# ML Aggarwal Algebraic Expression and Identities Exe-10.3 Class 8 ICSE Ch-10 Maths Solutions

ML Aggarwal Algebraic Expression and Identities Exe-10.3 Class 8 ICSE Ch-10 Maths Solutions. We Provide Step by Step Answer of  Exe-10.3 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-10.3 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-10.3 Questions Edition 2023-2024

### Algebraic Expression and Identities Exe-10.3

ML Aggarwal Class 8 ICSE Maths Solutions

Page-178

#### Question 1. Multiply:

(i) (5x – 2) by (3x + 4)
(ii) (ax + b) by (cx + d)
(iii) (4p – 7) by (2 – 3p)
(iv) (2x2 + 3) by (3x – 5)
(v) (1.5a – 2.5b) by (1.5a + 2.56)
(vi) {(3/7)p2+4q2} by 7{p2 – (3/4)q2}

(i) (5x – 2) by (3x + 4)

= (5x – 2) × (3x + 4)

= 5x (3x + 4) – 2 (3x + 4)

= 15x2 + 20x – 6x – 8

= 15x2+ 14x – 8

(ii) (ax + b) by (cx + d)

= (ax + b) × (cx + d)

= ax (cx + d) + b (cx + d)

= acx2 + adx + bcx + bd

(iii) (4p – 7) by (2 – 3p)

= (4p – 7) × (2 – 3p)

= 4p(2 – 3p) -7(2 – 3p)

= 8p – 12p2 – 14 + 21p

= 29p – 12p2 – 14

(iv) (2x2 + 3) by (3x – 5)

= (2x2 + 3) (3x – 5)

= 2x2(3x – 5) + 3(3x – 5)

= 6x3 – 10x2 + 9x – 15

(v) (1.5a – 2.5b) by (1.5a + 2.5b)

= (1.5a – 2.5b) (1.5a + 2.5b)

= 1.5a(1.5 + 2.5b) – 2.5b(1.5a + 2.5b)

= 2.25a2 + 3.75ab – 3.75a6 – 6.25b2

= 2.25a2 – 6.25b2

(vi) {(3/7)p2+4q2} by 7{p2 – (3/4)q2}

#### Question 2. Multiply:

(i) (x – 2y + 3) by (x + 2y)
(ii) (3 – 5x + 2 × 2) by (4x – 5)

(i) (x – 2y + 3) by (x + 2y)

= (x – 2y + 3) × (x + 2y)

= x (x + 2y) – 2y(x + 2y) + 3 (x + 2y)

= x2 + 2xy – 2xy – 4y2 + 3x + 6y

= x2 – 4y2 + 3x + 6y

(ii) (3 – 5x + 2x2) by (4x – 5)

= (4x – 5) (3 – 5x + 2x2)

= 4x(3 – 5x + 2x2) – 5(3 – 5x + 2x2)

= 12x – 20x2 + 8x3 – 15 + 25x – 10x2

= 8x3 – 30x2 + 37x – 15

#### Question 3. Multiply:

(i) (3x2 – 2x – 1) by (2x2 + x – 5)
(ii) (2 – 3y – 5y2) by (2y – 1 + 3y2)

(i) (3x2 – 2x – 1) by (2x2 + x – 5)

= (3x2 – 2x – 1) (2x2 + x – 5)

= 3x2(2x2 + x – 5) – 2x(2x2 + x – 5) -1(2x2 + x – 5)

= 6x4 + 3x3 – 15x2 – 4x3 – 2x2 + 10x – 2x2 – x + 5

= 6x4 – x3 – 19x2 + 9x + 5

(ii) (2 – 3y – 5y2) by (2y- 1 + 3y2)

= (2 – 3y – 5y2) × (2y- 1 + 3y2)

= 2(2y – 1 + 3y2 ) – 3y (2y – 1 + 3y2) -5y2(2y – 1 + 3y2)

= 4y – 2 + 6y2 – 6y2 + 3y – 9y3 – 10y3 + 5y2 – 15y4

= -15y4 – 19y3 + 5y2 + 7y – 2

#### Question 4. Simplify:

(i) (x2 + 3) (x – 3) + 9
(ii) (x + 3) (x – 3) (x + 4) (x – 4)
(iii) (x + 5) (x + 6) (x + 7)
(iv) (p + q – 2r) (2p – q + r) – 4qr
(v) (p + q) (r + s) + (p – q)(r – s) – 2(pr + qs)
(vi) (x + y + z) (x – y + z) + (x + y – z) (-x + y + z) – 4zx

(i) (x2 + 3) (x – 3) + 9

= x2 (x – 3) + 3(x – 3) + 9

= x2 – 3x2 + 3x – 9 + 9

= x3 – 3x2 + 3x

(ii) (x + 3) (x – 3) (x + 4) (x – 4)

= {(x + 3) (x – 3)} × {(x + 4) (x – 4)}

= {x (x – 3) + 3 (x – 3)} {x (x – 4) + 4 (x – 4)}

= (x2 – 3x + 3x – 9) {x2 – 4x + 4x – 16}

= (x2 – 9) (x2 – 16)

= x2 (x2 – 16) – 9 (x2 – 16)

= x4 – 16x2 – 9x2 + 144

= x4 – 25x2 + 144

(iii) (x + 5) (x + 6) (x + 7)

= {(x + 5) × (x + 6)} (x + 7)

= (x2 + 6x + 5x + 30) (x + 7)

= (x2 + 11x + 30) (x + 7)

= x(x2+ 11x + 30) + 7(x2+ 11x + 30)

= x3 + 11x2 + 30x + 7x2 + 77x + 210

= x3 + 18x2 + 107x + 210

(iv) (p + q – 2r)(2p – q + r) – 4qr

= p(2p – q + r) + q(2p – q + r) – 2r(2p – q + r) – 4qr

= 2p2 – pq + pr + 2pq – q2 + qr – 4pr + 2qr – 2r2 – 4qr

= 2p2 – q2 – 2r2 + pq – 3pr – 2qr

(v) (p + q)(r + s) + (p – q) (r – s) – 2(pr + qs)

= (pr + ps + qr + qs) + (pr – ps – qr + qs) – 2pr – 2qs

= 0

(vi) (x + y + z)(x – y + z) + (x + y – z)(-x + y + z) – 4zx

= x2 – xy + xz + xy – y2 + yz + xz – yz + z2 – x2 + xy + xz

– xy + x2 + yx + xz – yz – z2 – 4zx

= 0

#### Question 5. If two adjacent sides of a rectangle are 5x2 + 25xy + 4y2 and 2x2 – 2xy + 3y2, find its area.

The adjacent sides of a rectangle are 5x2 + 25xy + 4y2 and 2x2 – 2xy + 3y2

Area of rectangle = Product of two adjacent sides

= (5x2 + 25xy + 4y2) (2x2 – 2xy + 3y2)

= 10x4– 10x3y+ 15x2y2 + 50x3y – 50x2y2 + 75xy3 + 8x2y2 – 8xy3 + 12y4

= 10x4 + 40x3y – 27x2y2 + 67xy3 + 12y4

Hence, the area of the rectangle is 10x4 + 40x3y – 27x2y2 + 67xy3 + 12y4.

— End of Algebraic Expression and Identities Exe-10.3 Class 8 ICSE Maths Solutions :–