ML Aggarwal Algebraic Expression and Identities Exe-10.4 Class 8 ICSE Ch-10 Maths Solutions

ML Aggarwal Algebraic Expression and Identities Exe-10.4 Class 8 ICSE Ch-10 Maths Solutions. We Provide Step by Step Answer of  Exe-10.4 Questions for Algebraic Expression and Identities as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

ML Aggarwal Algebraic Expression and Identities Exe-10.4 Class 8 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 8th
Chapter-10 Algebraic Expression and Identities
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-10.4 Questions
Edition 2023-2024

Algebraic Expression and Identities Exe-10.4

ML Aggarwal Class 8 ICSE Maths Solutions

Page-180

Question 1. Divide:

(i) – 39pq2r5 by – 24p3q3r
(ii) –34a2b3 by 6a3b2

Answer:

(i) – 39pq2r5 (÷) – 24p3q3r

= – 39pq2r5/ – 24p3q3r

ml class 8 chapter 10.3 img 2

(ii) –34a2b3 by 6a3b2

ml class 8 chapter 10.3 img 3


Algebraic Expression and Identities Exe-10.4

ML Aggarwal Class 8 ICSE Maths Solutions

Page-181

Question 2. Divide:

(i) 9x4 – 8x3 – 12x + 3 by 3x
(ii) 14p2q3 – 32p3q2 + 15pq2 – 22p + 18q by – 2p2q.

Answer:

(i) 9x4 – 8x3 – 12x + 3 by 3x

ml class 8 chapter 10.3 img 4

(ii) 14p2q3 – 32p3q2 + 15pq2 – 22p + 18q by – 2p2q.

ml class 8 chapter 10.3 img 5

Question 3. Divide:

(i) 6x2 + 13x + 5 by 2x + 1
(ii) 1 + y3 by 1 + y
(iii) 5 + x – 2x2 by x + 1
(iv) x3 – 6x2 + 12x – 8 by x – 2

Answer:

(i) 6x2 + 13x + 5 by 2x + 1

(i) 6x2 + 13x + 5 by 2x + 1

∴ Quotient = 3x + 5 and remainder = 0

(ii) 1 + y3 by 1 + y

ml class 8 chapter 10.3 img 7

∴ Quotient = y2 – y + 1 and remainder = 0

(iii) 5 + x – 2x2 by x + 1

On arranging the terms of dividend in descending order of powers of x and then dividing,

– 2x2 + x + 5 ÷ x + 1

ml class 8 chapter 10.3 img 8

∴ Quotient = – 2x + 3 and remainder = 2

(iv) x3 – 6x2 + 12x – 8 by x – 2

ml class 8 chapter 10.3 img 8

∴ Quotient = x2 – 4x + 4 and remainder = 0

Question 4. Divide:

(i) 6x3 + x2 – 26x – 25 by 3x – 7
(ii) m3 – 6m2 + 7 by m – 1

Answer:

(i) 6x3 + x2 – 26x – 25 by 3x – 7

(i) 6x3 + x2 – 26x – 25 by 3x – 7

∴ Quotient = 2x2 + 5x + 3 and remainder = – 4
(ii) m3 – 6m2 + 7 by m – 1

(ii) m3 – 6m2 + 7 by m – 1

∴ Quotient = m2 – 5m – 5 and remainder = 2.

Question 5. Divide:

(i) a3 + 2a2 + 2a + 1 by a2 + a + 1
(ii) 12x3 – 17x2 + 26x – 18 by 3x2 – 2x + 5

Answer:

(i) a3 + 2a2 + 2a + 1 by + a2 + a + 1

(i) a3 + 2a2 + 2a + 1 by + a2 + a + 1

∴ Quotient = a + 1 and remainder = 0.
(ii) 12x3 – 17x2 + 26x – 18 by 3x2 – 2x + 5

(ii) 12x3 – 17x2 + 26x – 18 by 3x2 – 2x + 5

∴ Quotient = 4x – 3 and remainder = -3

Question 6. If the area of a rectangle is 8x2 – 45y2 + 18xy and one of its sides is 4x + 15y, find the length of the adjacent side.

Answer:

Area of rectangle = 8x2 – 45y2 + 18xy

And, one side = 4x + 15y

∴ Second (adjacent) side = Area of rectangle/ One side

= 8x2 – 45y2 + 18xy ÷ 4x + 15y

Question 6. If the area of a rectangle is 8x2 – 45y2 + 18xy and one of its sides is 4x + 15y, find the length of the adjacent side.

Thus, length of the adjacent side is 2x – 3y.

Question 7. What polynomial be subtracted from the polynomial x4 + 2x3 + x2 – 18x – 15 so that the resulting polynomial is exactly divisible by x2 – 1 – x.

Answer:

x4 + 2x3 + x2 – 18x – 15 by x2 – 1 – x.

Question 7. What polynomial be subtracted from the polynomial x4 + 2x3 + x2 - 18x - 15 so that the resulting polynomial is exactly divisible by x2 - 1 - x.

Hence, Quotient =x2 + 3x + 5 and Remainder = 10x 10

— End of Algebraic Expression and Identities Exe-10.4 Class 8 ICSE Maths Solutions :–

Return to : ML Aggarwal Maths Solutions for ICSE Class -8

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