{"id":46881,"date":"2022-06-01T06:30:45","date_gmt":"2022-06-01T01:00:45","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=46881"},"modified":"2023-01-25T11:18:36","modified_gmt":"2023-01-25T05:48:36","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-4-chapter-test","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-4-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test<\/h2>\n

Factorize the following (1 to 12)
\nQuestion 1.
\n15(2x – 3)3<\/sup> – 10(2x – 3)
\nAnswer:
\n(i) 15(2x – 3)3<\/sup> – 10(2x – 3)
\n= 5(2x – 3) [3(2x – 3)2<\/sup> – 2]
\n= 5(2x – 3)[3(2x – 3)2<\/sup> – 2]<\/p>\n

(ii) a(b – c) (b + c) – d (c – b)
\nAnswer:
\na(b – c) (b + c) – d (c – b)
\n= a(b – c) (b + c) + d (b – c)
\n= (b – c) [a (b + c) + d]
\n= (b – c) (ab + ac + d)<\/p>\n

Question 2.
\n(i) 2a2<\/sup>x – bx + 2a2<\/sup> – b
\nAnswer:
\n(i) 2a2<\/sup>x – bx + 2a2<\/sup> – b
\n= 2a2<\/sup>x + 2a2<\/sup> – bx – b
\n= 2a2<\/sup> (x + 1) – b (x + 1)
\n= (x + 1) (2a2<\/sup> – b)<\/p>\n

(ii) p2<\/sup> – (a + 2b) p + 2ab
\nAnswer:
\np2<\/sup> – (a + 2b) p + 2ab
\n= p2<\/sup> – ap – 2bp + 2ab
\n= p(p – a) – 2b (p – a)
\n= (p – a) (p – 2b)<\/p>\n

Question 3.
\n(i) (x2<\/sup> – y2<\/sup>) z + (y2<\/sup> – z2<\/sup>) x
\nAnswer:
\n(i) (x2<\/sup> – y2<\/sup>) z + (y2<\/sup> – z2<\/sup>) x
\n= x2<\/sup>z – y2<\/sup>z + xy2<\/sup> – z2<\/sup> x
\n= x2<\/sup>z – z2<\/sup>x + xy2<\/sup> – y2<\/sup>z
\n= xz (x – z) + y2<\/sup> (x – z)
\n= (x – 2) (xz + y2<\/sup>)<\/p>\n

(ii) 5a4<\/sup> – 5a3<\/sup> + 30a2<\/sup> – 30a
\nAnswer:
\n5a4<\/sup> – 5a3<\/sup> + 30a2<\/sup> – 30a
\n= 5a [a3<\/sup> – a2<\/sup> + 6a – 6]
\n= 5a [a2<\/sup> (a – 1) + 6 (a – 1)]
\n= 5a (a – 1) (a2<\/sup> + 6)<\/p>\n

Question 4.
\n(i) b(c – d)2<\/sup> + a(d – c) + 3c – 3d
\nAnswer:
\nb(c – d)2<\/sup> + a(d – c) + 3c – 3d
\n= b(c – d)2<\/sup> – a(c – d) + 3c – 3d
\n= b(c – d)2<\/sup> – a (c – d) + 3 (c – d)
\n= (c – d) [b (c – d) – a + 3]
\n= (c – d) (bc – bd – a + 3)<\/p>\n

(ii) x3<\/sup> – x2<\/sup> – xy + x + y – 1
\nAnswer:
\nx3<\/sup> – x2<\/sup> – xy + x + y – 1
\n= x3<\/sup> – x2<\/sup> – xy + y + x – 1.
\n= x2<\/sup> (x – 1) – y (x – 1) + 1(x – 1)
\n= (x – 1) (x2<\/sup> – y + 1)<\/p>\n

Question 5.
\n(i) x(x + z) – y (y + z)
\nAnswer:
\nx (x + z) – y(y + z)
\n= x2<\/sup> + xz – y2<\/sup> – yz
\n= x2<\/sup> – y2<\/sup> + xz – yz
\n= (x + y)(x – y) + z(x – y) {\u2235 x2<\/sup> – y2<\/sup> = (x + y)(x – y)}
\n= (x – y)(x + y + z)<\/p>\n

(ii) a12<\/sup>x4<\/sup> – a4<\/sup>x12<\/sup>
\nAnswer:
\na12<\/sup>x4<\/sup> – a4<\/sup>x12<\/sup> = a4<\/sup>x4<\/sup> (a8<\/sup> – x8<\/sup>)
\n= a4<\/sup>x4<\/sup> {(a4<\/sup>)2<\/sup> – (x4<\/sup>)2<\/sup>}
\n= a4<\/sup>x4<\/sup> (a4<\/sup> + x4<\/sup>) (a4<\/sup> – x4<\/sup>) (\u2235 a2<\/sup> – b2<\/sup> = (a + b) (a – b)}
\n= a4<\/sup>x4<\/sup> (a4<\/sup> + x4<\/sup>) {(a2<\/sup>)2<\/sup> – (x2<\/sup>)2<\/sup>}
\n= a4<\/sup>x4<\/sup> (a4<\/sup> + x4<\/sup>) (a2<\/sup> + x2<\/sup>) (a2<\/sup> – x2<\/sup>)
\n= a4<\/sup>x4<\/sup> (a4<\/sup> + x4<\/sup>) (a2<\/sup> + x2<\/sup>) (a + x) (a – x)<\/p>\n

Question 6.
\n(i) 9x2<\/sup> + 12x + 4 – 16y2<\/sup>
\nAnswer:
\n9x2<\/sup> + 12x + 4 – 16y2<\/sup>
\n\u21d2 (3x)2<\/sup> + 2 \u00d7 3x \u00d7 2 + (2)2<\/sup> – 16y2<\/sup>
\n\u21d2 (3x + 2)2<\/sup> + (4y)2<\/sup>
\n\u21d2 (3x + 2 + 4y) (3x + 2 – 4y)<\/p>\n

(ii) x4<\/sup> + 3x2<\/sup> + 4
\nAnswer:
\nx4<\/sup> + 3x2<\/sup> + 4
\n= (x2<\/sup>)3<\/sup> + 3(x2<\/sup>) + 4
\n= (x2<\/sup>)2<\/sup> + (2)2<\/sup> + 4x2<\/sup> – x2<\/sup>
\n= (x2<\/sup> + 2)2<\/sup> – (x)2<\/sup> {\u2235 a2<\/sup> – b2<\/sup> = (a + b) (a – b)}
\n= (x2<\/sup> + 2 + x) (x2<\/sup> + 2 – x)
\n= (x2<\/sup> + x + 2) (x2<\/sup> – x + 2)<\/p>\n

Question 7.
\n(i) 21x2<\/sup> – 59xy + 40y2<\/sup>
\nAnswer:
\n(i) 21x2<\/sup> – 59xy + 40y2<\/sup>
\n\u2235 21 \u00d7 40 = 840
\n\u2234 840 = (- 35) (- 29)
\nand – 59 = – 35 – 24
\n= 7x (3x – 5y) – 8y (3x – 5y)
\n= (3x – 5y) (7x – 8y)<\/p>\n

(ii) 4x3<\/sup>y – 44x2<\/sup>y + 112xy
\nAnswer:
\n4x3<\/sup>y – 44x2<\/sup>y + 112xy
\n= 4xy (x2<\/sup> – 11x + 28)
\n= 4xy [x2<\/sup> – 7x – 4x + 28]
\n\u2235 – 11 = -7 – 4
\n28 = (-7) (-4)
\n= 4xy [x (x – 7) – 4 (x – 7)]
\n= 4xy(x – 7) (x – 4)<\/p>\n

Question 8.
\n(i) x2<\/sup>y2<\/sup> – xy – 72
\nAnswer:
\nx2<\/sup>y2<\/sup> – xy – 72
\n= x2<\/sup>y2<\/sup> – 9xy + 8xy – 72 { \u2235 – 72 = -9 \u00d7 8l
\n-1 = -9 + 8 }
\n= xy(xy – 9) + 8 (xy – 9)
\n= (xy – 9) (xy + 8)<\/p>\n

(ii) 9x3<\/sup>y + 41x2<\/sup>y2<\/sup> + 20xy3<\/sup>
\nAnswer:
\n9x3<\/sup>y + 41x2<\/sup>y2<\/sup> + 20xy3<\/sup>
\n= xy (9x2<\/sup> + 41xy + 20y2<\/sup>)
\n= xy {9x2<\/sup> + 36xy + 5xy + 2oy2<\/sup>}
\n{\u2235 9 \u00d7 20 = 180
\n180 = 36 \u00d7 5
\n41 = 36 + 5}
\n= xy {9x (x + 4y) + 5y (x + 4y)}
\n= xy (x + 4y) (9x + 5y)<\/p>\n

Question 9.
\n(i) (3a – 2b)2<\/sup> + 3 (3a – 2b) – 10
\nAnswer:
\n(3a – 2b)2<\/sup> + 3 (3a – 2b) – 10
\n= x2<\/sup> + 3x – 10 where x = (3a- 2b)
\n= (x2<\/sup> + 5x) – (2x +10)
\n= x (x + 5) – 2 (x + 5)
\n= (x + 5) (x – 2)
\n= (3a – 2b + 5) (3a – 2b – 2)<\/p>\n

(ii) (x2<\/sup> – 3x) (x2<\/sup> – 3x + 7) + 10
\nAnswer:
\n(x2<\/sup> – 3x) (x2<\/sup> – 3x + 7) + 10
\nPut x2<\/sup> – 3x = 3
\nthen, y(y + 7) + 10
\n= y2<\/sup> + 7y + 10
\n= y2<\/sup> + 5y + 2y + 10
\n= y(y + 5) + 2(y + 5)
\n= (y + 5) (y + 2)
\n= (x2<\/sup> – 3x + 5) (x2<\/sup> – 3x + 2)<\/p>\n

Question 10.
\n(i) (x2<\/sup> – x) (4x2<\/sup> – 4x – 5) – 6
\nAnswer:
\n(x2<\/sup> – x) (4x2<\/sup> – 4x – 5) – 6
\n= (x2<\/sup> -x) [4 (x2<\/sup> – x) – 5] – 6
\n= y(4y – 5) – 6 where y = x2<\/sup> – x
\n= 4y2<\/sup> – 5y – 6
\n= 4y – 8y + 3y – 6
\n= 4y (y – 2) + 3 (y – 2)
\n= (4y + 3)(y – 2)
\n= (4x2<\/sup> – 4x + 3) (x2<\/sup> – x – 2)
\n= (x2<\/sup> – x – 2) (4x2<\/sup> – 4x + 3)
\n= (x2<\/sup> – 2x + x – 2) (4x2<\/sup> – 4x + 3)
\n= [x(x – 2) + 1 (x – 2)] (4x2<\/sup> – 4x + 3)
\n= (x – 2) (x +1) (4x2<\/sup> – 4x + 3)<\/p>\n

(ii) x4<\/sup> + 9x2<\/sup>y2<\/sup> + 81y4<\/sup>
\nAnswer:
\nx4<\/sup> + 9x2<\/sup>y2<\/sup> + 81y4<\/sup>
\n= (x4<\/sup> + 18x2<\/sup>y2<\/sup> + 81y4<\/sup>) – 9x2<\/sup>y2<\/sup>
\n= (x2<\/sup> + 9y2<\/sup>)2<\/sup> – (3xy)2<\/sup>
\n= (x2<\/sup> + 9y2<\/sup> + 3xy) (x2<\/sup> + 9y2<\/sup> – 3xy)<\/p>\n

Question 11.
\n(i) \\(\\frac{8}{27} x^{3}-\\frac{1}{8} y^{3}\\)
\nAnswer:
\n\"ML<\/p>\n

(ii) x6<\/sup> + 63x3<\/sup> – 64
\nAnswer:
\nx6<\/sup> + 63x3<\/sup> – 64 = x6<\/sup> + 64x3<\/sup> – x – 64
\n= x3<\/sup> (x3<\/sup> + 64) – 1 (x3<\/sup> + 64)
\n= (x3<\/sup> + 64) (x3<\/sup> – 1)
\n= [(x)3<\/sup> + (4)3<\/sup>][(x)3<\/sup> – (1)3<\/sup>]
\n= (x + 4) [(x)2<\/sup> – (4) (x) + (4)2<\/sup>] (x – 1) [(x)2<\/sup> + (1)(x) + (1)2<\/sup>]
\n= (x + 4) (x2<\/sup> – 4x + 16) (x – 1) (x2<\/sup> + x + 1)<\/p>\n

Question 12.
\n(i) \\(x^{3}+x^{2}-\\frac{1}{x^{2}}+\\frac{1}{x^{3}}\\)
\nAnswer:
\n\"ML<\/p>\n

(ii) (x + 1)6<\/sup> – (x – 1)6<\/sup>
\nAnswer:
\n(x + 1)6<\/sup> – (x – 1)6<\/sup>
\n[(X + 1)3<\/sup>]2<\/sup> – [(x – 1)3<\/sup>]2<\/sup> [(a2<\/sup> – b2<\/sup> = (a + b) (a – b)
\n= [(x + 1)3<\/sup> + (x – 1)3<\/sup>] [(x + 1)3<\/sup> – (x – 1)3<\/sup>]
\n= [(x + 1) + (x – 1)]
\n[(x+ 1)2<\/sup> – (x – 1)(x- 1) + (x- 1)2<\/sup>] [(x + 1) – (x – 1)] [(x + 1)2<\/sup> + (x + 1) (x – 1) + (x – 1)2<\/sup>]
\n= (x + 1 + x – 1) [x2<\/sup> + 2x + 1 – x2<\/sup> + 1 + x2<\/sup> + 1 – 2x (x + 1) – x + 1) [x2<\/sup> + 2x + 1 + x2<\/sup> – 1 + x2<\/sup> – 2x + 1 ]
\n= 2x (x2<\/sup> + 3) (2) (3x2<\/sup> + 1)
\n= 4x (x2<\/sup> + 3) (3x2<\/sup> + 1)<\/p>\n

Question 13.
\nShow that (97)3<\/sup> + (14)3<\/sup> is divisible by 111.
\nAnswer:
\n(97)3<\/sup> + (14)3<\/sup>
\n= (97 + 14) [(97)2<\/sup> – (97) (14) + (14)2<\/sup>]
\n= 111 \u00d7 [(97)2<\/sup> – (97)(14) + (14)2<\/sup>]
\nClearly given expression is divisible by 111.<\/p>\n

Question 14.
\nIf a + b = 8 and ab = 15, find the value of a4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup>
\nSolution:
\na4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup>
\n= a4<\/sup> + 2a2<\/sup>b2<\/sup> + b4<\/sup> – a2<\/sup>b2<\/sup>
\n= (a2<\/sup>)2<\/sup> + 2a2<\/sup>b2<\/sup> + (b2<\/sup>)2<\/sup> – (ab)2<\/sup>
\n= (a2<\/sup> + b2<\/sup>)2<\/sup> – (ab)2<\/sup>
\n= (a2<\/sup> + b2<\/sup> + ab) (a2<\/sup> + b – ab)
\nBut a + b = 8, ab = 15
\n\u2234 (a + b)22<\/sup> = 82<\/sup>
\n\u21d2 a2<\/sup> + b2<\/sup> + 2ab = 64
\n\u21d2 a2<\/sup> + b2<\/sup> + 2 \u00d7 15 = 64
\n\u21d2 a2<\/sup> + b2<\/sup> + 30 = 64
\na2<\/sup> + b2<\/sup> = 64 – 30 = 34
\nNow a4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup>
\n= (a2<\/sup> + b2<\/sup> + ab) (a2<\/sup> + b2<\/sup> – ab)
\n= (34 + 15) (34 – 15)
\n= 49 \u00d7 19 = 931<\/p>\n

ML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test Factorize the following (1 to 12) Question 1. 15(2x – 3)3 – 10(2x – 3) Answer: (i) 15(2x – 3)3 – 10(2x – 3) = 5(2x – 3) [3(2x – 3)2 – 2] = 5(2x – 3)[3(2x – 3)2 – 2] (ii) … 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":"\nML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-4-chapter-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test Factorize the following (1 to 12) Question 1. 15(2x – 3)3 – 10(2x – 3) Answer: (i) 15(2x – 3)3 – 10(2x – 3) = 5(2x – 3) [3(2x – 3)2 – 2] = 5(2x – 3)[3(2x – 3)2 – 2] (ii) ... 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