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) –³⁄₄a²b³ by ⁶⁄₇  a³b²

Answer:

(i) – 39pq²r⁵ (÷) – 24p³q³r

= – 39pq²r⁵/ – 24p³q³r

(ii) –³⁄₄a²b³ by ⁶⁄₇  a³b²

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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

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

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

∴ Quotient = 3x + 5 and remainder = 0

(ii) 1 + y³ by 1 + y

∴ 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

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

(iv) x³ – 6x² + 12x – 8 by x – 2

∴ 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

∴ Quotient = 2x² + 5x + 3 and remainder = – 4
(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

∴ Quotient = a + 1 and remainder = 0.
(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

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.

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

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

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