

(v)Given:\(12a^8b^8 ÷ (-6a^6b^4)\)

\(=\frac{6\times 2 \times a \times a \times a \times a \times a \times a \times a \times a \times b \times b \times b \times b \times b \times b \times b \times b }{-1\times 6\times a \times a \times a \times a \times a \times a \times b \times b \times b \times b}\)

\(=\frac{2 \times a \times a \times b \times b \times b \times b}{-1}=-2a^2b^4\)

2.Divide the given polynomial by the given monomial,
(i) \((5x^2 – 6x) ÷ 3x\)

(ii) \((3y^8 – 4y^6 + 5y^4) ÷ y^4\)

(iii) \(8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x^2y^2z^2\)

(iv) \((x^3 + 2x^2 + 3x) ÷ 2x\)

(v) \((p^3q^6 – p^6q^3 – p^6q^3) ÷ p^3q^3\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given: \((5x^2 – 6x) ÷ 3x\)

\(=\frac{5x^2 – 6x}{3x}=\frac{5x^2}{3x}-\frac{6x}{3x}\)

\(=\frac{5}2x-2=\frac{1}3(5x-6)\)

(ii)Given:\((3y^8 – 4y^6 + 5y^4) ÷ y^4\)

\(=\frac{3y^8 – 4y^6 + 5y^4}{y^4}=\frac{3y^8}{y^4}-\frac{4y^6 }{y^4}+\frac{5y^4 }{y^4}\)

\(=3y^4-4y^2+5\)

(iii)Given:\(8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x^2y^2z^2\)

\(=\frac{8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)}{4x^2y^2z^2}=\frac{8\times x^3y^2z^2 \times (x+y+z)}{4x^2y^2z^2}\)

\(=2(x+y+z)\)

(iv)Given: \((x^3 + 2x^2 + 3x) ÷ 2x\)

\(=\frac{x^3 + 2x^2 + 3x}{2x}=\frac{x \times (x^2+2x+3)}{2\times x}\)

\(=\frac{1}2(x^2+2x+3)\)

(v)Given:\((p^3q^6 – p^6q^3 – p^6q^3) ÷ p^3q^3\)

\(=\frac{p^3q^6 – p^6q^3 – p^6q^3}{p^3q^3}=\frac{p^3q^3(q^3-p^3)}{p^3q^3}\)

\(=q^3-p^3\)

3.Work out the following divisions.
(i) \((10x – 25) ÷ 5\)

(ii) \((10x – 25) ÷ (2x – 5)\)

(iii) \(10y (6y + 21) ÷ 5 (2y + 7)\)

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

(v) \(96abc (3a – 12) (56 – 30) ÷ 144 (a – 4) (b – 6)\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given:\((10x – 25) ÷ 5\)

\(=\frac{10x-25}5=\frac{5\times (2x-5)}5\)

\(=2x-5\)

(ii)Given:\((10x – 25) ÷ (2x – 5)\)

\(=\frac{10x-25}{2x – 5}=\frac{5\times (2x-5)}{2x – 5}\)

\(=5\)

(iii)Given:\(10y (6y + 21) ÷ 5 (2y + 7)\)

\(=\frac{10y (6y + 21)}{ 5 (2y + 7)}=\frac{5\times 2\times y\times 3\times (2y + 7)}{ 5\times (2y + 7)}\)

\(=\frac{2\times y \times 3}1=6y\)

(iv)Given: \(9x^2y^2 (3z – 24) ÷ 27xy (z – 8)\)

\(=\frac{9x^2y^2 (3z – 24)}{ 27xy (z – 8)}=\frac{9\times x^2y^2 \times 3\times (z-8)}{ 9\times 3\times xy \times (z -8)}\)

\(=\frac{xy}1=xy\)

(v)Given: \(96abc (3a – 12) (56 – 30) ÷ 144 (a – 4) (b – 6)\)
\(=\frac{96abc (3a – 12) (56 – 30)}{ 144 (a – 4) (b – 6)}=\frac{48\times 2 \times abc \times 3\times (a-4)\times 5 \times (b-6)}{48\times 3\times (a-4) \times (b-6)}\)

\(=\frac{2 \times abc \times 5}1=10abc\)

4.Divide as directed.
(i) \(5(2x + 1)(3x + 5) ÷ (2x + l)\)

(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) \(x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)Given: \(5(2x + 1)(3x + 5) ÷ (2x + 1)\)

\(=\frac{5(2x + 1)(3x + 5)}{2x + 1}=\frac{5\times (2x+3)\times(3x + 5)}{(2x + 1)}\)

\(=\frac{5\times (3x+5)}1=5(3x+5)\)

(ii)Given: \(26xy (x + 5) (y – 4) ÷ 13x (y – 4)\)

\(=\frac{26xy (x + 5) (y – 4)}{13x (y – 4)}=\frac{13\times 2\times x \times y \times (x+5) \times (y-4)}{13\times x\times (y-4)}\)

\(=\frac{2\times y \times (x+5))}1=2y(x+5)\)

(iii)Given: \(52pqr (p + q) (q + r) (r + p) ÷104pq (q + r) (r + p)\)

\(=\frac{52pqr (p + q) (q + r) (r + p)}{104pq (q + r) (r + p)}=\frac{52\times p \times q \times r \times (p+q) \times (q+r) \times (r+p)}{52\times 2 \times p \times q \times (q+r) \times (r+p)}\)

\(=\frac{1}2r(p+q)\)

(iv)Given: \(20 (y + 4) (y^2 + 5y + 3) ÷ 5 (y + 4)\)

\(=\frac{20 (y + 4) (y^2 + 5y + 3) }{5 (y + 4)}=\frac{5\times 4 \times (y+4) \times (y^2+5y+3)}{13\times x\times (y-4)}\)

\(=\frac{4\times (y^2 + 5y + 3)}1=4(y^2 + 5y + 3)\)

(v)Given: \(x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)\)

\(=\frac{x (x + 1) (x + 2) (x + 3) }{x (x + 1)}=\frac{z \times (x+1) \times (x+2) \times (x+3)}{x \times (x+1)}\)

\(=\frac{(x+2)(x+3)}1=(x+2)(x+3)\)

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) \(39y^3 (50y^2 – 98)÷ 26y^2 (5y + 7)\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

(i)We have, \(y^2 +7y +10=y^2 +2y +5y +10\)

\(=y(y+2)+5(y+2)\)

\(=(y+2)(y+5)\)....(1)

∴\(\frac{y^2 +7y +10}{y + 5}=\frac{(y+2)(y+5)}{(y + 5)}\quad\)[Substituting (1)]

\(=y+5\)

(ii)We have, \(m^2 – 14m – 32=m^2 +2m – 16m – 32\)

\(=m(m+2)-16(m+2)\)

\(=(m+2)(m-16)\)....(1)

∴\(\frac{m^2 – 14m – 32}{m + 2}=\frac{(m+2)(m-16)}{(m + 2)}\quad\)[Substituting (1)]

\(=m-16\)

(iii)We have, \(5p^2 – 25p + 20=5(p^2 – 5p +4)\)

\(=5(p^2 -p – 4p +4)\)

\(=5(p(p-1)-4(p-1))\)

\(=5(p-1)(p-4)\)....(1)

∴\(\frac{5p^2 – 25p + 20}{p-1}=\frac{5(p-1)(p-4)}{(p – 1))}\quad\)[Substituting (1)]

\(=5(p-4)\)

(iv)We have, \(4yz (z^2 + 6z – 16)=4yz(z^2 + 8z-2z – 16)\)

\(=4yz(z^2 + 8z-2z – 16)\)

\(=4yz(z(z+8)-2(z+8))\)

\(=4yz(z+8)(z-2)\)....(1)

∴\(\frac{4yz (z^2 + 6z – 16)}{2y (z + 8)}=\frac{4yz(z+8)(z-2)}{2y (z + 8)}\quad\)[Substituting (1)]

\(=\frac{2z(z-2)}1=2z(z-2)\)

(v)We have, \(5pq (p^2 – q^2)=5pq(p+q)(p-q)\)....(1)

∴\(\frac{5pq (p^2 – q^2)}{2p (p + q)}=\frac{5\times p \times q \times (p+q) \times (p-q)}{2p \times (p + q)}\quad\)[Substituting (1)]

\(=\frac{5}2q(p-q)\)

(vi)We have, \(12xy (9x^2 – 16y^2)=12xy((3x)^2 – (4y)^2)\)

\(=12xy(3x-4y)(3x+4y)\)....(1)

∴\(\frac{12xy (9x^2 – 16y^2)}{4xy (3x + 4y)}=\frac{12xy(3x-4y)(3x+4y)}{4xy (3x + 4y)}\quad\)[Substituting (1)]

\(=3(3x-4y)\)

(vii)We have, \(39y^3 (50y^2 – 98)=39y^3 \times 2(25y^2 – 49\)

\(=78y^2(5y^2-7^2)\)

\(=78y^2(5y+7)(5y-7)\)....(1)

∴\(\frac{39y^3 (50y^2 – 98)}{26y^2 (5y + 7)}=\frac{78y^2(5y+7)(5y-7)}{26y^2 (5y + 7)}\quad\)[Substituting (1)]

\(=3(5y-7)\)

NCERT solutions for class 8 Maths Chapter 14 Factorization Exercise 14.4

1.Find and correct the errors in the following mathematical statements.
1. \(4 (x – 5) = 4x – 5\)

2. \(x(3x + 2) = 3x^2 + 2\)

3. \(2x + 3y = 5xy\)

4.\( x + 2x + 3x = 5xy\)

5. \(5y + 2y + y – 7y = 0\)

6. \(3x + 2x = 5x^2\)

7. \((2x)^2 + 4(2x) + 7 = 2x^2 + 8x + 7\)

8. \((2x)^2 + 5x = 4x + 5x = 9x\)

9. \((3x + 2)^2 = 3x^2 + 6x + 4\)

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 \((-3)2 + 5(-3) = -9 – 15 = -24\)

11.\( (y – 3)^2 = y^2 – 9\)

12.\( (z + 5)^2 = z^2 + 25\)

13.\( (2a + 3b)(a – b) = 2a^2 – 3b^2\)

14.\( (a + 4)(a + 2) = a^2 + 8\)

15.\( (a – 4)(a – 2) = a^2 – 8\)

16. \(\frac { 3{ x }^{ 2 } }{ 3{ x }^{ 2 } } =0\)

17. \(\frac { 3{ x }^{ 2 }+1 }{ 3{ x }^{ 2 } } =1+1=2\)

18.\( \frac { 3x }{ 3x+2 } =\frac { 1 }{ 2 } \)

19.\( \frac { 3 }{ 4x+3 } =\frac { 1 }{ 4x } \)

20. \(\frac { 4x+5 }{ 4x } =5\)

21. \(\frac { 7x+5 }{ 5 } =7x\)



NCERT Solutions for Class 8 Maths Chapter 14 Factorization


Answer :

1. We have LHS=\(4(x-5)=4\times x -4 \times 5 =4x-20\)

We should have RHS=LHS

Therefore,\(4(x-5)=4x-20\) is the correct statement.

2. We have LHS=\(x(3x + 2)=x \times 3x + x \times 2 =3x^2+2x\)

We should have RHS=LHS

Therefore,\(x(3x + 2)=3x^2+2x\) is the correct statement.

3.The given statement is incorrect only liked terms can be grouped together.Therefore, 2x+3y=2x+3y may be the correct statement

4.Using distributive property of multiplivation over addition:

\(LHS=x+2x+3x=6x\neq\) given RHS

So x+2x+3x=6x is the correct statement

5.Using distributive property of multiplivation over addition:

\(LHS=5y+2y+y-7y=(8-7)y=7y\neq\) given RHS

So 5y+2y+y-7y=7y is the correct statement

6.Using distributive property of multiplivation over addition:

\(LHS=3x + 2x=5x\neq\) given RHS

So 3x + 2x=5x is the correct statement

7.We have LHS:\((2x)^2 + 4(2x) + 7\)

\(=(2)^2\times (x)^2 + 4(2x) + 7= 4x^2 + 8x + 7\neq\) given RHS

So \((2x)^2 + 4(2x) + 7= 4x^2 + 8x + 7\) is the correct statement

8.We have LHS:\((2x)^2 + 5x\)

\(=(2)^2\times (x)^2 +5x= 4x^2 + 5x\neq\) given every value in RHS

So \((2x)^2 + 5x= 4x^2 +5x \) is the correct statement

9.We have LHS:\((3x + 2)^2\)

\(=(3x)^2+2\times (3x)\times(2)+2^2)=9x^2 +12x + 4\neq\) given RHS

So \((3x + 2)^2= 9x^2 +12x + 4\) is the correct statement

10.(a)Substituting x=-3 in \(x^2 +5x + 4\)

\(=(-3)^2+5(-3)+4=9-15+4=-2 \neq \) given RHS

So for x=-3, the value of \(x^2 +12x + 4 =-2\)

(b)Substituting x=-3 in \(x^2 -5x + 4\)

\(=(-3)^2-5(-3)+4=9+15+4=28 \neq \) given RHS

So for x=-3, the value of \(x^2 -5x + 4 =28\)

(c)Substituting x=-3 in \(x^2 +5x\)

\(=(-3)^2+5(-3)=9-15=-6 \neq \) given RHS

So for x=-3, the value of \(9x^2 +5x =62\)

11.We have, We have LHS:\((y-3)^2\)

\(=(y)^2-2\times (y)\times(3)+3^2)=y^2 -6y + 9\neq\) given RHS

So \((y-3)^2= y^2 -6y + 9\) is the correct statement

12.We have, We have LHS:\((z+5)^2\)

\(=(z)^2+2\times (z)\times(5)+5^2)=z^2 +10z + 25\neq\) given RHS

So \((z+5)^2= z^2 +10z + 25\) is the correct statement

13. Here we have LHS=\((2a+3b)(a-b)\)

\(=2a(a-b)+3b(a-b)=2a \times a - 2a times b+3b \times a - 3a \times b\)

\(=2a^2+ab-3b^2\neq \) given RHS

So \((2a+3b)(a-b)=2a^2+ab-3b^2\) is the correct statement

14.Here we have LHS=\((a+4)(a+2)\)

\(=a(a+2)+4(a+2)=a \times a + a times 2+4 \times a + 4 \times 2\)

\(=a^2+6a+8\neq \) given RHS

So \((a+4)(a+2)=a^2+6a+8\) is the correct statement

15.Here we have LHS=\((a-4)(a-2)\)

\(=a(a-2)-4(a-2)=a \times a - a \times 2-4 \times a + 4 \times 2\)

\(=a^2-6a+8\neq \) given RHS

So \((a-4)(a-2)=a^2-6a+8\) is the correct statement

16.We have LHS=\(\frac{3x^2}{3x^2}=1 \neq\) given RHS

So we have \(\frac{3x^2}{3x^2}=1\) as the correct statement

17.We have LHS=\(\frac{3x^2+1}{3x^2}=\frac{3x^2}{3x^2}+\frac{1}{3x^2}=1+\frac{1}{3x^2} \neq\) given RHS

So we have \(\frac{3x^2+1}{3x^2}=1+\frac{1}{3x^2}\) as the correct statement

18.We have LHS=\(\frac{3x}{3x+2}\neq \frac{1}2 \)

So we have \(\frac{3x^2}{3x^2}=\frac{3x}{3x+2}\) as the correct statement

19.Clearly in LHS we can see that, \(\frac{3}{4x+3} \neq \frac{1}{4x}\)

So we have \(\frac{3}{4x+3}=\frac{3}{4x+3}\) as the correct statement

20.Clearly in LHS we can see that, \(\frac{4x+5}{4x}=\frac{4x}{4x}+\frac{5}{4x}=1+\frac{5}{4x} \neq 5\)

So we have \(\frac{4x+5}{4x}=1+\frac{5}{4x}\) as the correct statement

21.Clearly in LHS we can see that, \(\frac{7x+5}{5}=\frac{7x}{5}+\frac{5}{5}=1+\frac{7x}{5} \neq 7x\)

So we have \(\frac{7x+5}{5}=1+\frac{7x}{5}\) as the correct statement



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