{"id":44473,"date":"2022-05-30T15:30:29","date_gmt":"2022-05-30T10:00:29","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44473"},"modified":"2023-01-25T11:36:54","modified_gmt":"2023-01-25T06:06:54","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-10-ex-10-3\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.3"},"content":{"rendered":"
Question 1. (ii) (ax + b) by (cx + d) (iii) (4p – 7) by (2 – 3p) (iv) (2x2<\/sup> + 3) by (3x – 5) (v) (1.5a – 2.5b) by (1.5a + 2.5b) Question 2. (ii) (3 – 5x + 2x2<\/sup>) by (4x – 5) Question 3. (ii) (2 – 3y – 5y2<\/sup>) by (2y- 1 + 3y2<\/sup>) Question 4. (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 Question 5. ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.3 Question 1. Multiply: (i) (5x – 2) by (3x + 4) (ii) (ax + b) by (cx + d) (iii) (4p – 7) by (2 – 3p) (iv) (2×2 + 3) by (3x – 5) (v) (1.5a – 2.5b) … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\n
\nMultiply:
\n(i) (5x – 2) by (3x + 4)
\n(ii) (ax + b) by (cx + d)
\n(iii) (4p – 7) by (2 – 3p)
\n(iv) (2x2<\/sup> + 3) by (3x – 5)
\n(v) (1.5a – 2.5b) by (1.5a + 2.56)
\n(vi) \\(\\left(\\frac{3}{7} p^{2}+4 q^{2}\\right) \\text { by } 7\\left(p^{2}-\\frac{3}{4} q^{2}\\right)\\)
\nSolution:
\n(i) (5x – 2) by (3x + 4)
\n= 5x (3x + 4) – 2 (3x + 4)
\n15x2<\/sup> + 20x – 6x – 8
\n= 15x2<\/sup>+ 14x – 8<\/p>\n
\n= ax (cx + d) + b (cx + d)
\n= acx2<\/sup> + adx + bcx + bd<\/p>\n
\n= (4p – 7) (2 -3p)
\n= 4p(2 – 3p) -7(2 – 3p)
\n= 8p – 12p2<\/sup> – 14 + 21p
\n= 29p – 12p2<\/sup> – 14<\/p>\n
\n= (2x2<\/sup> + 3) (3x – 5)
\n= 2x2<\/sup>(3x – 5) + 3(3x – 5)
\n= 6x3<\/sup> – 10x2<\/sup> + 9x – 15<\/p>\n
\n= (1.5a – 2.5b) (1.5a + 2.5b)
\n= 1.5a(1.5 + 2.5b) – 2.5b(1.5a + 2.5b)
\n= 2.25a2<\/sup> + 3.75ab – 3.75a6 – 6.25b2<\/sup>
\n= 2.25a2<\/sup> – 6.25b2<\/sup>
\n<\/p>\n
\nMultiply:
\n(i) (x – 2y + 3) by (x + 2y)
\n(ii) (3 – 5x + 2 \u00d7 2) by (4x – 5)
\nSolution:
\n(i) (x – 2y + 3) by (x + 2y)
\n= x (x + 2y) – 2y(x + 2y) + 3 (x + 2y)
\n= x2<\/sup> + 2xy – 2xy – 4y2<\/sup> + 3x + 6y
\n= x2<\/sup> – 4y2<\/sup> + 3x + 6y<\/p>\n
\n= (4x – 5) (3 – 5x + 2x2<\/sup>)
\n= 4x(3 – 5x + 2x2<\/sup>) – 5(3 – 5x + 2x2<\/sup>)
\n= 12x – 20x2<\/sup> + 8x3<\/sup> – 15 + 25x – 10x2<\/sup>
\n= 8x3<\/sup> – 30x2<\/sup> + 37x – 15<\/p>\n
\nMultiply:
\n(i) (3x2<\/sup> – 2x – 1) by (2x2<\/sup> + x – 5)
\n(ii) (2 – 3y – 5y2<\/sup>) by (2y – 1 + 3y2<\/sup>)
\nSolution:
\n(i) (3x2<\/sup> – 2x – 1) by (2x2<\/sup> + x – 5)
\n= (3x2<\/sup> – 2x – 1) (2x2<\/sup> + x – 5)
\n= 3x2<\/sup>(2x2<\/sup> + x – 5) – 2x(2x2<\/sup> + x – 5) -1(2x2<\/sup> + x – 5)
\n= 6x4<\/sup> + 3x3<\/sup> – 15x2<\/sup> – 4x3<\/sup> – 2x2<\/sup> + 10x – 2x2<\/sup> – x + 5
\n= 6x4<\/sup> – x3<\/sup> – 19x2<\/sup> + 9x + 5<\/p>\n
\n= 2(2y – 1 + 3y2<\/sup> )- 3y (2y – 1 + 3y2<\/sup>) -5y2<\/sup>(2y – 1 + 3y2<\/sup>)
\n= 4y – 2 + 6y2<\/sup> – 6y2<\/sup> + 3y – 9y3<\/sup> – 10y3<\/sup> + 5y2<\/sup> – 15y4<\/sup>
\n= -15y4<\/sup> – 19y3<\/sup> + 5y2<\/sup> + 7y – 2<\/p>\n
\nSimplify:
\n(i) (x2<\/sup> + 3) (x – 3) + 9
\n(ii) (x + 3) (x – 3) (x + 4) (x – 4)
\n(iii) (x + 5) (x + 6) (x + 7)
\n(iv) (p + q – 2r) (2p – q + r) – 4qr
\n(v) (p + q) (r + s) + (p – q)(r – s) – 2(pr + qs)
\n(vi) (x + y + z) (x – y + z) + (x + y – z) (-x + y + z) – 4zx
\nSolution:
\n(i) (x2<\/sup> + 3) (x – 3) + 9
\n= x2<\/sup> (x – 3) + 3(x – 3) + 9
\n= x2<\/sup> – 3x2<\/sup> + 3x – 9 + 9
\n= x3<\/sup> – 3x2<\/sup> + 3x<\/p>\n
\n= {(x + 3 (x – 3)} {(x + 4) (x – 4)}
\n= {x (x – 3) + 3 (x – 3)} {x (x – 4) + 4 (x – 4)}
\n= (x2<\/sup> – 3x + 3x – 9) {x2<\/sup> – 4x + 4x – 16}
\n= (x2<\/sup> – 9) (x2<\/sup> – 16)
\n= x2<\/sup> (x2<\/sup> – 16) – 9 (x2<\/sup> – 16)
\n= x4<\/sup> – 16x2<\/sup> – 9x2<\/sup> + 144
\n= x4<\/sup> – 25x2<\/sup> + 144<\/p>\n
\n= (x2<\/sup> + 6x + 5x + 30) (x + 7)
\n= (x2<\/sup> + 11x + 30) (x + 7)
\n= x(x2<\/sup>+ 11x + 30) + 7(x2<\/sup>+ 11x + 30)
\n= x3<\/sup> + 11x2<\/sup> + 30x + 7x2<\/sup> + 77x + 210
\n= x3<\/sup> + 18x2<\/sup> + 107x + 210<\/p>\n
\n= p(2p – q + r) + q(2p – q + r) – 2r(2p – q + r) – 4qr
\n= 2p2<\/sup> – pq + pr + 2pq – q2<\/sup> + qr – 4pr + 2qr – 2r2<\/sup> – 4qr
\n= 2p2<\/sup> – q2<\/sup> – 2r2<\/sup> + pq – 3pr – 2qr<\/p>\n
\n= pr + ps + qr + qs + pr – ps – qr + qs – 2pr – 2qs = 0<\/p>\n
\n= x2<\/sup> – xy + xz + xy – y2<\/sup> + yz + xz – yz + z2<\/sup> – x2<\/sup> + xy + xz
\n– xy + x2<\/sup> + yx + xz – yz – z2<\/sup> – 4zx = 0<\/p>\n
\nIf two adjacent sides of a rectangle are 5x2<\/sup> + 25xy + 4y2<\/sup> and 2x2<\/sup> – 2xy + 3y2<\/sup>, find its area.
\nSolution:
\nAdjacents sides of a rectangle are
\n5x2<\/sup> + 25xy + 4y2<\/sup> and
\n2x2<\/sup> – 2xy + 3y2<\/sup>
\n\u2234 Area of rectangle = Product of two sides
\n= (5x2<\/sup> + 25xy + 4y2<\/sup>) (2x2<\/sup> – 2xy + 3y2<\/sup>)
\n= 10x4<\/sup>– 10x3<\/sup>y+ 15x2<\/sup>y2<\/sup> + 50x3<\/sup>y – 50x2<\/sup>y2<\/sup> + 75xy3<\/sup> + 8x2<\/sup>y2<\/sup> – 8xy3<\/sup> + 12y4<\/sup>
\n= 10x4<\/sup> + 40x3<\/sup>y – 27x2<\/sup>y2<\/sup> + 67xy3<\/sup> + 12y4<\/sup><\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"