NCERT Solutions for Class 8 Maths<\/a> Chapter 14 Factorization Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.1<\/h2>\n Question 1. \nFind the common factors of the given terms. \n(i) 12x, 36 \n(ii) 2y, 22xy \n(iii) 14pq, 28p2<\/sup>q2<\/sup> \n(iv) 2x, 3x2<\/sup>, 4 \n(v) 6 abc, 24ab2<\/sup>, 12 a2<\/sup>b \n(vi) 16x3<\/sup>, – 4x2<\/sup>, 32x \n(vii) 10 pq, 20 qr, 30rp \n(viii) 3x2<\/sup>y3<\/sup>, 10x3<\/sup>, 6x2<\/sup>y2<\/sup>z \nAnswer: \n(i) 12x = 2 x 2 x 3 x x \n36 = 2 x 2 x 3 x 3 \n\u2234 The common factor = 2 x 2 x 3 = 12<\/p>\n(ii) 2y = 2 x y \n22y = 2 x 11 x y \n\u2234 The common factor = 2 x y = 2y<\/p>\n
(iii) 14pq = 2 x 7 x p x q \n28 p2<\/sup>q2<\/sup> = 2 x 2 x 7 x p x p x q x q \n\u2234 The common factor \n= 2 x 7 x p x q = 14pq<\/p>\n <\/p>\n
(iv) 2x = 2 x x \n3x2<\/sup> = 3 x x x x \n4 = 2 x 2 \n\u2234 Common factor = 1 \n(1 is a factor of every term)<\/p>\n(v) 6abc = 2 x 3 x a x b x c \n24ab2<\/sup> = 2 x 2 x 2 x 3 x a x b x b \n12a2<\/sup>b = 2 x 2 x 3 x a x a x b \n\u2234 The common factor \n= 2 x 3 x a x b = 6ab<\/p>\n(vi) 16x3<\/sup> = 2 x 2 x 2x 2 x x x x x x \n-4x2<\/sup> = -1 x 2 x 2 x x x x \n32x = 2 x 2 x 2 x 2 x 2 x x \n\u2234 The common factor = 2 x 2 x x = 4x<\/p>\n(vii) 10pq = 2 x 5 x p x q \n20qr = 2 x 2 x 5x q x r \n30rp = 2 x 3 x 5 x r x p \n\u2234\u00a0 The common factor = 2 x 5 = 10<\/p>\n
(viii) 3x2<\/sup>y3<\/sup> = 3 x x x x x y x y x y \n10x3<\/sup>y2<\/sup> = 2 x 5 x x x x x x x y x y \n6x2<\/sup>y2<\/sup>z = 2 x 3 x x x x x y x y x z \n\u2234\u00a0 The common factor = x x x x y x y \n= x2<\/sup>y2<\/sup><\/p>\nQuestion 2. \nFactorise the following expressions. \n(i) 7x – 42 \n(ii) 6p – 12q \n(iii) 7a2<\/sup> + 14a \n(iv) – 16 z + 20 z3<\/sup> \n(v) 20 l2<\/sup> m + 30 alm \n(vi) 5x2<\/sup>y – 15xy2<\/sup> \n(vii) 10a2<\/sup> + 15 b2<\/sup> + 20c2<\/sup> \n(viii) – 4a2<\/sup> + 4ab – 4ca \n(ix) x2<\/sup>yz + xyz2<\/sup> xy2<\/sup>z xyz2<\/sup> \n(x) ax2<\/sup>y + bxy2<\/sup> + cxyz \nAnswer: \n(i) 7x – 42 \n=7 x x – 2 x 3 x 7 \n= 7 (x – 2 x 3) \n= 7 (x – 6)<\/p>\n(ii) 6p – 12q \n= 2 x 3p – 2 x 2 x 3 x q \n= 2 x 3 [p – 2 x q] \n= 6 (p – 2q)<\/p>\n
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(iii) 7a2<\/sup> + 14a \n=7 x a x a + 2 x 7 x a \n= 7a (a + 2)<\/p>\n(iv) – 16z + 20 z3<\/sup> \n= -2 x 2 x 2 x 2 x z + 2 x 2 x 5 x z x z x z \n= 2 x 2 x z [-2x2<\/sup> + 5 x z x z] \n= 4z (-4 + 5z2<\/sup>)<\/p>\n(v) 20 l2<\/sup>m + 30 alm \n= 2 x 2 x 5 x l x l x m + 2 x 3 x 5 x a x l x m \n= 2 x 5 x l x m[2 x l + 3a] \n= 10lm (2l + 3a)<\/p>\n(vi) 5x2<\/sup>y-15xy2<\/sup> \n=5 x x x x x y – 3 x 5 x x x y x x \n= 5xy (x – 3y)<\/p>\n(vii) 10a2<\/sup> – 15b2<\/sup> + 20c2<\/sup> \n= 2 x 5 x a x a – 3 x 5 x b x b + 2 x 2 x 5 x c x c \n= 5[2 x a x a – 3 x b x b + 2 x 2 x c x c] \n= 5 (2a2<\/sup> – 3b2<\/sup> + 4c2<\/sup>)<\/p>\n(viii) -4a2<\/sup> + 4ab – 4ca \n= -2 x 2 x a x a + 2 x 2 x a x b – 2 x 2 x c x a \n= 2 x 2 x a (-a + b – c) \n= 4a (-a + b – c)<\/p>\n(ix) x2<\/sup>yz + xy2<\/sup>z + xyz2<\/sup> \n= x x x x y x z + x x y x y x z + x x y x z x z \n= x x y x z[x + y + z] \n= xyz (x + y + z)<\/p>\n(x) ax2<\/sup>y + bxy2<\/sup> + cxyz \n= a x x x x x y + b x x x y x y + c x x x y x z \n= x x y[a x x + b x y + c x z] \n= xy (ax + by + cz)<\/p>\n <\/p>\n
Question 3. \nFactorise \n(i) x2<\/sup> + xy + 8x + 8y \n(ii) 15xy – 6x + 5y – 2 \n(iii) ax + bx – ay – by \n(iv) 15pq + 15 + 9q – 25p \n(v) z – 7 + 7xy – xyz \nAnswer: \n(i) x2<\/sup> + xy + 8x + 8y \n= x(x + y) + 8 (x + y) \n= (x + y) (x + 8)<\/p>\n(ii) 15xy – 6x + 5y – 2 \n= 3x (5y – 2) + 1 (5y – 2) \n= (5y – 2) (3x + 1)<\/p>\n
(iii) ax + bx – ay – by \n= x (a + b) – y (a + b) \n= (a + b) (x – y)<\/p>\n
(iv) 15pq + 15 + 9q + 25p \n= 15pq + 25p + 9q + 15 \n= 5p (3q + 5) + 3 (3q + 5) \n(Re-arranging the terms) \n= (3q + 5) (5p + 3)<\/p>\n
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(v) z – 7 + 7xy – xyz \n= z – xyz + 7xy – 7 \n= z( 1 – xy) + (7 (xy – 1) \n(Re-arranging the terms) \n= z (1 – xy) – 7 (1 – xy) \n= (1 – xy) (z- 7)<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.1 Question 1. Find the common factors of the given terms. (i) 12x, 36 (ii) 2y, 22xy (iii) 14pq, 28p2q2 (iv) 2x, …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1<\/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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n