{"id":27907,"date":"2021-07-02T14:08:39","date_gmt":"2021-07-02T08:38:39","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27907"},"modified":"2021-07-02T14:08:39","modified_gmt":"2021-07-02T08:38:39","slug":"ncert-solutions-for-class-8-maths-chapter-9-ex-9-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-9-ex-9-4\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4<\/h2>\n

Question 1.
\nMultiply the binomials.
\n(i) (2x + 5) and (4x – 3)
\n(ii) (y – 8) and (3y – 4)
\n(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
\n(iv) (a+ 3b) and (x + 5)
\n(v) (2pq + 3q2<\/sup>) and (3pq -2q2<\/sup>)
\n(vi) \\(\\left(\\frac{3}{4} a^{2}+3 b^{2}\\right)\\) and \\(\\left(a^{2}-\\frac{2}{3} b^{2}\\right)\\)
\nAnswer:
\n(2x + 5) (4x – 3) = 2x (4x – 3) + 5 (4x – 3)
\n= 2x \u00d7 4x + 2x (-3) + 5 \u00d7 4x + 5 (- 3)
\n= 8x2<\/sup> – 6x + 20x – 15
\n= 8x2<\/sup> + 14x – 15<\/p>\n

(ii) (y – 8) (3y – 4) = y (3y – 4) – 8 (3y – 4)
\n= y \u00d7 3y + y (-4) – [8 \u00d7 3y + 8 (-4)]
\n= 3y2<\/sup> – 4y – (24y – 32)
\n= 3y2<\/sup> – 4y – 24y + 32
\n= 3y2<\/sup> – 28y + 32<\/p>\n

(iii) (2.5l – 0.5m) (2.5l + 0.5m)
\n= 2.5l (2.5l + 0.5m) – 0.5m (2.5l + 0.5m)
\n= 2.5l \u00d7 2.5l + 2.5l \u00d7 0.5m – [0.5m \u00d7 2.5l + 0.5m \u00d7 0.5m]
\n= 6.25l2<\/sup> + 1.25lm – 1.25lm – 0.25m2<\/sup>
\n= 6.25l2<\/sup> – 0.25m2<\/sup><\/p>\n

(iv) (a + 3b) (x + 5) = a \u00d7 (x + 5) + 3b \u00d7 (x + 5)
\n= ax + 5a + 3bx + 15b
\n= ax + 5a + 3bx + 15b<\/p>\n

(v) (2pq + 3q2<\/sup>) \u00d7 (3pq – 2q2<\/sup>)
\n= 2pq \u00d7 (3pq – 2q2<\/sup>) + 3q2<\/sup> \u00d7 (3pq – 2q2<\/sup>)
\n= 6p2<\/sup>q2<\/sup> – 4pq3<\/sup> + 9pq3<\/sup> – 6q4<\/sup>
\n= 6p2<\/sup>q2<\/sup> + 5pq3<\/sup> – 6q4<\/sup>
\n\"NCERT
\n\"NCERT
\n= 3a4<\/sup> -2a2<\/sup>b2<\/sup> +12a2<\/sup>b2<\/sup> – 8b4<\/sup>
\n= 3a4<\/sup> + 10a2<\/sup>b2<\/sup> – 8b4<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind the product.
\n(i) (5 – 2x) (3 + x)
\n(ii) (x + 7y) (7x – y)
\n(iii) (a2<\/sup> + b) (a + b2<\/sup>)
\n(iv) (p2<\/sup> – q2<\/sup>) (2p + q)
\nAnswer:
\n(i) (5 – 2x) (3 + x) = 5 \u00d7 (3 + x) – 2x (3 + x)
\n= 5 \u00d7 3 + 5 \u00d7 x – (2x) \u00d7 3 – 2x (x)
\n= 15 + 5x – 6x – 2x2<\/sup> = 15 – x – 2x2<\/sup><\/p>\n

(ii) (x + 7y)(7x – y) = x \u00d7 (7x – y) + 7y \u00d7 (7x – y)
\n= x \u00d7 7x – x \u00d7 y + 7y \u00d7 7x – 7y \u00d7 y
\n= 7x2<\/sup> – xy + 49xy – 7y2<\/sup>
\n= 7x2<\/sup> + 48xy – 7y2<\/sup><\/p>\n

(iii) (a2<\/sup> + b)(a + b2<\/sup>) = a2<\/sup> \u00d7 (a + b2<\/sup>) + b \u00d7 (a + b2<\/sup>)
\n= a2<\/sup> \u00d7 a + a2<\/sup> \u00d7 b2<\/sup> + b \u00d7 a + b \u00d7 b2<\/sup>
\n= a3<\/sup + a2<\/sup>b2<\/sup> + ab + b3<\/sup><\/sup><\/p>\n

(iv) (p2<\/sup> – q2<\/sup>) (2p + q)
\n= p2<\/sup> \u00d7 (2p + q) – q2<\/sup> (2p + q)
\n= p2<\/sup> \u00d7 2p + p2<\/sup> \u00d7 q – q2<\/sup> \u00d7 2p – q2<\/sup> \u00d7 q
\n= 2p3<\/sup> + p2<\/sup>q – 2pq2<\/sup> – q3<\/sup><\/p>\n

Question 3.
\nSimplify
\n(i) (x2<\/sup> – 5) (x + 5) + 25
\n(ii) (a2<\/sup> + 5) (b3<\/sup> + 3) + 5
\n(iii) (t + s2<\/sup>) (t2<\/sup> – s)
\n(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
\n(v) (x + y) (2x + y) + (x + 2y) (x – y)
\n(vi) (x + y) (x2<\/sup> – xy + y2<\/sup>)
\n(vii) (1.5x – 4y) (1.5x + 4y + 3) – 45x + 12y
\n(viii) (a + b + c) (a + b – c)
\nAnswer:
\n(i) (x2<\/sup> – 5) (x + 5) + 25
\n= x2<\/sup> \u00d7 (x + 5) – 5 (x + 5) + 25
\n= x2<\/sup>\u00d7 x + x2<\/sup> \u00d7 5 – 5 \u00d7 x – 5 \u00d7 5 + 25
\n= x3<\/sup> + 5x2<\/sup> – 5x – 25 + 25
\n= x3<\/sup> + 5x2<\/sup> – 5x<\/p>\n

(ii) (a2<\/sup> + 5) (b3<\/sup> + 3) + 5
\n= a2<\/sup> \u00d7 (b3<\/sup> + 3) + 5 \u00d7 (b3<\/sup> + 3) + 5
\n= a2<\/sup>b3<\/sup> + 3a2<\/sup> + 5b3<\/sup> + 15 + 5
\n= a2<\/sup>b3<\/sup> + 3a2<\/sup> + 53<\/sup> + 20<\/p>\n

(iii) (t + s2<\/sup>) (t2<\/sup> – s) = t \u00d7 (t2<\/sup> – s) + s2<\/sup> \u00d7 (t2<\/sup> – s)
\n= t \u00d7 t2<\/sup> – t \u00d7 s + s2<\/sup> \u00d7 t2<\/sup> – S2<\/sup> \u00d7 s
\n= t3<\/sup> – ts + s2<\/sup>t2<\/sup> – s3<\/sup><\/p>\n

(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
\n= a \u00d7 (c – d) + b \u00d7 (c – d) + a \u00d7 (c + d) – b \u00d7 (c + d ) + 2 \u00d7 ac + 2 \u00d7 bd
\n= a \u00d7 c – a \u00d7 d+ b \u00d7 c – b \u00d7 d + a \u00d7 c + a \u00d7 d – b \u00d7 c – b \u00d7 d + 2ac + 2bd
\n= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
\n= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2bd
\n= 4ac – 2bd + 2bd = 4ac<\/p>\n

(v) (x + y) (2x + y) + (x + 2y) (x – y)
\n= x \u00d7 (2x + y) + y (2x + y) + x \u00d7 (x – y) + 2y \u00d7 (x – y)
\n= x \u00d7 2x + x \u00d7 y + y \u00d7 2x + y \u00d7 y + x \u00d7 x + x \u00d7 (-y) + 2y \u00d7 x + 2y \u00d7 (-y)
\n= 2x2<\/sup> + xy + 2xy + y2<\/sup> + x2<\/sup> – xy + 2xy – 2y2<\/sup> = 2x2<\/sup> + x2<\/sup> + xy + 2xy + 2xy – xy + y2<\/sup> – 2y2<\/sup>
\n= 3x2<\/sup> + 5xy – xy – y2<\/sup>
\n= 3x2<\/sup> + 4xy – y2<\/sup><\/p>\n

\"NCERT<\/p>\n

(vi) (x + y) (x2<\/sup> – xy + y2<\/sup>)
\n= x \u00d7 (x2<\/sup> – xy + y2<\/sup>) + y \u00d7 (x2<\/sup> – xy + y2<\/sup>)
\n= x \u00d7 x2<\/sup> + x \u00d7 (-xy) + x \u00d7 y2<\/sup> + y \u00d7 x2<\/sup> + y \u00d7 (-xy) + y \u00d7 y2<\/sup>
\n= x3<\/sup> – x2<\/sup>y + xy2<\/sup> + x2<\/sup>y – xy2<\/sup> + y3<\/sup>
\n= x3<\/sup> – x2<\/sup>y + x2<\/sup>y + xy2<\/sup> – xy2<\/sup> + y3<\/sup>
\n= x3<\/sup> + y3<\/sup><\/p>\n

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
\n= 1.5x \u00d7 (1.5x + 4y + 3) – 4y \u00d7 (1.5x + 4y + 3) – 4.5x + 12y
\n= 1.5x \u00d7 1.5x + 1.5x \u00d7 4y + 1.5x \u00d7 3 – 4y \u00d7 1.5x – 4y \u00d7 4y – 4y \u00d7 3 – 4.5x + 12y
\n= 2.25xz + 6xy + 4.5x – 6xy – 16y2
\n– 12y – 4.5x + 12y
\n= 2.25x2<\/sup> + 6xy – 6xy + 4.5x – 4.5x – 16y2<\/sup> – 12y + 12y
\n= 2.25x2<\/sup> + 0 + 0 – 16y2<\/sup> + 0
\n= 2.25x2<\/sup> – 16y2<\/sup><\/p>\n

(viii) (a + b + c) (a + b – c)
\n= a \u00d7 (a + b – c) + b \u00d7 (a + b – c) + c \u00d7 (a + b – c)
\n= a \u00d7 a + a \u00d7 b + a \u00d7 -c + b \u00d7 a + b \u00d7 b – b \u00d7 c + c \u00d7 a + b \u00d7 c – c \u00d7 c
\n= a2<\/sup> + ab – ac + ab + b2<\/sup> – bc + ac + bc – c2<\/sup>
\n= a2<\/sup> + ab + ab – ac + ac – bc + bc + b2<\/sup> – c2<\/sup>
\n= a2<\/sup> + 2ab + o + o + b2<\/sup> – c2<\/sup>
\n= a2<\/sup> + b2<\/sup> – c2<\/sup> + 2ab<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4 Question 1. Multiply the binomials. (i) (2x + 5) and (4x – 3) (ii) (y …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-9-ex-9-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4 Question 1. Multiply the binomials. 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