3<\/sup><\/p>\nQuestion 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 \n= 2x – 5<\/p>\n
<\/p>\n
(ii) (10x – 25) \u00f7 (2x – 5) \n \n <\/p>\n
(iii) 10y (6y + 21) \u00f7 5 (2y + 7) \n \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 \n= x2-1<\/sup> x y2-1<\/sup> = xy.<\/p>\n(v) 96abc (3a – 12) (5b – 30) \u00f7 144 (a – 4) (b – 6) \n \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 <\/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 <\/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 <\/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":{"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":"","adv-header-id-meta":"","stick-header-meta":"","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 14.3 - 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