Algebraic Expressions and Identities

ALGEBRAIC EXPRESSIONS AND IDENTITIES Monomial : An algebraic expression containing only one term is called a monomial. CONTENTS  Constant & Variable Eg. –5,3y,7xy,  Algebraic Expressions monomials.   Multiplication of Algebraic Expressions   2 2 x yz, 3 5 2 3 a bc etc. are all 3  Binomial : An algebraic expression containing two terms is called a binomial. Eg. The expression 2x – 3, 3x + 2y, xyz –5 etc. are all binomials. Identities  Trinomial : An algebraic expression containing three terms is called a trinomial. Eg. The expressions a – b + 2, x2 + y2 – xy, Constant : A symbol having a fixed numerical value is called a constant. Variable : A symbol which takes various numerical values is called a variable. Eg. We know that the perimeter P of a square of side s is given by P = 4 × s. Here, 4 is a constant and P and s are variables. Eg. The perimeter P of a rectangle of sides l and b is given by P = 2 (l + b). Here, 2 is a constant and l and b are variables. A combination of constants and variables connected by the signs of fundamental operation of addition, subtraction, multiplication and division is called an algebraic expression.  Terms : Various parts of an algebraic expression which are separated by the signs of + or – are called the ‘terms’ of the expression. x3 – 2y3 – 3x2y2z etc. are trinomial.  Factors : Each terms in an algebraic expression is a product of one or more number(s) and / or literal(s). These number(s) and liteal(s) are known as the factors of that terms. A constant factor is called a numerical factor, while a variable factor is known as a literal factor.  Coefficient : In a term of an algebraic expression any of the factors with the sign of the term is called the coefficient of the other factors. Eg. In – 5xy, the coefficient of x is – 5y; the coefficient of y is –5x and the coefficient of xy is –5. Eg. In –x, the coefficient of x is –1.  Constant Term : A term of the expression having no literal factor is called a constant term. Eg. In the algebraic expression x2 – xy + yz – 4, the constant term is –4. Eg. 2x2 – 3xy + 5y2 is an algebraic expression consisting of three terms, namely, 2x2,–3xy and 5y2.  Like and Unlike Terms : The terms having the same literal factors are called like or similar terms, otherwise they are called unlike terms. Eg. The expression 2x3 – 3x2 + 4x – 7 is an algebraic expression consisting of four terms, namely, 2x3, –3x2 , 4x and – 7. Eg. In the algebraic expression 2a2b + 3ab2 – 7ab –4ba2, 2 2 we have 2 a b and –4ba as like terms, whereas 2 3ab and –7ab are unlike terms. - 13 Edubull  EXAMPLES  Ex.1 Add : 7x2 – 4x + 5, – 3x2 + 2x – 1 and 5x2 –x + 9. Sol. We have, Required sum = (7x2 – 4x +5) + (–3x2 + 2x–1) + (5x2 – x+9) = 7x2 – 3x2 + 5x2 – 4x + 2x – x + 5 – 1 + 9 [Collecting like terms] = (7 – 3 + 5)x2 + (–4 + 2 – 1)x +(5 – 1 + 9) [Adding like terms] = 9x2 – 3x + 13 1 5 1 1 1 x + , – x2 + x – and 3 2 2 2 3 1 1 – 2x2 + x– . 5 6 Ex.2 Add : 5x2 – Sol. Required sum Ex.3 1 5 =  5x 2 – x   +  – 1 x 2  1 x – 1  3 2  2 2 3  +   2 x 2  1 x – 1   5 6 Find the product of the following pairs of polynomials : (i) 4, 7x (ii) – 4a, 7a (iii) – 4x,7xy (iv) 4x3, – 3xy (v) 4x, 0 1 1 1 1 5 = 5x – x2 – 2x2 – x + x + x + 2 3 2 5 2 2 – Sol. 1 1 – [Collecting like terms] 3 6 We have, (i) 4 × 7x = (4 × 7) × x = 28 × x = 28 x (ii) (–4a) × (7a) = (–4 × 7) × (a × a) = –28a2 1    1 1 1 5 1 1 =  5 – – 2 x2 +  –    x +  – –  2    3 2 5  2 3 6 (iii) (–4x) × (7xy) = (–4 × 7) × (x × xy) = –28x1+1y = –28x2y [Adding like term]  10 – 1 – 4  2  10  15  6   15 – 2 – 1 = x + x +  2  30     6  = 5 2 11 x + x+2 2 30 (iv) (4x3) × (–3xy) = (4 × –3) × (x3 × xy) = – 12 (x3+1 y) = –12x4y (v) 4x × 0 = (4 × 0) × x = 0 × x = 0 - 14 Ex.4 Find the areas of rectangles with the following pairs of monomials as their length and breadth respectively : (i) (x, y) (ii) (10x, 5y) (iii) (2x2, 5y2) (iv) (4a, 3a2) (ii) We have,  3  (4x2yz) ×  – x 2 yz 2   2  3  =  4    × (x2 × x2 × y × y × z × z2) 2  (v) (3mn, 4np) Sol. = –6x2+2y1+1z1+2 = – 6x4y2z3 We know that the area of a rectangle is the product of its length and breadth. (iii) We have, Length Breadth Length × Breadth = Area (i) x y x×y = xy (ii) 10 x 5y 10x × 5y = 50xy (iii) 2x 2 2 2 5y 2 2x × 5y  8 2 3  3 2   – x yz  ×  – xy z   5   4  3  8 =  –  –  × (x2 × x × y × y2 × z3 × z) 5 4  = (2 × 5) × (x2 × y2) = = 10x2y2 3a2 (iv) 4a 4a × 3a2 = (4 × 3) 6 2 1 1 2 31 6 3 3 4 x y z = x y z 5 5 (iv) We have, 2 × (a × a )  3 2  7 4   x y ×  x y   14  2  = 12 a3 (v) 3mn 4np 3mn × 4np = (3 × 4)  3 7 =    × (x2 × x4 × y × y)  14 2  × (m × n × n × p) = 12 mn2p Ex.5 Multiply : 2 3 = 3 2 (i) 3ab c by 5a b c (ii) 4x2yz by – 3 2 2 x yz 2 (v) We have, (2.1a2bc) × (4ab2) = (2.1 × 4) × (a2 × a × b × b2 × c) 8 3 (iii) – x2yz3 by – xy2z 5 4 (iv) 3 2 7 x y by x 4 y 14 2 (v) 2.1a2bc by 4ab2 Sol. 3 2+4 1+1 3 6 2 x y = xy 4 4 = 8.4a2+1b1+2c = 8.4a3b3c Ex.6 Multiply : (i) –6a2bc, 2a2b and – (i) We have, 4 5 2 a b , 10a3b and 6 9 (3ab2c3) × (5a3b2c) (ii) = (3 × 5) × (a × a3 × b2 × b2 × c3 × c) (iii) 3.15x and – 23x2y = 15a 1 4 1+3 2+2 3+1 b 4 4 4 = 15a b c c (iv) – x, x2yz and – 3 xyz2 7 Sol. (i) We have, Sol. (i) We have, (3xyz) × (5x) × 0  1 (– 6a2bc) × (2a2b) ×  –   4 = (3×5×0)× (x×x×y×z) = 0 × x2yz = 0 1  =  – 6  2    × (a2 × a2 × b × b × c) 4  = 3a 2+2 1+1 b 4 2 c = 3a b c  6   5   12   ab  ×  bc  ×  abc  5  6   9   6 5 12  × (a × a × b × b × b × c × c)     5 6 9  (ii) We have, 4 5 2 3  a b  × (10a b) × (6) 9  = 4  =   10  6  × (a5 × a3× b2 × b) 9  = (ii) We have, 80 5+3 2+1 80 8 3 a b = ab 3 3 12 1+1 1+1+1 1+1 4 2 3 2 a b c = abc 9 3 (iii) We have, 3 2 2 2 2 2 3  x yz  × (0.5xy z ) × (1.16x yz ) × (2xyz) 4  =  3  0.5  1.16  2  × (x2 × x × x2 × x × 4 2 (iii) We have, (3) × (15x) × (–23x y) = (3 × 15 – 23) × (x × x2 × y) y × y × y × y × z2 × z2 × z3 × z) =  3  5  116  2  × (x2+1+2+1 × y1+2+1+1  4 10 100 = –1035x1+2y = –1035x3y. (iv) We have, –3  xyz 2  (–x) × (x yz) ×  7   =  – 1  – 3  × (x × x2 × x × y × y × z × z2) = Ex.7 7  × z2+2+3+1) 100 (iv) We have, (20x10y20z30) × (10xyz)2 = (20x10y20z30) × (10xyz) × (10xyz) = (20 × 10 × 10) × (x10 × x × x × y20 × y × y 3 1+2+1 1+1 1+2 3 4 2 3 x y z = xyz 7 7 Multiply each of the following monomials : (i) 3xyz, 5x, 0  = 87 x6y5z8 2   2 (ii) 6 5 12 ab, bc, abc 5 6 9 3 (iii) x2yz2, 0.5xy2z2, 1.16x2yz3, 2xyz 4 × z30 × z × z) = 2000x10+1+1y20+1+1z30+1+1 = 2000x12y22z32 (v) We have, (–3x2y) × (4xy2z) × (–xy2z2) ×  4 z  5  =  – 3  4  –1  4  × (x2 × x × x × y × y2  (vi) 20x10 y20z30, (10xyz)2 4  (v) (–3x2y), (4xy2z), (–xy2z2) and  z  5  5 × y2 × z×z2 ×z) = 48 2+1+1 1+2+2 1+2+1 48 4 5 4 x y z = xyz 5 5 Edubull Ex.8 Express the following product as a monomial: (x3) × (7x5) ×  1 x 2  × (–6x4) 5 Multiplication of a Monomial & a Binomial Ex.10 Multiply : 2x by (3x + 5y)  Sol. Verify the product for x = 1 2x × (3x + 5y) = 2x × 3x + 2x × 5y = 6x2 + 10xy We have, Sol. (x3) × (7x5) ×  1 x 2  × (–6x4) 5 Ex.11 Multiply : (7xy + 5y) by 3xy  = 1  7  1  –6  × (x3 × x5 × x2 × x4) 5   =– Sol. 42 3+5+2+4 42 14 x =– x 5 5 = 7xy × 3xy + 5y × 3xy = 21x1+1y1+1+15xy1+1 = 21x2y2 + 15xy2 1  L.H.S. = (x ) × (7x ) ×  x 2  × (–6x4) 5  Ex.12 Multiply : – 1 = (1)3 × {7 × (15)} ×   (1) 2  × {–6 × (1)4} Sol. 3 5 5 =1×7× Ex.9 Sol.  We have,  3ab 2  2a  3ab 2  × =  – –  – 5  3 5   42 42 × (1)14 = – 5 5  L.H.S. = R.H.S. Find the value of (5a6) × (–10ab2) × (–2.1a2b3) 1 for a = 1 and b = 2 We have, (5a6) × (–10ab2) × (–2.1a2b3) = (5 × –10 × – 2.1) × (a6 × a × a2 × b2 × b3) =  5  –10  – 21  × (a6 × a × a2 × b2 × b3) =– 6+1+2 2+3 4 2   1 Ex.13 Multiply :  3x – y x  by xy. 5 2   Sol. Horizontal method We have, 4   1  3x – y 2 x  × xy 5   2 = 105 a b = 105a b Putting a = 1 and b = 1 , we have 2 105a9b5 = 105 × (1)9 ×  1  2 = 105 × 1 × = 3x × 9 5 1 105 = 32 32 5   ×b   3 2 3 2 3 × a1+1b2 + ab2+1 = – a2b2 + ab3 5 3 5 5 5 10   3ab 2  2a  by  – b 5  3   3ab 2   2a  –  × – b   5   3   1 42 ×–6 =– 5 5 and, R.H.S. = –  We have, (7xy + 5y) × 3xy Verification : For x = 1, we have  We have,  = 3  Column method We have, 3x – 1 4 1 xy – y2x × xy 2 5 2 1  ×x×x×y– 2 × 1 xy 2 3 2 2 x y – x 2 y3 2 5 4 1    × y2 × y × x × x 5 2 = 4 2 yx 5 3 2 2 3 2 x y – y3x2 = x2y – x2 y3 2 5 2 5 Edubull Ex.14 Determine each of the following products and find the value of each for x = 2, y = 1.15, z = 0.01. (i) 27x2 (1 – 3x) (ii) xz(x2 + y2) 2 (iii) z (x – y) (iv) (2z – 3x) × (– 4y) Sol. (i) We have, Ex.15 Simplify the expression and evaluate them as directed : (i) x(x – 3) + 2 for x = 1 (ii) 3y (2y – 7) – 3(y–4) – 63 for y = – 2 Sol. (i) We have, 2 27x (1 – 3x) x(x – 3) + 2 = x2 – 3x + 2 = 27x2 × (1 – 3x) For x = 1, we have = 27x2 × 1 – 27x2 × 3x x2 – 3x + 2 = (1)2 – 3 × 1 + 2 = 1 – 3 + 2 [Expanding the bracket] = 3 –3 = 0 2 = 27x – 81x3 (ii) We have, Putting x = 2, we have 27x2 (1 – 3x) 3y(2y – 7) – 3 (y – 4) – 63 = 27× (2)2 × (1 – 3×2) = 27 × 4 × (1 – 6) = (6y2 – 21y) – (3y – 12) – 63 = 27 × 4 × – 5 = – 540 = 6y2 – 21y – 3y + 12 – 63 (ii) We have, xz(x2 + y2) = 6y2 – 24y – 51 = xz × (x2 + y2) For y = –2, we have = xz × x2 + xz × y2 = x3z + xy2z 6y2 – 24y – 51 = 6 × (–2)2 – 24(–2) – 51 Putting x = 2, y = 1.15 and z = 0.01, we get = 6 × 4 + 24 × 2 – 51 = 24 + 48 – 51 xz (x2 + y2) = 72 – 51= 21 = 2 × 0.01 × {(2)2 + (1.15)2} = 0.02 × (4 + 1.3225) = 0.02 × 5.3225 = 0.106450 Ex.16 Subtract 3pq (p – q) from 2pq(p + q) Sol. (i) We have, 3pq (p – q) = 3p2q – 3pq2 (iii) We have, z2(x – y) and, 2pq (p + q) = 2p2q + 2pq2 = z2 × (x – y) Subtraction : = z2 × x – z2 × y = z2x – z2y 2p2q + 2pq2 3p2q – 3pq2 Putting x = 2 y = 1.15 and z = 0.01, we get – – p q + 5pq2 z (x – y) = (0.01)2 × (2 – 1.15) Ex.17 Add : (i) p(p – q), q (q – r) and r(r – p) (ii) 2x (z – x – y) and 2y(z – y – x) = (0.0001) × (0.85) = 0.000085 (vi) We have, + ____ + 2 2 Sol. (i) We have, p (p – q) + q (q – r) + r (r – p) (2z – 3x) × (– 4y) = p2 – pq + q2 – qr + r2 – rp = (2z) × (– 4y) – 3x × (– 4y) = – 8zy + 12xy = p2 + q2 + r2 – pq – qr – rp Putting x = 2, y= 1.15 and z = 0.01, we have (ii) We have, (2z – 3x) × – 4y 2x(z – x – y) + 2y (z – y – x) = [(2 × 0.01) – (3 × 2)] × (–4 × 1.15) = 2xz – 2x2 – 2xy + 2yz – 2y2 – 2xy = (0.02 – 6) × (–4.6) = – 5.98 × –4.6 = 27.508 = 2xz – 2x2 – 4xy + 2yz – 2y2 Ex.20 Multiply (2x + 3y) and (4x – 5y) Ex.18 Simplify each of the following expressions : (i) 15a2 – 6a (a – 2) + a(3 + 7a) Sol. Sol. (ii) x2(1 – 3y2) + x(xy2 – 2x) – 3y(y – 4x2y) (2x + 3y) × (4x – 5y) (iii) 4st(s – t) – 6s2(t – t2) – 3t2(2s2 – s) + 2st(s– t) = 2x × (4x – 5y) + 3y × (4x – 5y) = (2x × 4x – 2x × 5y) + (3y × 4x – 3y × 5y) (i) We have, 2 = 8x2 – 10xy + 12xy – 15y2 = 15a2 – 6a2 + 12a + 3a + 7a2 2 2 = 8x2 + 2xy – 15y2 2 2 = 15a – 6a + 7a + 12a + 3a = 16a + 15a (ii) We have, Ex.21 Multiply (7a + 3b) and (2a + 3b) by column method. Sol. x2(1 – 3y2) + x(xy2 – 2x) – 3y(y – 4x2y)  2a  3b × y + 3y ×4 x2y 2 2 2 2 2 2 2 2 14 a 2  6ab 2 2 = x – 3x y + x y – 2x – 3y + 12x y 2 2 2 2 2 2 =(x – 2x ) + (–3x y + x y + 12x y ) – 3y 2 2 2 2 (iii) 4st(s – t) – 6s (t – t ) – 3t (2s – s) + 2st(s – t) 2 2 = 4st × s – 4st × t – 6s × t + 6s × t – 3t2 × 2s2 + 3t2 × s + 2st × s – 2st × t = 4s2t – 4st2 – 6s2t + 6s2t2 – 6s2t2 2 2 2 + 3st + 2s t – 2st 2 2 2 2 2 2 = – 3st Multiplication of Two Binomials Ex.19 Multiply (3x + 2y) and (5x + 3y). We have, (3x + 2y) × (5x + 3y) = 3x × (5x + 3y) + 2y × (5x + 3y) = (3x × 5x + 3x × 3y) + (2y × 5x + 2y × 3y) = 15x2 + 9xy + 10xy + 6y2 = 15x2 + 19xy + 6y2 7x – 3y  4x – 5y 28x 2 –12 xy – 35 xy  15 y 2 Multiplying7 x – 3 y by 4 x Multiplying 7 x – 3 yby  5 y 28x 2 – 47 xy  15 y 2 Adding theliketerms 2 2 + (6s t – 6s t ) = (15x2 + 9xy) + (10xy + 6y2) We have, 2 = (4s t – 6s t + 2s t) + (–4st + 3st – 2st ) 2 2 Ex.22 Multiply (7x – 3y) by (4x – 5y) by column method. Sol. 2 Multiplying 7a  3b by 2a  21ab  9b 2 Multiplyin g 7 a  3b by 3b 14 a 2  27 ab  9 b 2 Adding the like term 2 = –x2 + 10x2y2 – 3y2 We have, 7a  3b = x2 × 1 – 3y2 × x2 + x × xy2 – x × 2x – 3y 2 2 = (8x – 10xy) + (12xy – 15y ) 15a2 – 6a(a – 2) + a(3 + 7a) Sol. We have, Ex.23 Multiply (0.5x – y) by (0.5x + y) Sol. Horizontal Method : We have, (0.5x – y) × (0.5x + y) = 0.5x (0.5x + y) – y (0.5x + y) = 0.5x × 0.5x + 0.5x × y – y × 0.5x – y × y = 0.25x2 + 0.5xy – 0.5xy – y2 2 2 = 0.25x – y Column method: We have, 0.5x – y  0.5x  y 0.25x 2 – 0.5xy Multiplying 0.5x – y by 0.5x  0.5xy – y 2 Multiplying 0.5x – y by y 0.25x 2 – y 2 Adding the like terms 3y  4y    Ex.24 Multiplying  4x   and  3x –  5 5     Sol. (ii) We have, (3m – 2n) (2m – 3n) Horizontal Method : = 3m (2m – 3n) – 2n (2m – 3n) 3y   4y    4x   ×  3x –  5   5   = 3m × 2m – 3m × 3n – 2n × 2m + 2n × 3n 4 y  3y  4y   = 4x ×  3x – ×  3x – +  5  5  5   = 6m2 – 13mn + 6n2 = 4x × 3x – 4x × = 12x2 – 4y 3y 3y 4 y + × 3x – × 5 5 5 5 16 9 12 2 xy + xy – y 5 5 25 7 12 2 = 12x – xy – y 5 25 2 Column method: When m = 1, n = –1, we get (3m – 2n) (2m – 3n) = 6m2 – 13mn + 6n2 = 6 × (1)2 – 13×1× (–1) + 6× (–1)2 = 6 + 13 + 6 = 25 (iii) We have (4a2 + 3b) (4a2 + 3b) = 4a2 × (4a2 + 3b) + 3b × (4a2 + 3b) We have, 3y 5 4y  3x – 5 9 2 12x  xy 5 16 12 2 – xy – y 5 25 7 12 2 12x 2 – xy – y 5 25 = 6m2 – 9mn – 4mn + 6n2 = 4a2 × 4a2 + 4a2 × 3b + 3b × 4a2 + 3b × 3b 4x  = 16a4 + 12a2b + 12a2b + 9b2 = 16a4 + 24a2b + 9b2 When, a = 1, b = 2, we get 3y by 3x. 5 3y 4y Multiplying 4 x  by – . 5 5 (4a2 + 3b) (4a2 + 3b) Adding the like terms = 16 × (1)4 + 24 × (1)2 × 2 + 9 × (2)2 Multiplying 4x  Ex.25 Find the value of the following products: (i) (x + 2y) (x – 2y) at x = 1, y = 0 = 16a4 + 24a2b + 9b2 = 16 + 48 + 36 = 100 Ex.26 Simplify the following : (ii) (3m – 2n) (2m – 3n) at m = 1, n = –1 (i) (2x + 5) (3x – 2) + (x + 2) (2x – 3) (iii) (4a2 + 3b) (4a2 + 3b) at a = 1, b =2 (ii) (3x + 2) (2x + 3) – (4x – 3) (2x – 1) Sol. (i) We have, (x + 2y) (x – 2y) (iii) (2x + 3y) (3x + 4y) – (7x + 3y) (x + 2y) Sol. (i) We have, = x(x – 2y) + 2y (x – 2y) (2x + 5) (3x – 2) + (x + 2) (2x – 3) = x × x – x × 2y + 2y × x – 2y × 2y = 2x(3x – 2) + 5(3x – 2) + x (2x – 3) + 2 (2x – 3) = x2 – 2xy + 2yx – 4y2 = 6x2 – 4x + 15x – 10 + 2x2 – 3x + 4x – 6 = x2 – 4y2 When x = 1, y = 0, we get (x + 2y) (x – 2y) = x2 – 4y2 = (1)2 – 4 × (0)2 = 1 – 0 = 1. = (6x2 + 2x2) + (–4x + 15x – 3x + 4x) + (–10 – 6) = 8x2 + 12x – 16 (ii) We have, (3x + 2) (2x + 3) – (4x – 3) (2x – 1) = {3x(2x+3) + 2 (2x + 3)} – {4x(2x –1) – 3(2x –1)} = (6x2 + 9x + 4x + 6) – (8x2 – 4x – 6x + 3) = (6x2 + 13x + 6) – (8x2 – 10x + 3) 6x2 + 13x + 6 – 8x2 + 10x – 3 = (9x3 + 15x2) + (– 15x2 – 25x) + (6x + 10) = – 2x2 + 23x + 3 = 9x3 + 15x2 – 15x2 – 25x + 6x + 10 (iii) We have, 3 = 9x – 19x + 10 (2x + 3y) (3x + 4y) – (7x + 3y) (x + 2y) = {2x(3x + 4y) + 3y (3x + 4y) – 7x(x + 2y) + 3y (x + 2y)} = (6x2 + 8xy + 9xy + 12y2) – (7x2 + 14xy + 3xy + 6y2) = (6x2 + 17xy + 12y2) – (7x2 + 17xy + 6y2) (5 – x) (3 – 2x) (4 – 3x) = {(5 – x) (3 – 2x)} × (4 – 3x) = {5(3 – 2x) – x (3 – 2x)} × (4 – 3x) = (15 – 10x – 3x + 2x2) × (4 – 3x) = 6x2 +17xy + 12y2 – 7x2 – 17xy – 6y2 = (2x2 – 13x + 15) + (4 – 3x) = 6x2 – 7x2 + 17xy – 17xy + 12y2 – 6y2 = 2x2 × (4 – 3x) 2 – 13x × (4 – 3x) + 15 × (4 – 3x) = 8x – 6x – 52x + 39x2 + 60 – 45x = –x2 + 6y2. Ex.27 Multiply : (2x2 – 3x + 5) by (5x + 2). Sol. (ii) We have, 3 = – 6x3 + 47x2 – 97x + 60 Horizontal method: We have, (2x3 – 3x + 5) × (5x + 2) Identities = (2x2 – 3x + 5) × 5x + (2x2 – 3x + 5) × 2 = (10x3 – 15x2 + 25x) + (4x2 – 6x + 10) = 10x3 – 11x2 + 19x + 10 Column Method: We have, 2x 2 – 3x  5  5x  2 10 x 3 – 15 x 2  25 x  4x 2 – 6x  10 3 10 x – 11x 2  19 x  10 Multiplying 2x 2 – 3x  5 by 5x Multiplying 2x 2 – 3x  5 by 2 Adding the like terms Ex.28 Simplify : (i) (3x – 2) (x – 1) (3x + 5) (ii) (5 – x) (3 – 2x) (4 – 3x)  Identity An indentity is an equality which is true for all values of the variables (s).  Standard Identities : Identity 1. (a + b)2 = a2 + 2ab + b2 Identity 2. (a – b)2 = a2 – 2ab + b2 Identity 3. (a + b) (a – b) = a2 – b2 Other identity (x + a) (x + b) = x2 + x (a + b) + ab Ex.29 Evaluate : 2 (i) (2x + 3y) (ii) (2x – 3y)2 (iii) (2x + 3y) (2x – 3y) Sol. (i) We have, (3x – 2) (x – 1) (3x + 5) Sol. (i) We have, (2x + 3y)2 = (2x)2 + 2 × (2x) × (3y) + (3y)2 = {(3x – 2) (x – 1)} × (3x + 5) [Using: (a + b)2 = a2 + 2ab + b2] [By Associativity of Multiplication] = {3x(x –1) – 2 (x – 1)} × (3x + 5) 2 = (3x – 3x – 2x + 2) × (3x + 5) = 4x2 + 12xy + 9y2 (ii) We have, (2x – 3y)2 = (2x)2 – 2 × (2x) × (3y) + (3y)2 [Using: (a – b)2 = a2 – 2ab + b2] = (3x2 – 5x + 2) × (3x + 5) = 4x2 – 12xy + 9y2 = 3x2 ×(3x + 5) – 5x(3x + 5) + 2 × (3x + 5) (iii)We have, Edubull 2 (2x + 3y) (2x – 3y) = (2x) – (3y) 2 2 [Using : (a + b) = a2 + 2ab + b2] [Using : (a + b) (a – b) = a2 – b2] 2 2 = 4x – 9y . Ex.30 Write down the squares of each of the following binomials :  1 y2    (ii)  5b –  (iii)  y   2 2    a  (i)  x   2  Sol. (i) We have, 2 a a  a 2 +  x   = x + 2 × x × 2 2  2  16 4 x + 8x2 + 9 9 (ii) We have, 2 2  2   x  5y 2   x 2  5y 2  3  3  2  =  x 2  5y 2  3   2 [Using : (a + b)2 = a2 + 2ab + b2] a 4 = (ii) We have, 2 1 1  1 2 +    5b –  = (5b) – 2 × 5b × 2 2  2 Ex.32 If x + 2 [Using : (a –b)2 = a2 – 2ab + b2] 1 4 4 4 20 2 2 x + x y + 25y4 9 3 1 = 4, find the values of x (i) x2 + x+ 2 2  2   2  y  y  = y2 + 2 × y × y +  y   2  2  2   2 (ii) x4 + 1 x4 1 =4 x On squaring both sides, we get 2 1  2 x   = 4 x  y4 4 2 Ex.31 Find the product of the following binomials : Sol. 1 x2 Sol. (i) We have, (iii) We have, = y2 + y3 + [ a.a = a2] 2 2  =  x 2  + 2 × x2× 5y2 + (5y2)2 3 3   2 = 25b2 – 5b + 2 2 [Using : (a + b)2 = a2 + 2ab + b2] = x2 + xa + =  x2 + 2 × x × 4  4  (i)  x 2  3   x 2  3  3 3      x2 + 2 + 2  2  (ii)  x 2  5y 2   x 2  5y 2  3  3   x2 + 1 1 +   = 16 x x 1 = 16 x2 1 = 16 – 2 x2 [On transposing 2 on RHS] (i) We have, 4 2  4 2   x  3  x  3 3 3     4  =  x 2  3 3  2  x2 + 1 = 14 x2 (ii) We have, 2 2 [ a.a = a ] 4 4  =  x 2  + 2 × x2 × 3 + (3)2 3 3  x2 + 1 = 14 x2 On squaring both sides, we get Edubull 2 2  2 1  2  x  2  = 14 x   1 1    x   = x2 + 2 + 2 x x   2 2 1  1   (x2)2 +  2  + 2 × x2 × 2 = 196 x x   x4 + 1 + 2 = 196 x4  x4 + 1 = 196 – 2 x4 [On transposing 2 on RHS]  x4 + 1 = 194 x4 1    x   = 27 + 2 x   2 1   x  x 2  27(given)    2 1    x   = 29 x  x+ 1 1 = 9, find the value of x2 + 2 . x x We have, 1 =± x [Taking square root of both sides] Ex.33 If x – Sol. 29 (ii) We have, 2 1 1  1 2 +  x –  = x – 2 × x × x x   x 1 x– =9 x On squaring both sides, we get 2 1 1    x –  = x2 – 2 + 2 x x  2 1   x –  = 81 x  2 1 1    x –  = x2 + 2 – 2 x x  2  x2 – 2 × x × 1 1 +   = 81 x x 2 1    x –  = 27 – 2 x  1  x2 – 2 + 2 = 81 x x2 +  x2 +  2 1   x  x 2  27(given)    1 = 81 + 2 x2 [On transposing – 2 on RHS] 2 1    x –  = 25 x  1 = 83 x2 2 1    x –  = 52 x  1 Ex.34 If x + 2 = 27, find the values of each of x the following : 2 1 (i) x + x 2 x– 1 (ii) x – x 1 =±5 x Ex.35 If 3x + 2y = 12 and xy = 6, find the value of 9x2 + 4y2. Sol. (i) We have, 2 1 1 1  2 + 2 x   = x + 2 × x × x x x  2 1 1    x   = x2 + 2 + 2 x x   Sol. We have, (3x + 2y)2 = (3x)2 + (2y)2 + 2 × 3x × 2y  (3x + 2y)2 = 9x2 + 4y2 + 12xy  122 = 9x2 + 4y2 + 12 × 6 Edubull [Putting 3x + 2y = 12 and xy = 6] = (2 x) 2 – (3y) 2 (4x2 + 9y2)  144 = 9x2 + 4y2 + 72 2 [Using : (a + b) (a – b) = a2 – b2] 2  144 – 72 = 9x + 4y 2 = (4x2 – 9y2) (4x2 + 9y2) 2  9x + 4y = 72 Ex.36 If 4x2 + y2 = 40 and xy = 6, find the value of 2x + y. Sol. We have, = (4x2)2 – (9y2)2 [Using : (a + b) (a – b) = a2 – b2] = 16x4 – 81y4. (2x + y)2 = (2x)2 + y2 + 2 × 2x × y  (2x + y)2 = (4x2 + y2) + 4xy (iii) We have, (x – 1) (x + 1) (x2 + 1) (x4 + 1) 2  (2x + y) = 40 + 4 × 6 [Using 4x2 + y2 = 40 and xy = 6]  (2xy + y)2 = 64  2x + y = ± = {(x – 1) (x + 1)} (x2 + 1) (x4 + 1) = (x2 – 1) (x2 + 1) (x4 + 1) 64 = ( x 2 – 1)( x 2  1) + (x4 + 1)  2x + y = ± 8 [Taking square root of both sides] Ex.37 Find the continued product : = ( x 2 ) 2 – 12 (x4 + 1) = (x4 – 1) (x4 + 1) 2 (i) (x + 2) (x – 2) (x + 4) (ii) (2x + 3y) (2x – 3y) (4x2 + 9y2) = ( x 4 ) 2 – 12 (iii) (x – 1) (x + 1) (x2 + 1) (x4 + 1) = x8 – 1 1  1  1  1   (iv)  x –   x    x 2  2   x 4  4  x  x  x  x   y   y   (v)  x – – 1  x   1 5   5   (x + 2) (x – 2) (x2 + 4) = {(x + 2)(x – 2)} (x2 + 4) [By associativity of multiplication] 2 = (x – 2 ) (x2 + 4) [ (a + b) (a – b) = a2 – b2] = (x2 – 4) (x2 + 4) = (x2)2 – 42 1  1  2 1   4 1   x –  x   x  2  x  4  x x  x   x      1   1  1  1   =  x –  x    x 2  2   x 4  4  x  x   x   x   Sol. (i) We have, 2 (iv) We have [ (a + b) (a – b) = a 2 – b2] = x4 – 16 1   1   1   = x2 – 2  x2  2   x4  4  x   x   x   2  1   1    = ( x 2 ) 2 –  2    x 4  4  x   x     1   1   = x4 – 4  x4  4  x   x    1  = (x ) –  4  x  4 2 (ii) We have, (2x + 3y) (2x – 3y) (4x2 + 9y2) = {(2x + 3y) (2x – 3y)} (4x + 9y ) 1 x8 = {(2x + 3y) (2x – 3y)} (4x2 + 9y2) (v) We have, 2 2 = x8 – 2 Edubull 6x 40  6 = 6 6  x = 40 (ii) We have, 4x = 982 – 882 y   y    x – – 1  x   1 5 5        y     y  = x –   1 x    1   5     5  y  = x2 –   1 5  2   4x = (98 + 88) × (98 – 88) [Using : a2 – b2 = (a + b) (a – b)]  4x = 186 × 10  y 2 2y  = x –    1  25 5  2 = x2 –  y2 2y – –1 25 5 Sol. (ii) (99)2 (iii) (93)2  25x = (536 + 136) × (536 – 136) [Using : (a2 – b2) = (a + b) (a – b)] (i) (101)2 = (100 + 1)2  25x = 672 × 400 = (100)2 + 2 × 100 × 1 + (1)2 25x 672 400 = [Dividing both sides by 25] 25 25  x = 672 × 16 [Using : (a + b)2 = a2 + 2ab + b2]  = 10000 + 200 + 1 = 10201 (ii) (99)2 = (100 – 1)2 = (100)2 – 2 × 100 × 1 + (1)2 [Using : (a – b)2 = a2 – 2ab + b2] = 10000 – 200 + 1 = 9801 2 (iii) (93) = (90 + 3)2 = (90)2 + 2 × 90 × 3 + (3)2 = 8100 + 540 + 9 = 8649 Ex.39 Find the value of x, if (ii) 4x = 982 – 882 (iii) 25x = 5362 – 1362 Sol. (i) We have, 6x = 232 – 172  6x = (23 + 17) × (23 – 17) [Using : a2 – b2 = (a + b) (a – b)]  6x = 40 × 6 [Dividing both sides by 4] 1860 4  x = 465 (iii) We have, 25x = 5362 – 1362 We have, (i) 6x = 232 – 172 4x 186  10 = 4 4 x= Ex.38 Using the formulae for squaring a binomial, evaluate the following : (i) (101)2 [Dividing both sides by 6]  x = 10752 Ex.40 What must be added to 9x2 – 24x + 10 to make it a whole square ? Sol. We have, 9x2 – 24x + 10 = (3x)2 – 2 × 3x × 4 + 10 It is evident from the above expression that First term = 3x and , Second term = 4 To make the given expression a whole square, we must have (4)2 = 16 in place of 10. Hence, we must add 6 to it to make a perfect square. Adding and subtracting 6, we get 9x2 – 24x + 10 + 6 – 6 = 9x2 – 24x + 16 – 6 = (3x – 4)2 –6 Ex.41 Find the following products: (i) (x + 2) (x + 3) (ii) (x + 7) (x – 2) (iii) (y – 4) (y – 3) (iv) (y – 7) (y + 3) (v) (2x – 3) (2x + 5) (iv) (3x + 4) (3x – 5) Edubull Sol. Using the identity : (x + a) (x + b) = x2 + (a + b)x + ab, we have (i) (x + 2) (x + 3) = x2 + (2 + 3)x + 2 × 3 = x2 + 5x + 6 (ii) (x + 7) (x – 2) = (x + 7) {x + (– 2)} = x2 + {7 + (–2)}x + 7 × 2 = x2 + 5x – 14 (iii) (y – 4) (y – 3) = {y + (– 4)}{y + (–3)} = y2 + {(–4) + (–3)}y + (–4) × (–3)} = y2 – 7y + 12 (iv) (y – 7) (y + 3) = {y + (– 7)} (y + 3) = y2 + {(–7) + 3} y + (–7) × 3 = y2 – 4y – 21 (v) (2x – 3) (2x + 5) = (y – 3) (y + 5), where y = 2x = {y + (–3)} (y + 5) = y2 + {(–3) + 5}y + (–3) × 5 = y2 + 2y – 15 Ex.42 Evaluate the following: (i) 107 × 103 (ii) 56 × 48 (iii) 95 × 97 Sol. Using the identity : (x + a) (x + b) = x2 + (a + b)x + ab we have (i) 107 × 103 = (100 + 7) × (100 + 3) = (100)2 + (7 + 3) × 100 + 7 × 3 = 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021 (ii) 56 × 48 = (50 + 6) × (50 – 2) = (50 + 6) × {50 + (–2)} = (50)2 + {6 + (–2)} × 50 + 6 × – 2 = 2500 + 4 × 50 – 12 = 2500 + 200 – 12 = 2700 – 12 = 2688 (iii) 95 × 97 = (100 – 5) × (100 – 3) = {100 + (–5)}× {100 + (–3)} = (100)2 + {(–5) + (–3)} × 100 + (–5) × (–3) = 10000 – 8 × 100 + 15 = 10000 – 800 + 15 = 9215