3<\/sup> – 36a<\/p>\n(v) (pq + qr + rp) \u00d7 0 \n= (pq \u00d7 0) + (qr \u00d7 0) + (rp \u00d7 0) \n= 0 + 0 + 0 \n= 0<\/p>\n
Question 2. \nComplete the table. \n \nSolution: \n(i) a \u00d7 (b + c + d) \n= (a \u00d7 b) + (a \u00d7 c) + (a \u00d7 d) \n= ab + ac + ad<\/p>\n
(ii) (x + y – 5) (5xy) \n= (x \u00d7 5xy) + (y \u00d7 5xy) – (5 \u00d7 5xy) \n= 5x2<\/sup>y + 5xy2<\/sup> – 25xy<\/p>\n(iii) p \u00d7 (6p2<\/sup> – 7p + 5) \n= (p \u00d7 6p2<\/sup>) + p \u00d7 (-7p) + p \u00d7 5 \n= 6p3<\/sup> + (-7p2<\/sup>) + 5p \n= 6p3<\/sup> – 7p2<\/sup> + 5p<\/p>\n(iv) 4p2<\/sup>q2<\/sup> \u00d7 (p2<\/sup> – q2<\/sup>) \n= (4p2<\/sup>q2<\/sup> \u00d7 p2<\/sup>) + (4p2<\/sup>q2<\/sup> \u00d7 -q2<\/sup>) \n= 4p4<\/sup>q2<\/sup> + (-4p2<\/sup>q4<\/sup>) \n= 4p4<\/sup>q2<\/sup> – 4p2<\/sup>q4<\/sup><\/p>\n(v) (a + b + c) \u00d7 abc \n= (a \u00d7 abc) + (b \u00d7 abc) + (c \u00d7 abc) \n= a2<\/sup>bc + ab2<\/sup>c + abc2<\/sup><\/p>\n <\/p>\n
Question 3. \nFind the product. \n(i) (a2<\/sup>) \u00d7 (2a22<\/sup>) \u00d7 (4a26<\/sup>) \n(ii) \\(\\left(\\frac{2}{3} x y\\right) \\times\\left(\\frac{-9}{10} x^{2} y^{2}\\right)\\) \n(iii) \\(\\left(-\\frac{10}{3} \\mathrm{pq}^{3}\\right) \\times\\left(\\frac{6}{5} \\mathrm{p}^{3} \\mathrm{q}\\right)\\) \n(iv) x \u00d7 x2<\/sup> \u00d7 x3<\/sup> \u00d7 x4<\/sup> \nSolution: \n(i) (a2<\/sup>) \u00d7 (2a22<\/sup>) \u00d7 (4a26<\/sup>) \n= (1 \u00d7 2 \u00d7 4) \u00d7 (a2<\/sup> \u00d7 a22<\/sup> \u00d7 a26<\/sup>) \n= 8 \u00d7 (a2+22+26<\/sup>) \n= 8a50<\/sup> \n \n(iv) x \u00d7 x2<\/sup> \u00d7 x3<\/sup> \u00d7 x4<\/sup> \n= x1+2+3+4<\/sup> \n= x10<\/sup><\/p>\nQuestion 4. \n(a) Simplify 3x(4x – 5) + 3 and find its values for \n(i) x = 3 \n(ii) x = \\(\\frac{1}{2}\\) \n(b) Simplify a (a2<\/sup> + a + 1) + 5 and find its value for \n(i) a = 0 \n(ii) a = 1 \n(iii) a = -1. \nSolution: \n(a) 3x (4x – 5) + 3 \n= (3x \u00d7 4x) + (3x \u00d7 -5) + 3 \n= 12x2<\/sup> – 15x + 3 \n(i) when x = 3 \n12x2<\/sup> – 15x + 3 \n= 12 \u00d7 (3)2<\/sup> – 15(3) + 3 \n= 12 \u00d7 9 – 45 + 3 \n= 108 – 45 + 3 \n= 111 – 45 \n= 66<\/p>\n(ii) When x = \\(\\frac{1}{2}\\) \n12x2<\/sup> – 15x + 3 \n <\/p>\n(b) a(a2<\/sup> + a + 1) + 5 \n= (a \u00d7 a2<\/sup>) + (a \u00d7 a) + (a \u00d7 1) + 5 \n= a3<\/sup> + a2<\/sup> + a + 5<\/p>\n(i) When a = 0 \na3<\/sup> + a2<\/sup> + a + 5 \n= (0)3<\/sup> + (0)2<\/sup> + 0 + 5 \n= 0 + 0 + 0 + 5 \n= 5<\/p>\n(ii) When a = 1 \na3<\/sup> + a2<\/sup> + a + 5 \n= (13<\/sup>) + (12<\/sup>) + 1 + 5 \n= 1 + 1 + 1 + 5 \n= 8<\/p>\n <\/p>\n
(iii) When a = -1 \na3<\/sup> + a2<\/sup> + a + 5 \n= (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 5 \n= (-1) + 1 – 1 + 5 \n= -1 + 1 – 1 + 5 \n= 4<\/p>\nQuestion 5. \n(a) Add: p(p – q), q(q – r) and r (r – p) \n(b) Add: 2x(z – x – y) and 2y(z – y – x) \n(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l) \n(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c (-a + b + c) \nSolution: \n(a) p(p – q) + q(q – r) + r(r – p) \n= p \u00d7 p – p \u00d7 q + q \u00d7 q + q(-r) + r \u00d7 r + r \u00d7 (-p) \n= p2<\/sup> – pq + q2<\/sup> – qr + r2<\/sup> – rp \n= p2<\/sup> + q2<\/sup> + r2<\/sup> – (pq + qr + rp)<\/p>\n(b) 2x (z – x – y) + 2y(z – y – x) \n= 2x \u00d7 z + 2x (-x) + 2x (-y) + 2yz + 2y (-y) + 2y(-x) \n= 2xz – 2x2<\/sup> – 2xy + 2yz – 2y2<\/sup> – 2xy \n= -2x2<\/sup> – 2y2<\/sup> – 4xy + 2yz + 2xz<\/p>\n(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n) \n= (4l \u00d7 10n) + (4l \u00d7 -3m) + (4l \u00d7 2l) + (-3l \u00d7 1) + (-3l \u00d7 -4m) + (-3l \u00d7 5n) \n= 40ln + (-12lm) + 8l2<\/sup> + (-3l2<\/sup>) + 12lm + (-15ln) \n= 40ln – 12lm + 8l2<\/sup> – 3l2<\/sup> + 12lm – 15ln \n= 40ln – 15ln – 12lm + 12lm + 8l2<\/sup> – 3l2<\/sup> \n= 25ln + 5l2<\/sup> \n= 5l2<\/sup> + 25ln<\/p>\n <\/p>\n
(d) Simplify the I part we get 3a (a + b + c) – 2b(a – b + c) \n= (3a \u00d7 a) + (3a \u00d7 b) + (3a \u00d7 c) – [(2b \u00d7 a) + 2b(-b) + (2b) \u00d7 c] \n= 3a2<\/sup> + 3ab + 3ac – (2ab – 2b2<\/sup> + 2bc) \n= 3a2<\/sup> + 3ab + 3ac – 2ab + 2b2<\/sup> – 2bc \n= 3a2<\/sup> + 2b2<\/sup> + 3ab – 2ab + 3ac – 2bc \n= 3a2<\/sup> + 2b2<\/sup> + ab + 3ac – 2bc \nSimplify the 2nd part 4c \u00d7 (-a + b + c) = (4c \u00d7 – a) + (4c \u00d7 b) + (4c \u00d7 c) \nAccording to the given question = -4ac + 4bc + 4c2<\/sup> \n2nd part – 1st part \n= -4ac + 4bc + 4c2<\/sup> – (3a2<\/sup> + 2b2<\/sup> + ab + 3ac – 2bc) \n= -4ac + 4bc + 4c2<\/sup> – 3a2<\/sup> – 2b2<\/sup> – ab – 3ac + 2bc \n= -4ac – 3ac + 4bc + 2bc + 4c2<\/sup> – 3a2<\/sup> – 2b2<\/sup> – ab \n= -7ac + 6bc + 4c2<\/sup> – 3a2<\/sup> – 2b2<\/sup> – ab \n= -3a2<\/sup> – 2b2<\/sup> + 4c2<\/sup> – ab + 6bc – 7ac<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Question 1. Carry out the multiplication of the expressions in each of the following pairs. …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3<\/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":"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 9 Algebraic Expressions and Identities Ex 9.3 - 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