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) – 39pq^{2}r^{5} by – 24p^{3}q^{3}r

(ii) –^{3}⁄_{4}a^{2}b^{3} by ^{6}⁄_{7 }a^{3}b^{2}

**Answer:**

**(i) – 39pq ^{2}r^{5} (÷) – 24p^{3}q^{3}r**

= – 39pq^{2}r^{5}/ – 24p^{3}q^{3}r

**(ii) – ^{3}⁄_{4}a^{2}b^{3} by ^{6}⁄_{7 }a^{3}b^{2}**

**Algebraic Expression and Identities Exe-10.4**

**ML Aggarwal Class 8 ICSE Maths Solutions**

Page-181

**Question 2. Divide:**

(i) 9x^{4} – 8x^{3} – 12x + 3 by 3x

(ii) 14p^{2}q^{3} – 32p^{3}q^{2} + 15pq^{2} – 22p + 18q by – 2p^{2}q.

**Answer:**

**(i) 9x ^{4} – 8x^{3} – 12x + 3 by 3x**

**(ii) 14p ^{2}q^{3} – 32p^{3}q^{2} + 15pq^{2} – 22p + 18q by – 2p^{2}q.**

**Question 3. Divide:**

(i) 6x^{2} + 13x + 5 by 2x + 1

(ii) 1 + y^{3} by 1 + y

(iii) 5 + x – 2x^{2} by x + 1

(iv) x^{3} – 6x^{2} + 12x – 8 by x – 2

**Answer:**

**(i) 6x ^{2} + 13x + 5 by 2x + 1**

∴ Quotient = 3x + 5 and remainder = 0

**(ii) 1 + y ^{3} by 1 + y**

∴ Quotient = y^{2} – y + 1 and remainder = 0

**(iii) 5 + x – 2x ^{2} by x + 1**

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

**– 2x ^{2} + x + 5 ÷ x + 1**

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

**(iv) x ^{3} – 6x^{2} + 12x – 8 by x – 2**

∴ Quotient = x^{2} – 4x + 4 and remainder = 0

**Question 4. Divide:**

(i) 6x^{3} + x^{2} – 26x – 25 by 3x – 7

(ii) m^{3} – 6m^{2} + 7 by m – 1

**Answer:**

**(i) 6x ^{3} + x^{2} – 26x – 25 by 3x – 7**

∴ Quotient = 2x^{2} + 5x + 3 and remainder = – 4

**(ii) m ^{3} – 6m^{2} + 7 by m – 1**

∴ Quotient = m^{2} – 5m – 5 and remainder = 2.

**Question 5. Divide:**

(i) a^{3} + 2a^{2} + 2a + 1 by a^{2} + a + 1

(ii) 12x^{3} – 17x^{2} + 26x – 18 by 3x^{2} – 2x + 5

**Answer:**

**(i) a ^{3} + 2a^{2} + 2a + 1 by + a^{2} + a + 1**

∴ Quotient = a + 1 and remainder = 0.

**(ii) 12x ^{3} – 17x^{2} + 26x – 18 by 3x^{2} – 2x + 5**

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

**Question 6. If the area of a rectangle is 8x**^{2} – 45y^{2} + 18xy and one of its sides is 4x + 15y, find the length of the adjacent side.

^{2}– 45y

^{2}+ 18xy and one of its sides is 4x + 15y, find the length of the adjacent side.

**Answer:**

Area of rectangle = 8x^{2} – 45y^{2} + 18xy

And, one side = 4x + 15y

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

= 8x^{2} – 45y^{2} + 18xy ÷ 4x + 15y

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

**Question 7. What polynomial be subtracted from the polynomial x**^{4} + 2x^{3} + x^{2} – 18x – 15 so that the resulting polynomial is exactly divisible by x^{2} – 1 – x.

^{4}+ 2x

^{3}+ x

^{2}– 18x – 15 so that the resulting polynomial is exactly divisible by x

^{2}– 1 – x.

**Answer:**

**x ^{4} + 2x^{3} + x^{2} – 18x – 15 **by

**x**

^{2}– 1 – x.**Hence, Quotient =x ^{2 }+ 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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