EXERCISE 14 · 192 NCERT Textual Exercises and Assignments Exercise – 14.1 1. (i) 12x = 2 × 2 3 × x 36 = 2 × 2 × 3 × 3 Hence, the common factors are 2, 2 and 3 = 2 × 2 ×

187

NCERT Textual Exercises and Assignm

entsEXERCISE 14.1

1. Find the common factors of the given terms. (i) 12x, 36 (ii) 2y, 22xy (iii) 14pq, 28p2q2 (iv) 2x, 3x2, 4 (v) 6abc, 24ab2, 12a2b (vi) 16x3, – 4x2, 32x (vii) 10pq, 20qr, 30rp (viii) 3x2y3, 10x3y2, 6x2y2z 2. Factorise the following expressions. (i) 7x – 42 (ii) 6p – 12q (iii) 7a2 + 14a (iv) –16z + 20z3

(v) 20l2m + 30alm (vi) 5x2y – 15xy2

(vii) 10a2 – 15b2 + 20c2 (viii) –4a2 + 4ab – 4ca (ix) x2yz + xy2z + xyz2 (x) ax2y + bxy2 + cxyz 3. Factorise. (i) x2 + xy + 8x + 8y (ii) 15xy – 6x + 5y – 2 (iii) ax + bx – ay – by (iv) 15pq + 15 + 9q + 25p (v) z – 7 + 7xy – xyz

Test Yourself (F-1) 1. Factorise the following expressions: (i) 3x – 45 (ii) 7x – 14y (iii) 5a2 + 35a (iv) –12y + 20y3

(v) 15a2b + 35ab (vi) pq – pqr (vii) 18m3 – 45mn2 (viii) 17l2 + 85m2

(ix) 6x3y – 12x2y + 15x4 (x) 2a5b3 – 14a2b2 + 4a3b 2. Factorise: (i) 2ab + 2b + 3a (ii) 6xy – 4y + 6 – 9x (iii) 2x + 3xy + 2y + 3y2 (iv) 15b2 – 3bx2 – 5b + x2

(v) a2x2 + axy + abx + by (vi) a2x + abx + ac + aby + b2y + bc (vii) ax3 – bx2 + ax – b (viii) mx – my – nx + ny (ix) 2m3 + 3m – 2m2 – 3 (x) a2 + 11b + 11ab + a

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NCERT Textual Exercises and Assignm

entsEXERCISE 14.2

1. Factorise the following expressions. (i) a2 + 8a + 16 (ii) p2 – 10p + 25 (iii) 25m2 + 30m + 9 (iv) 49y2 + 84yz + 36z2

(v) 4x2 – 8x + 4 (vi) 121b2 – 88bc + 16c2

(vii) (l + m)2 – 4lm (Hint: Expand (l + m)2 first) (viii) a4 + 2a2b2 + b4

2. Factorise. (i) 4p2 – 9q2 (ii) 63a2 – 112b2

(iii) 49x2 – 36 (iv) 16x5 – 144x3

(v) (l + m)2 – (l – m)2 (vi) 9x2y2 – 16 (vii) (x2 – 2xy + y2) – z2 (viii) 25a2 – 4b2 + 28bc – 49c2

3. Factorise the expressions. (i) ax2 + bx (ii) 7p2 + 12q2

(iii) 2x3 + 2xy2 + 2xz2 (iv) am2 + bm2 + bn2 + an2

(v) (lm + l) + m + 1 (vi) y(y + z) + 9(y + z) (vii) 5y2 – 20y – 8z + 2yz (viii) 10ab + 4a + 5b + 2 (ix) 6xy – 4y + 6 – 9x 4. Factorise. (i) a4 – b4 (ii) p4 – 81 (iii) x4 – (y + z)4 (iv) x4 – (x – z)4

(v) a4 – 2a2b2 + b4

5. Factorise the following expressions. (i) p2 + 6p + 8 (ii) q2 – 10q + 21 (iii) p2 + 6p – 16

Test Yourself (F-2) 1. Factorise: (i) a2 + 14a + 49 (ii) x2 – 12x + 36 (iii) 4p2 – 25q2 (iv) 25x2 – 20xy + 4y2

(v) 169m2 – 625n2 (vi) x x2 23

19

+ +

(vii) 121a2 + 154ab + 49b2 (viii) 3x3 – 75x (ix) 36 – 49x2 (x) 1 – 6x + 9x2

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NCERT Textual Exercises and Assignm

ents 2. Factorise: (i) x2 + 7x + 12 (ii) p2 – 6p + 8 (iii) m2 – 4m – 21 (iv) x2 – 14x + 45 (v) x2 – 24x + 108 (vi) a2 + 13a + 12 (vii) x2 – 5x + 6 (viii) x2 – 14xy + 24y2

(ix) m2 – 21m – 72 (x) x2 – 28x + 132 3. Factorise the following:

(i) xx

221 2+ + (ii) x x2 1

4+ +

(iii) (2m + 3n)2 – (3m + 2n)2 (iv) 63p2q2 – 7 (v) 5y5 – 405y (vi) x4 – (2y – 3z)2

EXERCISE 14.3 1. Carry out the following divisions. (i) 28x4 ÷ 56x (ii) –36y3 ÷ 9y2

(iii) 66pq2r3 ÷ 11qr2 (iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (–6a6b4) 2. Divide the given polynomial by the given monomial (i) (5x2 – 6x) ÷ 3x (ii) (3y8 – 4y6 + 5y4) ÷ y4

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2

(iv) (x3 + 2x2 + 3x) ÷ 2x (v) (p3q6 – p6q3) ÷ p3q3

3. Work out the following divisions. (i) (10x – 25) ÷ 5 (ii) (10x – 25) ÷ (2x – 5) (iii) 10y(6y + 21) ÷ 5(2y + 7) (iv) 9x2y2(3z – 24) ÷ 27xy(z – 8) (v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6) 4. Divide as directed. (i) 5(2x + 1) (3x + 5) ÷ (2x + 1) (ii) 26xy(x + 5) (y – 4) ÷ 13x(y – 4) (iii) 52pqr(p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p) (iv) 20(y + 4) (y4 + 5y + 3) ÷ 5(y + 4) (v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

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NCERT Textual Exercises and Assignm

ents 5. Factorise the expressions and divide them as directed. (i) (y2 + 7y + 10) ÷ (y + 5) (ii) (m2 – 14m – 32) ÷ (m + 2) (iii) (5p2 – 25p + 20) ÷ (p – 1) (iv) 4yz(z2 + 6z – 16) ÷ 2y(z + 8) (v) 5pq(p2 – q2) ÷ 2p(p + q) (vi) 12xy(9x2 – 16y2) ÷ 4xy(3x + 4y) (vii) 39y3(50y2 – 98) ÷ 26y2(5y + 7)

Test Yourself (F-3) 1. Simplify: (i) 6x4 ÷ 32x (ii) –42y3 ÷ 7y2

(iii) 30a3b3c3 ÷ 45abc (iv) (7m2 – 6m) ÷ m (v) (–72l4m5n8) ÷ (–8l2m2n3) 2. Work out the following divisions: (i) 5y3 – 4y2 + 3y ÷ y (ii) (9x5 – 15x4 – 21x4) ÷ (3x2) (iii) (5x3 – 4x2 + 3x) ÷ (2x) (iv) 4x2y – 28xy + 4xy2 ÷ (4xy) (v) (8x4yz – 4xy3z + 3x2yz4) ÷ (xyz) 3. Simplify the following expressions: (i) (x2 + 7x + 10) ÷ (x + 2) (ii) (a2 + 24a + 144) ÷ (a + 12) (iii) (m2 + 5m – 14) ÷ (m + 7) (iv) (25m2 – 4n2) ÷ (5m + 2n) (v) (4a2 – 4ab – 15b2) ÷ (2a – 5b) (vi) (a4 – b4) ÷ (a – b)

Maths VIII – Factorisation

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NCERT Textual Exercises and Assignments

Exercise – 14.1 1. (i) 12x = 2 × 2 3 × x 36 = 2 × 2 × 3 × 3 Hence, the common factors are 2, 2 and 3 = 2 × 2 × 3 = 12 (ii) 2y = 2 × y 22xy = 2 × 11 × x × y Hence the common factors are 2 and y = 2 × y = 2y (iii) 14pq = 2 × 7 × p × q 28p2q2 = 2 × 2 × 7 × p × p × q × q Hence, the common factors are 2 × 7 × p × q = 14pq (iv) 2x = 2 × x × 1 3x2 = 3 × x × x × 1 4 = 2 × 2 × 1 Hence the common factor is 1. (v) 6abc = 2 × 3 × a × b × c 24ab2 = 2 × 2 × 2 × 3 × a × b × b 12a2b = 2 × 2 × 3 × a × a × b Hence the common factors are 2 × 3 × a × b = 6ab (vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x –4x2 = (–1) × 2 × 2 × x × x 32x = × 2 × 2 × 2 × 2 × 2 × x Hence the common factors are 2 × 2 × x = 4x (vii) 10pq = 2 × 5 × p × q 20qr = 2 × 2 × 5 × q × r 30rp = 2 × 3 × 5 × r × p Hence the common factors are 2 × 5 = 10 (viii) 3x2y3 = 3 × x × x × y × y × y 10x2y3 = 3 × x × x × y × y × y 6x2y2z = 2 × 3 × x × x × y × y × z Hence, common factors are x × x × y × y = x2y2

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2. (i) 7x – 42 = 7 × x – 2 × 3 × 7 Taking common factors from each term, = 7(x – 2 × 3) = 7(x – 6) (ii) 6p – 12q = 2 × 3 × p – 2 × 2 × 3 × q Taking common factors from each term, = 2 × 3(p – 2q) = 6(p – 2q) (iii) 7a2 + 14a = 7 × a × a + 2 × 7 × a Taking common factors from each term, = 7 × a(a + 2) = 7a (a + 2) (iv) –16z + 20z3 = (–1) × 2 × 2 × 2 × 2 × z + 2 × 2 × 5 × z × z × z Taking common factors from each term, = 2 × 2 × z(– 2 × 2 + 5 × z × z) = 4z(– 4 + 5z2) (v) 20l2m + 30alm = 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m Taking common factors from each term, = 2 × 5 × l × m(2 × l + 3 × a) = 10lm(2l + 3a) (vi) 5x2y –15xy2 = 5 × x × x × y + 3 × 5 × x × y × y Taking common factors from each term, = 5 × x × y(x – 3y) = 5xy(x – 3y) (vii) 10a2 – 15b2 + 20c2 = 2 × 5 × a × a – 3 × 5 × b × b + 2 × 2 × 5 × c × c Taking common factors from each term, = 5 (2 × a × a – 3 × b × b + 2 × 2 × c × c) = 5(2a2 – 3b2 + 4c2) (viii) –4a2 + 4ab – 4ca = (–1) × 2 × 2 × a × a + 2 × 2 × a × b – 2 × 2 × c × a Taking common factors from each term, = 2 × 2 × a(–a + b –c) =4a(–a + b + c) (ix) x2yz + xy2 + xyz2 = x × x × y × z + x × y × z + x × y × z × z Taking common factors from each term, = x × y × z(x + y + z) = xyz(x + y + z)

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(x) ax2y + bxy2 + cxyz = a × x × x × y + b × x × y × y + c × x × y × z Taking common factors from each term, = x × y(a × x + b × y + c × z) = xy(ax + by + cz) 3. (i) x2 + xy + 8x + 8y = x(x + y) + 8 (x + y) = (x + y)(x + 8) (ii) 15xy – 6x + 5y – 2 = 3x(5y – 2) + 1(5y – 2) = (5y – 2)(3x + 1) (iii) ax + bx – ay – by = (ax + bx) – (ay + by) = x(a + b)–y(a + b) = (a + b) (x – y) (iv) 15pq + 15 + 9q + 25p = 15pq + 25p + 9q + 15 = 5p(3q + 5) + 3(3q + 5) = (3q + 5)(5p + 3) (v) z – 7 + 7 xy – xyz = 7xy – 7 – xyz + z = 7(xy – 1) – z(xy – 1) = (xy – 1)(7 – z) = (–1)(1 – xy) (–1)(z – 7) = (1 – xy)(z – 7)

Exercise – 14.2 1. (i) a2 + 8a + 16 = a2 + (4 + 4)a + 4 × 4 Using identity x2 + (a + b) x + ab = (x + a)(x + b), Here x = a, a = 4 and b = 4 a2 + 8a + 16 = (a + 4) (a + 4) = (a + 4)2

(ii) p2 – 10p + 25 = p2 + (–5 – 5)p + (– 5)(– 5) Using identity x2 + (a + b) x + ab = (x + a)(x + b) Here x = p, a = – 5 and b = – 5 p2 – 10p + 25 = (p – 5)(p – 5) = (p – 5) (iii) 25m2 + 30m + 9 = (5m)2 + 2 × 5m × 3 + (3)2

Using identity a2 + 2ab + b2 = (a + b)2, here a = 5m, b = 3 25m2 + 30m + 9 = (5m + 3)2

(iv) 49y2+ 84yz + 36z2 = (7y)2 + 2 × 7y × 6z + (6z)2

Using identity a2 + 2ab + b2 = (a + b)2, here a = 7y, b = 6z 49y2 + 84yz + 36z2 = (7y + 6z)2

(v) 4x2 – 8x + 4= (2x)2 – 2 × 2x × 2 + (2) 2 Using identity a2 – 2ab + b2 = (a – b)2, here a = 2x, b = 2 4x2 – 8x + 4 = (2x – 2)2 = (2)2 (x – 1)2 = 4(x – 1)2

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(vi) 121b2 – 88bc + 16c2 = (11b)2 – 2 × 11b × 4c + (4c)2

Using identity a2 – 2ab + b2 =(a – b)2 = (a – b)2, here a = 11b, b = 4c 121b2 – 88bc + 16c2 = (11b – 4c)2

(vii) (l + m)2 – 4lm = l2 + 2 × l × m + m2 – 4lm [ (a + b)2 = a2 + 2ab + b2] = l2 + 2lm + m2 – 4lm = l2 –2lm + m2

= (l – m)2 [ (a + b)2 = a2 – 2ab + b2] (viii) a4 + 2a2b2 + b4 = (a2)2 + 2 × a2 × b2 (b2)2

= (a2 + b2)2 [ (a + b)2 = a2 + 2ab + b2] 2. (i) 4p2 – 9q2 = (2p)2 – (3q)2

= (2p – 3q)(2p + 3q) [ a2 – b2 = (a – b) (a + b)] (ii) 63a2 – 112b2 = 7(9a2 – 16b2) = 7 [(3a)2 – (4b)2] =7(3a – 4b)(3a + 4b) [ a2 – b2 = (a – b) (a + b)] (iii) 49x2– 36 = (7x)2 – (6)2

= (7x – 6)(7x + 6) [ a2 – b2 = (a – b) (a + b)] (iv) 16x2 – 144x3 = 16x3 (x2 – 9) = 16x3 [(x)2 – (3)2] = 16x3 (x – 3)(x + 3) [ a2 – b2 = (a – b) (a + b)] (v) (l + m)2 – (l – m)2 = [(l + m) + (l – m)][(l + m) – (l – m)] [ a2 – b2 = (a – b) (a + b)] = (l + m + l – m) (l + m + l – m) = (2m)(2l) = 4lm (vi) 9x2y2 – 16 = (3xy)2 – (4)2

= (3xy – 4)(3xy + 4) [ a2 – b2 = (a – b) (a + b)] (vii) (x2– 2xy + y2) – z2 = (x – y)2 – z2 [ (a – b)2 = a2 – 2ab + b2] = (x – y – z) (x – y – z) [ a2 – b2 = (a – b) (a + b)] (viii) 25a2 – 4b2 + 28bc –49c2 = 25a2 = 25a2 – (4b2 – 28bc + 49c2) = 25a2 – [(2b)2 – 2× 2b × 7c + (7c)2] = 25a2 – (2b – 7c)2 [ (a – b)2 = a2 – 2ab + b2] = (5a) 2 – (2b – 7c)2

= [5a – (2b – 7c)][5a + (2b – 7c)] [ a2 – b2 = (a – b) (a + b)] = (5a – 2b + 7c) (5a + 2b – 7c) 3. (i) ax2 + bx = x (ax + b) (ii) 7p2 + 21q2 = 7(p2 + 3q2) (iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2) (iv) am2 + bm2 + bn2 + an2 = m2 (a + b) + n2(a + b)

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196

= (a + b) (m2 + n2) (v) (lm + l) + m + 1 = l(m + 1) + 1(m + 1) = (m + 1)(l + 1) (vi) y(y + z) + 9(y + z) = (y + z)(y + 9) (vii) 5y2 – 20y – 8z + 2yz = 5y2 – 20y + 2yz – 8z = 5y (y – 4) + 2z (y – 4) = (y – 4)(5y + 2z) (viii) 10ab + 4a + 5b + 2 = 2a(5b + 2) + 1 (5b + 2) = (5b + 2) (2a + 1) (ix) 6xy – 4y + 6 – 9x = 6xy – 9x – 4y + 6 = 3x(2y – 3)– 2(2y –3) = (2y –3)(3x – 2) 4. (i) a4 – b4 = (a2)2 – (b2)2

= (a2 – b2 ) (a2 + b2 ) [ a2 – b2 = (a – b) (a + b)] = (a – b ) (a + b )(a2 + b2 ) [ a2 – b2 = (a – b) (a + b)] (ii) p4 – 81 = (p2 – (9)2

= (p2 – 9) (p2 + 9) [ a2 – b2 = (a – b) (a + b)] = (p2 – 32) (p2 + 9) = (p – 3) (p + 3) (p2 + 9) [ a2 – b2 = (a – b) (a + b)] (iii) x4 – (y + z)4 = (x2)2 – [(y + z)2]2

= [x2 – (y + z)2][x2 + (y + z)2] [ a2 – b2 = (a – b) (a + b)] = [x – (y + z)][x + (y + z)][x2+ (y + z)2] [ a2 – b2 = (a – b) (a + b)] = (x – y + z)(x + y + z)[x + y + z][x2 + (y + z)2] (vi) x4 – (x + z)4 = (x2)2 – [(x – z)2]2

= [x2 – (x + z)2][x2 + (x + z)2] [ a2 – b2 = (a – b) (a + b)] = [x – (x – z)][x + (x + z)][x2 + (x – z)2] [ a2 – b2 = (a – b) (a + b)] = (x – x – z)(x + x + z)(x2 + x2 + z2) [ (a – b)2 = a2 – 2ab + b2] = x(2x + z)(2x2 – 2xz + z2) (v) a4 – 2a2b4 = (a2)2 – 2a2b2 + (b2)2

= (a2 – b2)2 [ (a – b)2 = a2 – 2ab + b2] = [(a – b)(a + b)]2 [ a2 – b2 = (a – b) (a + b)] = (a – b) (a + b)2 [ (xy)m =xm.ym] 5. (i) p2 + 6p + 8 = p2 + (4 + 2)p + 4 × 2 = p2 + 4p + 2p + 4 × 2 = p (p + 4) + 2 (p + 4) = (p + 4)(p + 2)

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(ii) q2 – 10q + 21 = q2 – (7 + 3) q + 7 × 3 = q2 – 7q – 3q + 7 × 3 = q(q – 7) – 3(q – 7) = (q – 7)(q – 3) (iii) p2 + 6p – 16 = p2 + (8 – 2)p – 8 × 2 = p2 + 8p – 2p – 8 × 2 = p(p + 8) – 2(p + 8) = (p + 8)(p – 2)

Exercise – 14.3

1. (i) 28 56 2856

2856

44 4

x x xx

xx

= 12

3x x x xm n m n

(ii) 36 9 369

369

3 23

2

3

2y y yy

yy

= – 4y x x xm n m n

(iii) 66 11 6611

6611

2 22 3

2

2 3

2pq qr pq rqr

pq rqr

= 6pqr x x xm n m n

(iv) 34 51 3451

3451

3 3 3 2 33 3 3

2 3

3 3 3

2 3x y z xy z x y zxy z

x y zxy z

= 23

2x y 8 5 62

3 2x x xx

− +

(v) 82

52

62

3 2xx

xx

xx

− +

= –2a2b4 4 5

232x x− +

2. (i) 5 6 3 5 63

22

x x x x xx

53

63

53

2 135 6

2xx

xx

x x

(ii) 3 4 5 3 4 58 6 4 48 6 4

4y y y y y y yy

3 4 5 3 4 58

4

6

4

4

44 2y

yy

yyy

y y

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198

(iii) 8 48

43 2 2 2 3 2 2 3 2 2 2 2

3 2 2 2 3 2 2 2 3

2x y z x y z x y z x y zx y z x y z x y z

x

yy z2 2

84

84

84

2 2 23 2 2

2 2 2

2 3 2

2 2 2

2 2 3

2 2 2

x y zx y z

x y zx y z

x y zx y z

x y z

= 2(x + y + z)

(iv) x x x x x x xx

3 23 2

2 3 2 2 32

xx

xx

xx

x x x x3 2 2

2

222

32 2

22

32

12

2 3

(v) p q p q p q p q p qp q

3 6 6 3 3 33 6 6 3

3 3

p qp q

p qp q

q p3 6

3 3

6 3

3 33 3

3. (i) 10 25 5 10 255

x x

5 2 55

2 5x

x

(ii) 10 25 2 5 10 252 5

x x xx

5 2 52 5

5x

x

(iii) 10 6 21 5 2 710 6 215 2 7

y y yy y

y

2 5 3 2 7

5 2 72 3 6

y yy

y y

(iv) 9 3 24 27 89 3 2427 8

2 22 2

x y z xy zx y zxy z

9

2738

xy xyxy z

xy

(v) 96 3 12 5 30 144 4 696 3 12 5 30144

abc a b a babc a b

a

44 6 b

12 4 2 3 4 5 6

12 4 3 4 610

abc a ba b

abc

4. (i) 5 2 1 3 5 2 15 2 1 3 5

2 1x x x

x xx

Maths VIII – Factorisation

199

= 5(3x +5)

(ii) 26 5 4 13 426 5 4

13 4xy x y x y

xy x yx y

13 2 5 413 4

2 5xy x yx y

y x

(iii) 52 104pqr p q q r r p pq q r r p

52104pqr p q q r r p

pq q r r p

5252 2

12

pqr p q q r r ppq q r r p

r p q

(iv) 20 4 5 3 420 4 5 3

5 42

2

y y y yy y y

y

= 4(y2 + 5y + 3)

(v) x x x x x xx x x x

x x

1 2 3 11 2 3

1

= (x + 2) (x + 3)

5. (i) y y y y yy

22

7 10 5 7 105

y yy

y y yy

2 22 5 2 55

2 5 2 55

y y

y2 55 [ x2 + (a + b) x + ab = (x + a) (x + b)]

(ii) m m m m mm

22

14 32 2 14 322

m m

m

2 16 2 16 22

m m

m16 2

2 [ x2 + (a + b) x + ab = (x + a) (x + b)]

= (m + 2)

(iii) 5 25 20 1 5 25 201

22

p p p p pp

5 20 5 201

5 4 5 41

2p p pp

p p pp

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200

5 5 4

15 1 4

1p p

pp p

p = 5(p – 4)

(iv) 4 6 16 2 84 6 16

2 82

2

yz z z y zyz z z

y z

4 8 2 8 2

2 8

2yz z zy z

4 2 8

2 8yz z z

y z [ x2 + (a + b) x + ab = (x + a) (x + b)]

= 2z(z – 2)

(v) 5 252

2 22 2

pq p q p p qpq p qp p q

5

2pq p q p q

p p q [ a2 – b2 = (a – b) (a + b)]

52q p q

(vi) 12 9 16 4 3 412 9 164 3 4

2 22 2

xy x y xy x yxy x yxy x y

12 3 4

4 3 4

2 2xy x y

xy x y

12 3 4 3 4

4 3 4xy x y x y

xy x y [ a2 – b2 = (a – b) (a + b)]

= 3(3x – 4y)

(vii) 39 50 98 26 5 739 50 9826 5 7

3 2 23 2

2y y y yy y

y y

39 2 25 4926 5 7

39 2 5 7

26 5 7

3 2

2

2 2 2

2

y yy y

y y

y y

39 2 5 7 5 7

26 5 7

2

2

y y yy y [ a2 – b2 = (a – b) (a + b)]

= 3y (5y – 7)

Maths VIII – Factorisation

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Exercise – 14.4 1. L.H.S. = 4 (x – 5) = 4x – 20 ≠ R.H.S. Hence the correct mathematical statement is 4(x – 5) = 4x – 20. 2. L.H.S. = x(3x + 2) = 3x2 + 2x ≠ R.Hs. Hence, the correct mathematical statement is x(3x + 2) = 3x2 + 2x 3. L.H.S. = 2x + 3y ≠ R.H.S. Hence the correct mathematical statement is 2x + 3y = 2x + 3y. 4. L.H.S. = x + 2x + 3x = 6x ≠ R.H.S. Hence the correct mathematical statement is x + 2x + 3x = 6x. 5. L.H.S. = 5y + 2y + y –7y = 8y – 7y = y ≠ R.H.S. Hence the correct mathematical statement is 5y + 2y + y – 7y = y. 6. L.H.S. = 3x + 2x = 5x ≠ R.H.S. Hence the correct mathematical statement is 3x + 2x = 5x. 7. L.H.S. = (2x)2 + 4 (2x) + 7 = 4x2 + 8x + 7 ≠ R.H.S. Hence the correct mathematical statement is (2x)2 + 4(2x) + 7 = 4x2 + 8x + 7. 8. L.H.S. = (2x)2 + 5x = 4x2 + 5x ≠ R.H.S. Hence the correct mathematical statement is (2x)2 + 5x = 4x2 + 5x. 9. L.H.S. = (3x + 2)2 (3x)2 + 2 × 3x × 2 + (2)2 = 9x2 + 12x + 4 ≠ R.H.S. Hence the correct mathematical statement is (3x + 2)2 = 9x2 + 12x + 4 . 10. (a) L.H.S. = x2 + 5x + 4 Putting x = – 3 in given expression, = (–3)2 + 5 (–3) + 4 = 9 – 15 + 4 = – 2 ≠ R.H.S. Hence x2 + 5x + 4 gives (–3)2 + 5 (–3) + 4 = 9 – 15 + 4 = – 2. (b) L.H.S. = x2 – 5x + 4 Putting x = – 3 in given expression, = (–3)2 – 5(–3) + 4 = 9 + 15 + 4 = 28 ≠ R.H.S. Hence x2 – 5x + 4 gives (–3)2 – 5(–3) + 4 = 9 + 15 + 4 = 28. (c) L.H.S. = x2 + 5x Putting x = – 3 in given expression, = (–3)2 + 5 (– 3) = 9 – 15 = – 6 ≠ R.H.S. Hence x2 + 5x gives (–3)2 +5 (–3) = 9 – 15 = – 6. 11. L.H.S. = (y – 3)2 = y2 – 2 × y × 3 + (3)2 [(a–b)2 = a2 –2ab + b2] = y2 – 6y + 9 ≠ R.H.S. Hence, the correct statement is (y – 3)2 = y2 – 6y + 9. 12. L.H.S. = (z + 5)2 = z2 + 2 × z × 5 + (5)2 = z2 + 10z + 25 [(a + b)2 = a2 + 2ab + b2]

Maths VIII – Factorisation

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Hence the correct statement is (z + 5)2 = z2 + 10z + 25. 13. L.H.S. = (2a + 3b) (a – b) = 2a(a – b) + 3b(a – b) = 2a2 – 2ab + 3ab – 3b2 = 2a2 + ab – 3b2 ≠

R.H.S. Hence the correct statement is (2a + 3b)(a – b) = 2a2 + ab – 3b2. 14. L.H.S. = (a + 4)(a + 2) = a(a + 2) + 4(a + 2) = a2 + 2a + 4a + 8 = a2 + 6a + 8 ≠ R.H.S. Hence the correct statement is (a + 4)(a + 2) = a2 + 6a + 8. 15. L.H.S. = (a – 4)(a – 2) = a(a – 2) – 4(a – 2) = a2 – 2a – 4a + 8 = a2 – 6a + 8 ≠ R.H.S. Hence the correct statement is (a – 4) (a – 2) = a2 6a + 8.

16. L.H.S. 33

111

2

2

xx

R.H.S.

Hence the correct statement = =33

12

2

xx

.

17. L.H.S. 3 13

33

13

1 13

2

2

2

2 2 2

xx

xx x x

R.H.S.

Hence the correct statement 3 13

1 13

2

2 2

xx x

.

18. L.H.S. 33 2

xx

R.H.S

Hence, the correct statement is 3 13

1 13

2

2 2

xx x

19. L.H.S. 34 3x

R.H.S.

Hence the correct statement is 3

4 33

4 3x x

20. L.H.S. = 4 54

44

54

1 54

xx

xx x x

R.H.S.

Hence the correct statement is 4 54

1 54

xx x

21. L.H.S. = 7 55

75

55

75

1x x x R.H.S

Hence the correct statement is 7 55

75

1x x .

Maths VIII – Factorisation