# NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.3

NCERT Solutions for Class 8 Maths Chapter 14, exercise 14.3 is prepared according to the latest and updated syllabus pattern. These NCERT Solutions for Class 8 crafted by subject experts help students to stick to a particular pattern of learning and score more. Best scores can be achieved in Maths subject by practising the textbook problems on a daily basis. The NCERT class 8 exercise solutions act as a great resource for practice and effective preparation for exams. Download now and start practising.

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### Access Other Exercise Solutions of Class 8 Maths Chapter 14 Factorisation

Exercise 14.1 Solutions : 3 Questions (Short answer type)
Exercise 14.2 Solutions : 5 Questions (Short answer type)
Exercise 14.4 Solutions : 21 Questions (Short answer type)

### Access Answers to NCERT Class 8 Maths Chapter 14 Factorisation Exercise 14.3 Page number 227

1. Carry out the following divisions.

(i) 28x4 Ã· 56x

(ii) â€“36y3 Ã· 9y2

(iii) 66pq2r3 Ã· 11qr2

(iv) 34x3y3z3 Ã· 51xy2z3

(v) 12a8b8 Ã· (â€“ 6a6b4)

Solution:

(i)28x4 = 2Ã—2Ã—7Ã—xÃ—xÃ—xÃ—x

56x = 2Ã—2Ã—2Ã—7Ã—x

2. Divide the given polynomial by the given monomial.

(i)(5x2â€“6x) Ã· 3x

(ii)(3y8â€“4y6+5y4) Ã· y4

(iii) 8(x3y2z2+x2y3z2+x2y2z3)Ã· 4x2 y2 z2

(iv)(x3+2x2+3x) Ã·2x

(v) (p3q6â€“p6q3) Ã· p3q3

Solution:

3. Work out the following divisions.

(i) (10xâ€“25) Ã· 5

(ii) (10xâ€“25) Ã· (2xâ€“5)

(iii) 10y(6y+21) Ã· 5(2y+7)

(iv) 9x2y2(3zâ€“24) Ã· 27xy(zâ€“8)

(v) 96abc(3aâ€“12)(5bâ€“30) Ã· 144(aâ€“4)(bâ€“6)

Solution:

(i) (10xâ€“25) Ã· 5 = 5(2x-5)/5 = 2x-5

(ii) (10xâ€“25) Ã· (2xâ€“5) = 5(2x-5)/( 2x-5) = 5

(iii) 10y(6y+21) Ã· 5(2y+7) = 10yÃ—3(2y+7)/5(2y+7) = 6y

(iv) 9x2y2(3zâ€“24) Ã· 27xy(zâ€“8) = 9x2y2Ã—3(z-8)/27xy(z-8) = xy

4. Divide as directed.

(i) 5(2x+1)(3x+5)Ã· (2x+1)

(ii) 26xy(x+5)(yâ€“4)Ã·13x(yâ€“4)

(iii) 52pqr(p+q)(q+r)(r+p) Ã· 104pq(q+r)(r+p)

(iv) 20(y+4) (y2+5y+3) Ã· 5(y+4)

(v) x(x+1) (x+2)(x+3) Ã· x(x+1)

Solution:

5. Factorise the expressions and divide them as directed.

(i) (y2+7y+10)Ã·(y+5)

(ii) (m2â€“14mâ€“32)Ã·(m+2)

(iii) (5p2â€“25p+20)Ã·(pâ€“1)

(iv) 4yz(z2+6zâ€“16)Ã·2y(z+8)

(v) 5pq(p2â€“q2)Ã·2p(p+q)

(vi) 12xy(9x2â€“16y2)Ã·4xy(3x+4y)

(vii) 39y3(50y2â€“98) Ã· 26y2(5y+7)

Solution:

(i) (y2+7y+10)Ã·(y+5)

First solve for equation, (y2+7y+10)

(y2+7y+10) = y2+2y+5y+10 = y(y+2)+5(y+2) = (y+2)(y+5)

Now, (y2+7y+10)Ã·(y+5) = (y+2)(y+5)/(y+5) = y+2

(ii) (m2â€“14mâ€“32)Ã· (m+2)

Solve for m2â€“14mâ€“32, we have

m2â€“14mâ€“32 = m2+2m-16mâ€“32 = m(m+2)â€“16(m+2) = (mâ€“16)(m+2)

Now, (m2â€“14mâ€“32)Ã·(m+2) = (mâ€“16)(m+2)/(m+2) = m-16

(iii) (5p2â€“25p+20)Ã·(pâ€“1)

Step 1: Take 5 common from the equation, 5p2â€“25p+20, we get

5p2â€“25p+20 = 5(p2â€“5p+4)

Step 2: Factorize p2â€“5p+4

p2â€“5p+4 = p2â€“p-4p+4 = (pâ€“1)(pâ€“4)

Step 3: Solve original equation

(5p2â€“25p+20)Ã·(pâ€“1) = 5(pâ€“1)(pâ€“4)/(p-1) = 5(pâ€“4)

(iv) 4yz(z2 + 6zâ€“16)Ã· 2y(z+8)

Factorize z2+6zâ€“16,

z2+6zâ€“16 = z2-2z+8zâ€“16 = (zâ€“2)(z+8)

Now, 4yz(z2+6zâ€“16) Ã· 2y(z+8) = 4yz(zâ€“2)(z+8)/2y(z+8) = 2z(z-2)

(v) 5pq(p2â€“q2) Ã· 2p(p+q)

p2â€“q2 can be written as (pâ€“q)(p+q) using identity.

5pq(p2â€“q2) Ã· 2p(p+q) = 5pq(pâ€“q)(p+q)/2p(p+q) = 5/2q(pâ€“q)

(vi) 12xy(9x2â€“16y2) Ã· 4xy(3x+4y)

Factorize 9x2â€“16y2 , we have

9x2â€“16y2 = (3x)2â€“(4y)2 = (3x+4y)(3x-4y) using identity: p2â€“q2 = (pâ€“q)(p+q)

Now, 12xy(9x2â€“16y2) Ã· 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)

(vii) 39y3(50y2â€“98) Ã· 26y2(5y+7)
st solve for 50y2â€“98, we have

50y2â€“98 = 2(25y2â€“49) = 2((5y)2â€“72) = 2(5yâ€“7)(5y+7)

Now, 39y3(50y2â€“98) Ã· 26y2(5y+7) =

CBSE Class 8 Maths Chapter 14 Factorisation Exercise 14.3, based on the topic, Division of Algebraic Expressions and subtopics like Division of a monomial by another monomial, Division of a polynomial by a monomial and Division of Algebraic Expressions Continued (Polynomial Ã· Polynomial). At the end of this section, students will get handy with the division of one algebraic expression by another and able to solve such problems without any help.