NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3
NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3
These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 8 Maths Chapter 14 Factorization 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)
Answer:
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) (p3q6– p6q3) ÷ p3q3
Answer:
(i) (5x2 – 6x) ÷ 3x
(ii) (3y8 – 4y6 + 5y4) ÷ y4
=3y4 – 4y6-4 + 5y4-4
=3y4 – 4y2 – 5y°
= 3y4 – 4y2 + 5
(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) 4x2y2z2
= [x3-2 x y2-2 x z2-2 + x2-2 x y3-2 x z2-2 + x2-2 x y2-2 x z3-2]
= 2 (x x y° x z° + x° x y x z° + x° x y° x z)
= 2 (x + y + z)
[using a° = 1]
(iv) x3 + 2x2 + 3x ÷ 2x
(v) (p3q6– p6q3) ÷ p3q3
= p3-3 x q6-3 – p6-3 x q3-3
= p0 x q3 – p3 x q0
= p3 x q3
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)
Answer:
(i) (10x – 25) ÷ 5
= 2x – 5
[Taking 3 as common from 6y + 21]
= 2 x y x 3
= 6y
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)
Answer:
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
= 5 (3x + 5)
(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
= (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) 4- (p – 1)
(iv) 4yz(z2 + 6z – 16) -f- 2y (z + 8)
(v) 5pq (p2 – q2) -f- 2p (p + q)
(vi) 12xy (9x2 – 16y2) -J- 4xy (3x + 4y)
(vii) 39y3 (50y2 – 98) + 26y2 (5y + 7)
Answer:
(i) y2 + 7y + 10
[10 = 2 x 5
[7 = 2 + 5]
= y2 + 5y + 2y + 10
= y( y + 5) + 2 (y + 5) = (y + 5) (y + 2)
∴ (y2 + 7y + 10) ÷ (y + 5)
(y+5)(y+2)/(y+5) = y + 2
(ii) m2 – 14m – 32
– 32 = -16 x 2
-14 = -16 + 2
= m (m – 16) + 2(m – 16)
= (m – 16) (m + 2)
(m2 – 14m – 32) ÷ (m + 2)
= (m−16)(m+2)/m+2
= m – 16
(iii) 5p2 – 25p + 20
= 5[p2 – 5p + 4]
= 5 [p2 – 4p – p + 4]
= 5[p(p – 4)- 1 (p – 4)
= 5(p – 4) (p – 1)
∴ 5p2 – 25 p + 20 ÷ (p – 1)
= 5(p−4)(p−1)/(p−1)
= 5 (p -4)
(iv) z2 + 6z – 16 = z2 + 8z – 2z – 16
-16 = 8 x – 2
6 = 8 + (-2)
= z(z + 8) – 2 (z – 2)
= (z + 8) (z – 2)
∴ 4yz (z2 + 6z – 16) 2y (z + 8)
= 3 (3x – 4y)
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