# NCERT Solutions for Class 8 Maths Chapter 14 Factorization

Written by Team Trustudies
Updated at 2021-05-07

## 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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

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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