{"id":28125,"date":"2021-07-03T15:18:52","date_gmt":"2021-07-03T09:48:52","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28125"},"modified":"2021-07-03T15:18:52","modified_gmt":"2021-07-03T09:48:52","slug":"ncert-solutions-for-class-8-maths-chapter-14-ex-14-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-14-ex-14-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 14 Factorization Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3<\/h2>\n

Question 1.
\nCarry out the following divisions.
\n(i) 28x4<\/sup> \u00f7 56x
\n(ii) – 36y3<\/sup> \u00f7 9y2<\/sup>
\n(iii) 66pq2<\/sup>r3<\/sup> \u00f7 11qr2<\/sup>
\n(iv) 34x3<\/sup>y3<\/sup>z3<\/sup> \u00f7 51xy2<\/sup>z3<\/sup>
\n(v) 12a8<\/sup>b8<\/sup> \u00f7 (- 6a6<\/sup>b4<\/sup>)
\nAnswer:
\n(i) 28x4<\/sup> \u00f7 56x
\n= \\(\\frac{2 \\times 2 \\times 7 \\times x^{4}}{2 \\times 2 \\times 2 \\times 7 \\times x}\\)
\n= \\(\\frac{x^{4-1}}{2}=\\frac{x^{3}}{2}=\\frac{1}{2} x^{2}\\)<\/p>\n

(ii) – 36y3<\/sup> \u00f7 9y2<\/sup>
\n= \\(\\frac{-2 \\times 2 \\times 3 \\times 3 \\times y^{3}}{3 \\times 3 \\times y^{2}}\\)
\n= – 4y3-2<\/sup>
\n= -4y<\/p>\n

\"NCERT<\/p>\n

(iii) 66pq2<\/sup>r3<\/sup> \u00f7 11qr2<\/sup>
\n= \\(\\frac{2 \\times 3 \\times 11 \\times p \\times q^{2} \\times r^{3}}{11 \\times q \\times r^{2}}\\)
\n= 2 x 3 x p x q2-1<\/sup> x r3-2<\/sup>
\n= 6 x p x q x r = 6pqr<\/p>\n

(iv) 34x2<\/sup>y2<\/sup>z2<\/sup> \u00f7 51xy2<\/sup>z3<\/sup>
\n= \\(\\frac{2 \\times 17 \\times x^{3} \\times y^{3} \\times z^{3}}{3 \\times 17 \\times x \\times y^{2} \\times z^{3}}\\)
\n= \\(\\frac{2}{3}\\) x x3-1<\/sup> x y3-2<\/sup> x z3-3<\/sup>
\n= \\(\\frac{2}{3}\\) x x2<\/sup> x y x z0<\/sup>
\n= \\(\\frac{2 x^{2} y}{3}\\)<\/p>\n

(v) 12a8<\/sup>b8<\/sup> \u00f7 (-6 a6<\/sup> b4<\/sup>)
\n= \\(\\frac{2 \\times 2 \\times 3 \\times a^{8} \\times b^{8}}{-2 \\times 3 \\times a^{6} \\times b^{4}}\\)
\n= -2 x a8-6<\/sup> x b8-4<\/sup>
\n= – 2 x a2<\/sup> x b4<\/sup>
\n= – 2a2<\/sup>b4<\/sup><\/p>\n

Question 2.
\nDivide the given polynomial by the given monomial.
\n(i) (5x2<\/sup> – 6x) \u00f7 3x
\n(ii) (3y8<\/sup> – 4y6<\/sup> + 5y4<\/sup>) \u00f7 y4<\/sup>
\n(iii) 8 (x3<\/sup>y2<\/sup>z2<\/sup> + x2<\/sup>y3<\/sup>z2<\/sup> + x2<\/sup>y2<\/sup>z3<\/sup>) \u00f7 4x2<\/sup>y2<\/sup>z2<\/sup>
\n(iv) (x3<\/sup> + 2x2<\/sup> + 3x) \u00f7 2x
\n(v) (p3<\/sup>q6<\/sup>– p6<\/sup>q3<\/sup>) \u00f7 p3<\/sup>q3<\/sup>
\nAnswer:
\n(i) (5x2<\/sup> – 6x) \u00f7 3x
\n\"NCERT<\/p>\n

(ii) (3y8<\/sup> – 4y6<\/sup> + 5y4<\/sup>) \u00f7 y4<\/sup>
\n\"NCERT
\n=3y4<\/sup> – 4y6-4<\/sup> + 5y4-4<\/sup>
\n=3y4<\/sup> – 4y2<\/sup> – 5y\u00b0
\n= 3y4<\/sup> – 4y2<\/sup> + 5<\/p>\n

\"NCERT<\/p>\n

(iii) 8(x3<\/sup>y2<\/sup>z2<\/sup> + x2<\/sup>y3<\/sup>z2<\/sup> + x2<\/sup>y2<\/sup>z3<\/sup>) 4x2<\/sup>y2<\/sup>z2<\/sup>
\n\"NCERT
\n= [x3-2<\/sup> x y2-2<\/sup> x z2-2<\/sup> + x2-2<\/sup> x y3-2<\/sup> x z2-2<\/sup> + x2-2<\/sup> x y2-2<\/sup> x z3-2<\/sup>]
\n= 2 (x x y\u00b0 x z\u00b0 + x\u00b0 x y x z\u00b0 + x\u00b0 x y\u00b0 x z)
\n= 2 (x + y + z)
\n[using a\u00b0 = 1]<\/p>\n

(iv) x3<\/sup> + 2x2<\/sup> + 3x \u00f7 2x
\n\"NCERT
\n\"NCERT<\/p>\n

(v) (p3<\/sup>q6<\/sup>– p6<\/sup>q3<\/sup>) \u00f7 p3<\/sup>q3<\/sup>
\n\"NCERT
\n= p3-3<\/sup> x q6-3<\/sup> – p6-3<\/sup> x q3-3<\/sup>
\n= p0<\/sup> x q3<\/sup> – p3<\/sup> x q0<\/sup>
\n= p3<\/sup> x q3<\/sup><\/p>\n

Question 3.
\nWork out the following divisions:
\n(i) (10x – 25) \u00f7 5
\n(ii) (10x-25) \u00f7 (2x-5)
\n(iii) 10y(6y + 21) \u00f7 5 (2y + 7)
\n(iv) 9x2y2(3z – 24) \u00f7 27xy (z – 8)
\n(v) 96abc (3a – 12) (5b – 30) \u00f7 144 (a – 4) (b – 6)
\nAnswer:
\n(i) (10x – 25) \u00f7 5
\n\"NCERT
\n= 2x – 5<\/p>\n

\"NCERT<\/p>\n

(ii) (10x – 25) \u00f7 (2x – 5)
\n\"NCERT
\n\"NCERT<\/p>\n

(iii) 10y (6y + 21) \u00f7 5 (2y + 7)
\n\"NCERT
\n[Taking 3 as common from 6y + 21]
\n= 2 x y x 3
\n= 6y<\/p>\n

(iv) 9x2y2(3z – 24) \u00f7 27xy (z – 8)
\n\"NCERT
\n= x2-1<\/sup> x y2-1<\/sup> = xy.<\/p>\n

(v) 96abc (3a – 12) (5b – 30) \u00f7 144 (a – 4) (b – 6)
\n\"NCERT
\nQuestion 4.
\nDivide as directed.
\n(i) 5(2x+1) (3x +5) \u00f7 (2x+1)
\n(ii) 26xy(x + 5) (y-4) \u00f7 13x(y-4)
\n(iii) 52pqr (p + q) (q + r) (r + p) \u00f7 104pq (q + r) (r + p)
\n(iv) 20 (y + 4) (y2<\/sup> + 5y + 3) \u00f7 5(y + 4)
\n(v) x(x + 1)(x + 2)(x + 3) \u00f7 x(x + 1)
\nAnswer:
\n(i) 5(2x + 1) (3x + 5) \u00f7 (2x + 1)
\n= \\(\\frac{5(2 x+1)(3 x+5)}{(2 x+1)}\\)
\n= 5 (3x + 5)<\/p>\n

\"NCERT<\/p>\n

(ii) 26xy (x + 5) (y – 4) \u00f7 13x (y – 4)
\n= \\(\\frac{2 \\times 13 \\times x \\times y \\times(x+5) \\times(y-4)}{13 \\times x \\times(y-4)}\\)
\n= 2 x y x (x + 5) = 2y (x + 5)<\/p>\n

(iii) 52pqr (p +q) (q + r) (r + p) \u00f7 104pq (q + r) (r + p)
\n= \\(\\frac{2 \\times 2 \\times 13 \\times p \\times q \\times r \\times(p+q) \\times(q+r) \\times(r+9)}{2 \\times 2 \\times 2 \\times 13 \\times p \\times q \\times(q+r) \\times(r+p)}\\)
\n_ 2x2xl3xpxqxrx(p + q)x(q + r)x(r + 9) 2x2x2xl3xpxqx(q + r)x(r + p)
\n= \\(\\frac{r(p+q)}{2}=\\frac{1}{2}\\)r(p+q)<\/p>\n

(iv) 20(y + 4) (y2<\/sup> + 5y + 3) \u00f7 5 (y + 4)
\n= \\(\\frac{4 \\times 5(y+4)\\left(y^{2}+5 y+3\\right)}{5 \\times(y+4)}\\)
\n= 4 (y2<\/sup> + 5y + 3)<\/p>\n

(v) x(x+ 1) (x + 2) (x + 3) \u00f7 x(x + 1)
\n= \\(\\frac{x(x+1)(x+2)(x+3)}{x(x}\\)
\n= (x + 2)(x + 3)<\/p>\n

Question 5.
\nFactorise the expressions and divide them as directed.
\n(i) (y2<\/sup> + 7y + 10) – (y + 5)
\n(ii) (m2<\/sup> – 14m – 32) + (m + 2)
\n(iii) (5p2<\/sup> – 25p + 20) 4- (p – 1)
\n(iv) 4yz(z2<\/sup> + 6z – 16) -f- 2y (z + 8)
\n(v) 5pq (p2<\/sup> – q2<\/sup>) -f- 2p (p + q)
\n(vi) 12xy (9x2<\/sup> – 16y2<\/sup>) -J- 4xy (3x + 4y)
\n(vii) 39y3 (50y2<\/sup> – 98) + 26y2<\/sup> (5y + 7)
\nAnswer:
\n(i) y2<\/sup> + 7y + 10
\n[10 = 2 x 5
\n[7 = 2 + 5]
\n= y2<\/sup> + 5y + 2y + 10
\n= y( y + 5) + 2 (y + 5) = (y + 5) (y + 2)
\n\u2234 (y2<\/sup> + 7y + 10) \u00f7 (y + 5)
\n\\(\\frac{(y+5)(y+2)}{(y+5)}\\) = y + 2<\/p>\n

(ii) m2<\/sup> – 14m – 32
\n– 32 = -16 x 2
\n-14 = -16 + 2
\n= m (m – 16) + 2(m – 16)
\n= (m – 16) (m + 2)
\n(m2<\/sup> – 14m – 32) \u00f7 (m + 2)
\n= \\(\\frac{(m-16)(m+2)}{m+2}\\)
\n= m – 16<\/p>\n

\"NCERT<\/p>\n

(iii) 5p2<\/sup> – 25p + 20
\n= 5[p2<\/sup> – 5p + 4]
\n= 5 [p2<\/sup> – 4p – p + 4]
\n= 5[p(p – 4)- 1 (p – 4)
\n= 5(p – 4) (p – 1)
\n\u2234 5p2<\/sup> – 25 p + 20 \u00f7 (p – 1)
\n= \\(\\frac{5(p-4)(p-1)}{(p-1)}\\)
\n= 5 (p -4)<\/p>\n

(iv) z2<\/sup> + 6z – 16 = z2<\/sup> + 8z – 2z – 16
\n-16 = 8 x – 2
\n6 = 8 + (-2)
\n= z(z + 8) – 2 (z – 2)
\n= (z + 8) (z – 2)
\n\u2234 4yz (z2<\/sup> + 6z – 16) 2y (z + 8)
\n= \\(\\frac{4 \\times y \\times z \\times(z+8) \\times(z-2)}{2 \\times y \\times(z+8)}\\)
\n4xyxzx(z + 8)x(z-2)<\/p>\n

(v) 5pq(p2<\/sup> – q2<\/sup>) \u00f7 2p(p + q)
\n= \\(\\frac{5 \\times p \\times q \\times(p+q)(p-q)}{2 \\times p \\times(p+q)}\\)
\n= \\(\\frac{5 q(p-q)}{2}\\)
\n= \\(\\frac { 5 }{ 2 }\\)q(p-q)<\/p>\n

(vi) 12xy (9x2<\/sup> – 16y2<\/sup>)
\n= 2 x 2 x 3 x x x y x [(3x)2<\/sup> – (4y)2<\/sup>]
\n= 2 x 2 x 3 x x x y x (3x + 4y) (3x – 4y)
\n12xy (9x2<\/sup> – 16y2<\/sup>) \u00f7 4xy (3x + 4y)
\n=\\(\\frac{2 \\times 2 \\times 3 \\times x \\times y \\times(3 x+4 y) \\times(3 x-4 y)}{2 \\times 2 \\times x \\times y \\times(3 x+4 y)}\\)
\n= 3 (3x – 4y)<\/p>\n

\"NCERT<\/p>\n

(vii) 39y3<\/sup> (50y2<\/sup> – 98)
\n= 3 x 13 x y3<\/sup> x 2 (25y2<\/sup> – 49)
\n= 3 x 13 x 2 x y3<\/sup> [(5y)2<\/sup> – 72<\/sup>]
\n= 3 x 13 x 2 x y3<\/sup> x (5y + 7) (5y – 7)
\n39y3<\/sup> (50y2<\/sup> – 98) + 26y2<\/sup> (5y + 7)
\n= \\(\\frac{3 \\times 13 \\times 2 \\times y^{3} \\times(5 y+7) \\times(5 y-7)}{2 \\times 13 \\times y^{2}(5 y+7)}\\)
\n= 3 x y3-2<\/sup> x (5y – 7)
\n= 3y(5y – 7)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3 Question 1. Carry out the following divisions. (i) 28×4 \u00f7 56x (ii) – 36y3 \u00f7 9y2 (iii) 66pq2r3 \u00f7 11qr2 (iv) …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3<\/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 14 Factorization Ex 14.3 - 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\/?p=28125\" \/>\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 14 Factorization Ex 14.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3 Question 1. Carry out the following divisions. (i) 28x4 \u00f7 56x (ii) – 36y3 \u00f7 9y2 (iii) 66pq2r3 \u00f7 11qr2 (iv) … NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 Read More »","og_url":"https:\/\/ncertsolutions.guru\/?p=28125","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2021-07-03T09:48:52+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"5 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/?p=28125#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/?p=28125#webpage","url":"https:\/\/ncertsolutions.guru\/?p=28125","name":"NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 - 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