NCERT solution class 8 chapter 14 Factorisation

Exercise 14.1   Exercise 14.2   Exercise 14.3   Exercise 14.4

Exercise 14.3

Question 1 

Carry out the following divisions.

(i) 28x4 ÷ 56x

(ii) −36y3 ÷ 9y2

(iii) 66pq2r3 ÷ 11qr2

(iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (−6a6b4)

Sol :

(i) 28x4 = 2 × 2 × 7 × x × x × x × x

56x = 2 × 2 × 2 × 7 × x

(ii) 36y3 = 2 × 2 × 3 × 3 × y × y × y

9y2 = 3 × 3 × y × y

(iii) 66 pq2 r3 = 2 × 3 × 11 × p × q × q × r × r × r

11qr2 = 11 × q × r × r

(iv) 34 x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z

51 xy2z3 = 3 ×17 × x × y × y ×z × z × z

(v) 12a8b8 = 2 × 2 × 3 × a8 × b8

6a6b4 = 2 × 3 × a6 × b4

= −2a2b4

 

 

Question 2

Divide the given polynomial by the given monomial.

(i) (5x2 − 6x) ÷ 3x

(ii) (3y8 − 4y6 + 5y4) ÷ y4

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2

(iv) (x3 + 2x2 + 3x) ÷ 2x

(v) (p3q6p6q3) ÷ p3q3

Sol :

(i) 5x2 − 6x = x(5x − 6)

(ii) 3y8 − 4y6 + 5y4 = y4(3y4 − 4y2 + 5)

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) = 8x2y2z2(x + y + z)

(iv) x3 + 2x2 + 3x = x(x2 + 2x + 3)

(v) p3q6p6q3 = p3q3(q3p3)

 

 

Question 3

Work out the following divisions.

(i) (10x − 25) ÷ 5

(ii) (10x − 25) ÷ (2x − 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

(iv) 9x2y2(3z − 24) ÷ 27xy(z − 8)

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

Sol :

(i)

(ii)

(iii)

(iv)

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

 

 

Question 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) (y2 + 5y + 3) ÷ 5(y + 4)

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

Sol :

(i) = 5(3x + 1)

(ii) = 2y (x + 5)

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

(iv) 20(y + 4) (y2 + 5y + 3) = 2 × 2 × 5 × (y + 4) (y2 + 5y + 3)

(v)

= (x + 2) (x + 3)

 

 

Question 5

Factorise the expressions and divide them as directed.

(i) (y2 + 7y + 10) ÷ (y + 5)

(ii) (m2 − 14m − 32) ÷ (m + 2)

(iii) (5p2 − 25p + 20) ÷ (p − 1)

(iv) 4yz(z2 + 6z − 16) ÷ 2y(z + 8)

(v) 5pq(p2q2) ÷ 2p(p + q)

(vi) 12xy(9x2 − 16y2) ÷ 4xy(3x + 4y)

(vii) 39y3(50y2− 98) ÷ 26y2(5y+ 7)

Sol :

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

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

= (y + 2) (y + 5)

(ii) m2 − 14m − 32 = m2 + 2m − 16m − 32

= m (m + 2) − 16 (m + 2)

= (m + 2) (m − 16)

(iii) 5p2 − 25p + 20 = 5(p2 − 5p + 4)

= 5[p2p − 4p + 4]

= 5[p(p −1) − 4(p −1)]

= 5(p −1) (p − 4)


​​​​​​​​​​​​​​​​​​​​​

(iv) 4yz(z2 + 6z −16) = 4yz [z2 − 2z + 8z − 16]

= 4yz [z(z − 2) + 8(z − 2)]

= 4yz(z − 2) (z + 8)

(v) 5pq(p2 q2) = 5pq (pq) (p + q)

(vi) 12xy(9x2 − 16y2) = 12xy[(3x)2 − (4y)2] = 12xy(3x − 4y) (3x + 4y)

(vii) 39y3(50y2 − 98) = 3 × 13 × y × y × y × 2[(25y2 − 49)]

= 3 × 13 × 2 × y × y × y × [(5y)2 − (7)2]

= 3 × 13 × 2 × y × y × y (5y − 7) (5y + 7)

26y2(5y + 7) = 2 × 13 × y × y × (5y + 7)

39y3(50y2 − 98) ÷26y2 (5y + 7)

 

Leave a Reply

Your email address will not be published. Required fields are marked *