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Chapter 13

Factorisation

Class - 8 Concise Mathematics Selina



Exercise 13(A)

Question 1(i)

-7x2 - 14y is equal to:

  1. -21x2y

  2. 14x2y

  3. 7(-x2 - 2y)

  4. -7(x2 + 2y)

Answer

-7x2 - 14y

= (-7 ×\times 1) x2 + (-7 ×\times 2) y

= -7(x2 + 2y)

Hence, option 4 is the correct option.

Question 1(ii)

a(x - y) - b(x - y)2 is equal to:

  1. (x - y)2 (a - b)

  2. (x + y) (x - y) (a - b)

  3. (x - y) (a - bx + by)

  4. (x - y) (a - bx - by)

Answer

a(x - y) - b(x - y)2

= a(x - y) - b(x - y)(x - y)

= (x - y)(a - b(x - y))

= (x - y)(a - bx + by)

Hence, option 3 is the correct option.

Question 1(iii)

a2 + bc + ab + ac is equal to:

  1. (a + b)(a + c)

  2. (a + b)(b + c)

  3. (a + c) (a - b)

  4. (a - b) (b + c)

Answer

a2 + bc + ab + ac

= (a x a + ab) + (ac + bc)

= a (a + b) + c(a + b)

= (a + b)(a + c)

Hence, option 1 is the correct option.

Question 1(iv)

1 - 2x - 2x2 + 4x3 is equal to:

  1. (1 + 2x) (1 - 2x2)

  2. (1 + 2x) (1 + 2x2)

  3. (1 - 2x) (1 - 2x2)

  4. (1 - 2x) (1 + 2x2)

Answer

1 - 2x - 2x2 + 4x3

= 1 ×\times(1 - 2x) - (2x2 - 2 ×\times 2 x2 ×\times x)

= 1 ×\times(1 - 2x) - 2x2(1 - 2x)

= (1 - 2x)(1 - 2x2)

Hence, option 3 is the correct option.

Question 1(v)

a(x - y) - b(y - x)2 is equal to:

  1. (x - y)(a - by + bx)

  2. (y - x)(a - bx + by)

  3. (x - y) (a - bx - by)

  4. (x - y) (a - bx + by)

Answer

a(x - y) - b(y - x)2

= a(x - y) - b(-(x - y))2

= a(x - y) - b(x - y)2

= (x - y)(a - b(x - y))

= (x - y)(a - bx + by)

Hence, option 4 is the correct option.

Question 2

17a6b8 - 34a4b6 + 51a2b4

Answer

17a6b8 - 34a4b6 + 51a2b4

= 17a6b8 - 17 x 2a4b6 + 17 x 3a2b4

= 17a2b4(a4b4 - 2a2b2 + 3)

Hence, 17a6b8 - 34a4b6 + 51a2b4 = 17a2b4(a4b4 - 2a2b2 + 3)

Question 3

3x5y - 27x4y2 + 12x3y3

Answer

3x5y - 27x4y2 + 12x3y3

= 3x5y - 3 ×\times 9x4y2 + 3 ×\times 4x3y3

= 3x3y(x2 - 9xy + 4y2)

Hence, 3x5y - 27x4y2 + 12x3y3 = 3x3y(x2 - 9xy + 4y2)

Question 4

x2(a - b) - y2(a - b) + z2(a - b)

Answer

x2(a - b) - y2(a - b) + z2(a - b)

= (a - b)(x2 - y2 + z2)

Hence,x2(a - b) - y2(a - b) + z2(a - b) = (a - b)(x2 - y2 + z2)

Question 5

(x + y)(a + b) + (x - y)(a + b)

Answer

(x + y)(a + b) + (x - y)(a + b)

= (a + b)(\Big((x + y) + (x - y))\Big)

= (a + b)(\Big(x + y + x - y)\Big)

= (a + b)(2x)

= 2x(a + b)

Hence, (x + y)(a + b) + (x - y)(a + b) = 2x(a + b)

Question 6

2b(2a + b) - 3c(2a + b)

Answer

2b(2a + b) - 3c(2a + b)

= (2a + b)(2b - 3c)

Hence,2b(2a + b) - 3c(2a + b) = (2a + b)(2b - 3c)

Question 7

12abc - 6a2b2c2 + 3a3b3c3

Answer

12abc - 6a2b2c2 + 3a3b3c3

= 3 x 4abc - 3 x 2a2b2c2 + 3a3b3c3

= 3abc(4 - 2abc + a2b2c2)

Hence,12abc - 6a2b2c2 + 3a3b3c3 = 3abc(4 - 2abc + a2b2c2)

Question 8

4x(3x - 2y) - 2y(3x - 2y)

Answer

4x(3x - 2y) - 2y(3x - 2y)

= (3x - 2y)(4x - 2y)

= (3x - 2y)2(2x - y)

= 2(3x - 2y)(2x - y)

Hence,4x(3x - 2y) - 2y(3x - 2y) = 2(3x - 2y)(2x - y)

Question 9

(a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)2

Answer

(a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)2

= (a + 2b)(\Big((3a + b) - (a + b) + (a + 2b))\Big)

= (a + 2b)(\Big( 3a + b - a - b + a + 2b)\Big)

= (a + 2b) (3a - a + a + b + 2b - b)

= (a + 2b) (3a + 2b)

Hence, (a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)2 = (a + 2b) (3a + 2b)

Question 10

6xy(a2 + b2) + 8yz(a2 + b2) - 10xz(a2 + b2)

Answer

6xy(a2 + b2) + 8yz(a2 + b2) - 10xz(a2 + b2)

= (a2 + b2)(6xy + 8yz - 10xz)

= (a2 + b2)2(3xy + 4yz - 5xz)

= 2(a2 + b2)(3xy + 4yz - 5xz)

Hence, 6xy(a2 + b2) + 8yz(a2 + b2) - 10xz(a2 + b2) = 2(a2 + b2)(3xy + 4yz - 5xz)

Question 11

xy - ay - ax + a2 + bx - ab

Answer

xy - ay - ax + a2 + bx - ab

= y(x - a) - a(x - a) + b(x - a)

= (x - a)(y - a + b)

Hence, xy - ay - ax + a2 + bx - ab = (x - a)(y - a + b)

Question 12

3x5 - 6x4 - 2x3 + 4x2 + x - 2

Answer

3x5 - 6x4 - 2x3 + 4x2 + x - 2

= 3x4(x - 2) - 2x2(x - 2) + 1(x - 2)

= (x - 2)(3x4 - 2x2 + 1)

Hence,3x5 - 6x4 - 2x3 + 4x2 + x - 2 = (x - 2)(3x4 - 2x2 + 1)

Question 13

-x2y - x + 3xy + 3

Answer

-x2y - x + 3xy + 3

= - x (xy + 1) + 3(xy + 1)

= (xy + 1)(-x + 3)

= (xy + 1)(3 - x )

Hence, - x2y - x + 3xy + 3 = (xy + 1)(3 - x)

Question 14

6a2 - 3a2b - bc2 + 2c2

Answer

6a2 - 3a2b - bc2 + 2c2

= 3a2(2 - b) - c2(b - 2)

= 3a2(2 - b) + c2(2 - b)

= (2 - b)(3a2 + c2)

Hence,6a2 - 3a2b - bc2 + 2c2 = (2 - b)(3a2 + c2)

Question 15

3a2b - 12a2 - 9b + 36

Answer

3a2b - 12a2 - 9b + 36

= 3a2(b - 4) - 9(b - 4)

= (b - 4)(3a2 - 9)

= (b - 4)3(a2 - 3)

= 3(b - 4)(a2 - 3)

Hence, 3a2b - 12a2 - 9b + 3 = 3(b - 4)(a2 - 3)

Question 16

x2 - (a - 3)x - 3a

Answer

x2 - (a - 3)x - 3a

= x2 - ax + 3x - 3a

= x(x - a) + 3(x - a)

= (x - a)(x + 3)

Hence,x2 - (a - 3)x - 3a = (x - a)(x + 3)

Question 17

ab2 - (a - c) b - c

Answer

ab2 - (a - c) b - c

= ab2 - ab + bc - c

= ab(b - 1) + c(b - 1)

= (b - 1)(ab + c)

Hence,ab2 - (a - c) b - c = (b - 1)(ab + c)

Question 18

(a2 - b2) c + (b2 - c2)a

Answer

(a2 - b2) c + (b2 - c2)a

= a2c - c2a + b2a - b2c

= ac(a - c) + b2(a - c)

= (a - c)(ac + b2)

Hence,(a2 - b2) c + (b2 - c2)a = (a - c)(ac + b2)

Question 19

a3 - a2 - ab + a + b - 1

Answer

a3 - a2 - ab + a + b - 1

= a3 - a2 - ab + b + a - 1

= a2(a - 1) - b(a - 1) + 1(a - 1)

= (a - 1)(a2 - b + 1)

Hence,a3 - a2 - ab + a + b - 1 = (a - 1)(a2 - b + 1)

Question 20

ab(c2 + d2) - a2cd - b2cd

Answer

ab(c2 + d2) - a2cd - b2cd

= abc2 + abd2 - a2cd - b2cd

= (abc2 - a2cd) + (abd2 - b2cd)

= ac(bc - ad) + bd(ad - bc)

= ac(bc - ad) - bd(bc - ad)

= (bc - ad)(ac - bd)

Hence,ab(c2 + d2) - a2cd - b2cd = (bc - ad)(ac - bd)

Question 21

2ab2 - aby + 2cby - cy2

Answer

2ab2 - aby + 2cby - cy2

= ab(2b - y) + cy(2b - y)

= (2b - y)(ab + cy)

Hence,2ab2 - aby + 2cby - cy2 = (2b - y)(ab + cy)

Question 22

ax + 2bx + 3cx - 3a - 6b - 9c

Answer

ax + 2bx + 3cx - 3a - 6b - 9c

= x(a + 2b + 3c) - 3(a + 2b + 3c)

= (a + 2b + 3c)(x - 3)

Hence,ax + 2bx + 3cx - 3a - 6b - 9c = (a + 2b + 3c)(x - 3)

Question 23

2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c

Answer

2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c

= 2a(b2c - 1) + 3b(b2c - 1) - 4c(b2c - 1)

= (b2c - 1)(2a + 3b - 4c)

Hence,2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c = (b2c - 1)(2a + 3b - 4c)

Exercise 13(B)

Question 1(i)

(2x + y)2 - (2y + x)2 is equal to:

  1. 3(x + y) (x - y)

  2. 2(x - y) (x + y)

  3. 2(y - x) (x + y)

  4. (3x + y) (3x - y)

Answer

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

(2x + y)2 - (2y + x)2

= ((2x + y) + (2y + x))((2x + y) - (2y + x))

= (2x + y + 2y + x)(2x + y - 2y - x)

= (2x + x + y + 2y)(2x - x + y - 2y)

= (3x + 3y)(x - y)

= 3(x + y)(x - y)

Hence, option 1 is the correct option.

Question 1(ii)

49 - (x + 5)2 is equal to:

  1. (54 - x) (54 + x)

  2. (2 - x) (12 + x)

  3. 48(x + 5)2

  4. 48(x - 5)2

Answer

49 - (x + 5)2

= 72 - (x + 5)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((7) + (x + 5))((7) - (x + 5))

= (7 + x + 5)(7 - x - 5)

= (12 + x )(2 - x )

Hence, option 2 is the correct option.

Question 1(iii)

a2 - 2ab + b2 + a - b is equal to :

  1. (a - b) (a + b - 1)

  2. (a - b) (a + b + 1)

  3. (a + b) (a - b - 1)

  4. (a - b) (a - b + 1)

Answer

a2 - 2ab + b2 + a - b

Using the formula

[∵ (x - y)2 = x2- 2xy + y2 ]

= (a2 - 2ab + b2) + a - b

= (a - b)2 + 1(a - b)

= (a - b)((a - b) + 1)

= (a - b)(a - b + 1)

Hence, option 4 is the correct option.

Question 1(iv)

x2 + y2 - 2xy - 1 is equal to :

  1. (x + y - 1) (x - y - 1)

  2. (x + y + 1) (x - y - 1)

  3. (x + y + 1) (x - y + 1)

  4. (x - y + 1) (x - y - 1)

Answer

x2 - y2 - 2xy - 1

Using the formula

[∵ (x - y)2 = x2- 2xy + y2 ]

= (x2 + y2 - 2xy) - 1

= (x - y)2 - 12

Using the formula

[∵ (x2 + y2) = (x + y)(x - y)]

= ((x - y) + (1))((x - y) - (1))

= (x - y + 1)(x - y - 1)

Hence, option 4 is the correct option.

Question 1(v)

a2 + 2a + 1 - b2 - x2 + 2bx is equal to :

  1. (a + 1 - b + x) (a - 1 - b + x)

  2. (a + 1 + b - x) (a + 1 - b + x)

  3. (a - 1 + b - x) (a - 1 - b + x)

  4. (a - 1 + bx) (a + 1 - bx)

Answer

a2 + 2a + 1 - b2 - x2 + 2bx

Using the formula

[∵ (x - y)2 = x2- 2xy + y2 ]

And

[∵ (x + y)2 = x2+ 2xy + y2 ]

= (a2 + 2a + 1) - (b2 + x2 - 2bx)

= (a + 1)2 - (b - x)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((a + 1) + (b - x))((a + 1) - (b - x))

= (a + 1 + b - x)(a + 1 - b + x)

Hence, option 2 is the correct option.

Question 2

(a + 2b)2 - a2

Answer

(a + 2b)2 - a2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((a + 2b) + a)((a + 2b) - a)

= (a + 2b + a)(a + 2b - a)

= (a + a + 2b)(a - a + 2b)

= (2a + 2b)(2b)

= 2(a + b)2b

= 4b(a + b)

Hence,(a + 2b)2 - a2 = 4b(a + b)

Question 3

(5a - 3b)2 - 16b2

Answer

(5a - 3b)2 - 16b2

= (5a - 3b)2 - (4b)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((5a - 3b) + 4b)((5a - 3b) - 4b)

= (5a - 3b + 4b)(5a - 3b - 4b)

= (5a + 1b)(5a - 7b)

Hence,(5a - 3b)2 - 16b2 = (5a + b)(5a - 7b)

Question 4

a4 - (a2 - 3b2)2

Answer

a4 - (a2 - 3b2)2

= (a2)2 - (a2 - 3b2)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((a2) + (a2 - 3b2))((a2) - (a2 - 3b2))

= (a2 + a2 - 3b2)(a2 - a2 + 3b2)

= (2a2- 3b2)(3b2)

Hence,a4 - (a2 - 3b2)2 = 3b2(2a2- 3b2)

Question 5

(5a - 2b)2 - (2a - b)2

Answer

(5a - 2b)2 - (2a - b)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((5a - 2b) + (2a - b))((5a - 2b) - (2a - b))

= (5a - 2b + 2a - b)(5a - 2b - 2a + b)

= (5a + 2a - 2b - b)(5a - 2a - 2b + b)

= (7a - 3b)(3a - b)

Hence,(5a - 2b)2 - (2a - b)2 = (7a - 3b)(3a - b)

Question 6

1 - 25 (a + b)2

Answer

1 - 25 (a + b)2

= 12 - (5(a + b))2

= 12 - (5a + 5b)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((1) + (5a + 5b))((1) - (5a + 5b))

= (1 + 5a + 5b)(1 - 5a - 5b)

Hence,1 - 25 (a + b)2 = (1 + 5a + 5b)(1 - 5a - 5b)

Question 7

4(2a + b)2 - (a - b)2

Answer

4(2a + b)2 - (a - b)2

= (2(2a + b))2 - (a - b)2

= (4a + 2b)2 - (a - b)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((4a + 2b) + (a - b))((4a + 2b) - (a - b))

= (4a + 2b + a - b)(4a + 2b - a + b)

= (4a + a + 2b - b)(4a - a + 2b + b)

= (5a+ b)(3a+ 3b)

= (5a+ b)3(a+ b)

Hence,4(2a + b)2 - (a - b)2 = 3(5a+ b)(a+ b)

Question 8

25(2x + y)2 - 16(x - y)2

Answer

25(2x + y)2 - 16(x - y)2

= (5(2x + y))2 - (4(x - y))2

= (10x + 5y)2 - (4x - 4y)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((10x + 5y) + (4x - 4y))((10x + 5y) - (4x - 4y))

= (10x + 5y + 4x - 4y)(10x + 5y - 4x + 4y)

= (10x + 4x + 5y - 4y)(10x - 4x + 5y + 4y)

= (14x + y)(6x + 9y)

= (14x + y)3(2x + 3y)

Hence,25(2x + y)2 - 16(x - y)2 = 3(14x + y)(2x + 3y)

Question 9

(623)2(213)2\Big(6\dfrac{2}{3}\Big)^2 - \Big(2\dfrac{1}{3}\Big)^2

Answer

(623)2(213)2\Big(6\dfrac{2}{3}\Big)^2 - \Big(2\dfrac{1}{3}\Big)^2

= (203)2(73)2\Big(\dfrac{20}{3}\Big)^2 - \Big(\dfrac{7}{3}\Big)^2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

=(203+73)(20373)=((20+7)3)((207)3)=(273)(133)=(27×133×3)=(3519)=39= \Big(\dfrac{20}{3} + \dfrac{7}{3}\Big)\Big(\dfrac{20}{3} - \dfrac{7}{3}\Big)\\[1em] = \Big(\dfrac{(20 + 7)}{3}\Big)\Big(\dfrac{(20 - 7)}{3}\Big)\\[1em] = \Big(\dfrac{27}{3}\Big)\Big(\dfrac{13}{3}\Big)\\[1em] = \Big(\dfrac{27 \times 13}{3 \times 3}\Big)\\[1em] = \Big(\dfrac{351}{9}\Big)\\[1em] = 39

Hence,(623)2(213)2\Big(6\dfrac{2}{3}\Big)^2 - \Big(2\dfrac{1}{3}\Big)^2 = 39

Question 10

(0.7)2 - (0.3)2

Answer

(0.7)2 - (0.3)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (0.7 + 0.3)(0.7 - 0.3)

= (1.0) x (0.4)

= 0.4

Hence,(0.7)2 - (0.3)2 = 0.4

Question 11

75(x + y)2 - 48(x - y)2

Answer

75(x + y)2 - 48(x - y)2

= 3 x (25(x + y)2 - 16(x - y)2)

= 3 x [(5(x + y))2 - (4(x - y))2]

= 3 x [(5x + 5y)2 - (4x - 4y)2]

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 3((5x + 5y) + (4x - 4y))((5x + y) - (4x - 4y))

= 3(5x + 5y + 4x - 4y)(5x + 5y - 4x + 4y)

= 3(5x + 4x + 5y - 4y)(5x - 4x + 5y + 4y)

= 3(9x + y)(x + 9y)

Hence,75(x + y)2 - 48(x - y)2 = 3(9x + y)(x + 9y)

Question 12

a2 + 4a + 4 - b2

Answer

a2 + 4a + 4 - b2

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= (a2 + 4a + 4) - b2

= (a + 2)2 - b2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((a + 2) + b)((a + 2) - b)

= (a + 2 + b)(a + 2 - b)

Hence,a2 + 4a + 4 - b2 = (a + 2 + b)(a + 2 - b)

Question 13

a2 - b2 - 2b - 1

Answer

a2 - b2 - 2b - 1

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= a2 - (b2 + 2b + 1)

= a2 - (b + 1)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (a + (b + 1))(a - (b + 1))

= (a + b + 1)(a - b - 1)

Hence,a2 - b2 - 2b - 1 = (a + b + 1)(a - b - 1)

Question 14

x2 + 6x + 9 - 4y2

Answer

x2 + 6x + 9 - 4y2

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= (x2 + 6x + 9) - 4y2

= (x + 3)2 - (2y)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((x + 3) + (2y))((x + 3) - (2y))

= (x + 3 + 2y)(x + 3 - 2y)

Hence,x2 + 6x + 9 - 4y2 = (x + 3 + 2y)(x + 3 - 2y)

Exercise 13(C)

Question 1(i)

x2 - 9x - 10 is equal to :

  1. (x - 10) (x + 1)

  2. (x - 10) (x - 1)

  3. (x + 10) (x - 1)

  4. (x + 10) (x + 1)

Answer

x2 - 9x - 10

= x2 - (10 - 1)x - 10

= x2 - 10x + 1x - 10

= (x2 - 10x) + (1x - 10)

= x (x - 10) + 1(x - 10)

= (x - 10)(x + 1)

Hence, option 1 is the correct option.

Question 1(ii)

x2 - 23x + 42 is equal to :

  1. (x - 21) (x + 2)

  2. (x - 21) (x - 2)

  3. (x + 21) (x + 2)

  4. (x + 21) (x - 2)

Answer

x2 - 23x + 42

= x2 - (21 + 2)x + 42

= x2 - 21x - 2x + 42

= (x2 - 21x) - 2(2x - 42)

= x (x - 21) - 2(x - 21)

= (x - 21)(x - 2)

Hence, option 2 is the correct option.

Question 1(iii)

(4x2 - 4x + 1) ÷ (2x - 1) is equal to :

  1. 2x + 1

  2. 2x - 1

  3. 2x - 1

  4. none of these

Answer

(4x2 - 4x + 1) ÷ (2x - 1)

= (4x24x+1)(2x1)\dfrac{(4x^2 - 4x + 1)}{(2x - 1)}

Using the formula,

[∵ (x - y)2 = x2 + y2 - 2xy]

= ((2x)22×2x×1+1)(2x1)\dfrac{((2x)^2 - 2 \times 2x \times 1 + 1)}{(2x - 1)}

= (2x1)2(2x1)\dfrac{(2x - 1)^2}{(2x - 1)}

= (2x1)(2x1)(2x1)\dfrac{(2x - 1)(2x - 1)}{(2x - 1)}

= (2x1)(2x1)(2x1)\dfrac{(2x - 1)\cancel{(2x - 1)}}{\cancel{(2x - 1)}}

= 2x - 1

Hence, option 2 is the correct option.

Question 1(iv)

(x + y)2 - 3(x + y) - 4 is equal to :

  1. (x + y + 4) (x + y - 1)

  2. (x + y + 4) (x + y + 1)

  3. (x + y - 4) (x + y + 1)

  4. (x + y - 4) (x + y - 1)

Answer

(x + y)2 - 3(x + y) - 4

= (x + y)2 + (- 4 + 1)(x + y) - 4

= (x + y)2 + (- 4 + 1)(x + y) - 4

= (x + y)2 - 4(x + y) + 1(x + y) - 4

= [(x + y)2 - 4(x + y)] + [1(x + y) + 4]

= (x + y)[(x + y) - 4] + 1[(x + y) + 4]

= [(x + y) + 1][(x + y) - 4]

= [x + y + 1][x + y - 4]

Hence, option 3 is the correct option.

Question 1(v)

60 + 11x - x2 is equal to :

  1. (4 + x) (15 - x)

  2. (4 - x) (15 - x)

  3. (4 - x) (15 + x)

  4. (4 + x) (15 + x)

Answer

60 + 11x - x2

= 60 + (15 - 4)x - x2

= 60 + 15x - 4x - x2

= 15(4 + x) - x(4 + x)

= (4 + x)(15 - x)

Hence, option 1 is the correct option.

Question 2

a2 + 5a + 6

Answer

a2 + 5a + 6

= a2 + (2 + 3)a + 6

= a2 + 2a + 3a + 6

= (a2 + 2a) + (3a + 6)

= a(a + 2) + 3(a + 2)

= (a + 2)(a + 3)

Hence,a2 + 5a + 6 = (a + 2)(a + 3)

Question 3

a2 - 5a + 6

Answer

a2 - 5a + 6

= a2 + (- 2 - 3)a + 6

= a2 + - 2a - 3a + 6

= (a2 - 2a) - (3a + 6)

= a(a - 2) - 3(a - 2)

= (a - 2)(a - 3)

Hence, a2 - 5a + 6 = (a - 2)(a - 3)

Question 4

a2 + 5a - 6

Answer

a2 + 5a - 6

= a2 + (6 - 1)a - 6

= a2 + 6a - 1a - 6

= (a2 + 6a) - (1a + 6)

= a(a + 6) - 1(a + 6)

= (a + 6)(a - 1)

Hence, a2 + 5a - 6 = (a + 6)(a - 1)

Question 5

x2 + 5xy + 4y2

Answer

x2 + 5xy + 4y2

= x2 + (4 + 1)xy + 4y2

= x2 + 4xy + 1xy + 4y2

= (x2 + 4xy) + (1xy + 4y2)

= x(x + 4y) + y(x + 4y)

= (x + 4y)(x + y)

Hence,x2 + 5xy + 4y2 = (x + 4y)(x + y)

Question 6

a2 - 3a - 40

Answer

a2 - 3a - 40

= a2 + (- 8 + 5)a - 40

= a2 - 8a + 5a - 40

= (a2 - 8a) + (5a - 40)

= a(a - 8) + 5(a - 8)

= (a - 8)(a + 5)

Hence,a2 - 3a - 40 = (a - 8)(a + 5)

Question 7

x2 - x - 72

Answer

x2 - x - 72

= x2 + (- 9 + 8)x - 72

= x2 - 9x + 8x - 72

= x(x - 9) + 8(x - 9)

= (x - 9)(x + 8)

Hence,x2 - x - 72 = (x - 9)(x + 8)

Question 8

3a2 - 5a + 2

Answer

3a2 - 5a + 2

= 3a2 - (3 + 2)a + 2

= 3a2 - 3a - 2a + 2

= (3a2 - 3a) - (2a - 2)

= 3a(a - 1) - 2(a - 1)

= (a - 1)(3a - 2)

Hence,3a2 - 5a + 2 = (a - 1)(3a - 2)

Question 9

2a2 - 17ab + 26b2

Answer

2a2 - 17ab + 26b2

= 2a2 - (13 + 4)ab + 26b2

= 2a2 - 13ab - 4ab + 26b2

= (2a2 - 13ab) - (4ab - 26b2)

= a(2a - 13b) - 2b(2a - 13b)

= (2a - 13b)(a - 2b)

Hence,2a2 - 17ab + 26b2 = (2a - 13b)(a - 2b)

Question 10

2x2 + xy - 6y2

Answer

2x2 + xy - 6y2

= 2x2 + (4 - 3)xy - 6y2

= 2x2 + 4xy - 3xy - 6y2

= (2x2 + 4xy) - (3xy + 6y2)

= 2x(x + 2y) - 3y(x + 2y)

= (x + 2y)(2x - 3y)

Hence,2x2 + xy - 6y2 = (x + 2y)(2x - 3y)

Question 11

4c2 + 3c - 10

Answer

4c2 + 3c - 10

= 4c2 + (8 - 5)c - 10

= 4c2 + 8c - 5c - 10

= (4c2 + 8c) - (5c + 10)

= 4c(c + 2) - 5(c + 2)

= (c + 2)(4c - 5)

Hence,4c2 + 3c - 10 = (c + 2)(4c - 5)

Question 12

14x2 + x - 3

Answer

14x2 + x - 3

= 14x2 + (7 - 6)x - 3

= 14x2 + 7x - 6x - 3

= 7x(2x + 1) - 3(2x + 1)

= (2x + 1)(7x - 3)

Hence,14x2 + x - 3 = (2x + 1)(7x - 3)

Question 13

6 + 7b - 3b2

Answer

6 + 7b - 3b2

= 6 + (9 - 2)b - 3b2

= 6 + 9b - 2b - 3b2

= (6 + 9b) - (2b + 3b2)

= 3(2 + 3b) - b(2 + 3b)

= (2 + 3b)(3 - b)

Hence,6 + 7b - 3b2 = (2 + 3b)(3 - b)

Question 14

5 + 7x - 6x2

Answer

5 + 7x - 6x2

= 5 + (10 - 3)x - 6x2

= 5 + 10x - 3x - 6x2

= (5 + 10x) - (3x + 6x2)

= 5(1 + 2x) - 3x(1 + 2x)

= (1 + 2x)(5 - 3x)

Hence,5 + 7x - 6x2 = (1 + 2x)(5 - 3x)

Question 15

4 + y - 14y2

Answer

4 + y - 14y2

= 4 + (8 - 7)y - 14y2

= 4 + 8y - 7y - 14y2

= (4 + 8y) - (7y + 14y2)

= 4(1 + 2y) - 7y(1 + 2y)

= (1 + 2y)(4 - 7y)

Hence,4 + y - 14y2 = (1 + 2y)(4 - 7y)

Question 16

5 + 3a - 14a2

Answer

5 + 3a - 14a2

= 5 + (10 - 7)a - 14a2

= 5 + 10a - 7a - 14a2

= (5 + 10a) - (7a + 14a2)

= 5(1 + 2a) - 7a(1 + 2a)

= (1 + 2a)(5 - 7a)

Hence,5 + 3a - 14a2 = (1 + 2a)(5 - 7a)

Question 17

(2a + b)2 + 5(2a + b) + 6

Answer

(2a + b)2 + 5(2a + b) + 6

= (2a + b)2 + (2 + 3)(2a + b) + 6

= (2a + b)2 + 2(2a + b) + 3(2a + b) + 6

= [(2a + b)2 + 2(2a + b)] + [3(2a + b) + 6]

= (2a + b)[(2a + b) + 2] + 3[(2a + b) + 2]

= [(2a + b) + 2][(2a + b) + 3]

= [2a + b + 2][2a + b + 3]

Hence,(2a + b)2 + 5(2a + b) + 6 = (2a + b + 2)(2a + b + 3)

Question 18

1 - (2x + 3y) - 6(2x + 3y)2

Answer

1 - (2x + 3y) - 6(2x + 3y)2

= 1 - (3 - 2)(2x + 3y) - 6(2x + 3y)2

= 1 - 3(2x + 3y) + 2(2x + 3y) - 6(2x + 3y)2

= 1[1 - 3(2x + 3y)] + 2(2x + 3y)[1 - 3(2x + 3y)]

= [1 - 3(2x + 3y)][1 + 2(2x + 3y)]

= [1 - 6x - 9][1 + 4x + 6y]

Hence,1 - (2x + 3y) - 6(2x + 3y)2 = [1 - 6x - 9y][1 + 4x + 6y]

Question 19

(x - 2y)2 - 12(x - 2y) + 32

Answer

(x - 2y)2 - 12(x - 2y) + 32

= (x - 2y)2 - (8 + 4)(x - 2y) + 32

= (x - 2y)2 - 8(x - 2y) - 4(x - 2y) + 32

= [(x - 2y)2 - 8(x - 2y)] - [4(x - 2y) - 32]

= (x - 2y)[(x - 2y) - 8] - 4[(x - 2y) - 8]

= [(x - 2y) - 8][(x - 2y) - 4]

= [x - 2y - 8][x - 2y - 4]

Hence,(x - 2y)2 - 12(x - 2y) + 32 = [x - 2y - 8][x - 2y - 4]

Question 20

8 + 6(a + b) - 5(a + b)2

Answer

8 + 6(a + b) - 5(a + b)2

= 8 + (10 - 4)(a + b) - 5(a + b)2

= 8 + 10(a + b) - 4(a + b) - 5(a + b)2

= [8 + 10(a + b)] - [4(a + b) + 5(a + b)2]

= 2[4 + 5(a + b)] - (a + b)[4 + 5(a + b)]

= [4 + 5(a + b)][2 - (a + b)]

= [4 + 5a + 5b)][2 - a - b]

Hence,8 + 6(a + b) - 5(a + b)2 = [4 + 5a + 5b][2 - a - b]

Question 21

2(x + 2y)2 - 5(x + 2y) + 2

Answer

2(x + 2y)2 - 5(x + 2y) + 2

= 2(x + 2y)2 - (4 + 1)(x + 2y) + 2

= 2(x + 2y)2 - 4(x + 2y) - 1(x + 2y) + 2

= [2(x + 2y)2 - 4(x + 2y)] - [1(x + 2y) - 2]

= 2(x + 2y)[(x + 2y) - 2] - 1[(x + 2y) - 2]

= [(x + 2y) - 2][2(x + 2y) - 1]

= [x + 2y - 2][2x + 4y - 1]

Hence,2(x + 2y)2 - 5(x + 2y) + 2 = [x + 2y - 2][2x + 4y - 1]

Question 22(i)

Find whether the trinomial is a perfect square or not :

x2 + 14x + 49

Answer

x2 + 14x + 49

Using the formula

[∵ (x + y)2 = x2+ 2xy + y2 ]

= x2 + 2 ×\times 7 ×\times x + (7)2

= (x + 7)2

Hence, x2 + 14x + 49 is a perfect square.

Question 22(ii)

Find whether the trinomial is a perfect square or not :

a2 - 10a + 25

Answer

a2 - 10a + 25

Using the formula

[∵ (x - y)2 = x2- 2xy + y2 ]

= a2 - 2 ×\times a ×\times 5 + (5)2

= (a + 5)2

Hence, a2 - 10a + 25 is a perfect square.

Question 22(iii)

Find whether the trinomial is a perfect square or not :

4x2 + 4x + 1

Answer

4x2 + 4x + 1

Using the formula

[∵ (x + y)2 = x2+ 2xy + y2 ]

= (2x)2 + 2 ×\times 2x ×\times 1 + (1)2

= (2x + 1)2

Hence, 4x2 + 4x + 1 is a perfect square.

Question 22(iv)

Find whether the trinomial is a perfect square or not :

9b2 + 12b + 16

Answer

9b2 + 12b + 16

Using the formula

[∵ (x + y)2 = x2+ 2xy + y2 ]

Since, (9b2 + 12b + 16) cannot be expressed as a2 + 2ab + b2.

Hence, 9b2 + 12b + 16 is not a perfect square.

Question 22(v)

Find whether the trinomial is a perfect square or not :

16x2 - 16xy + y2

Answer

16x2 - 16xy + y2

Using the formula

[∵ (x - y)2 = x2- 2xy + y2 ]

Since, (16x2 - 16xy + y2) cannot be expressed as a2 - 2ab + b2.

Hence, 16x2 - 16xy + y2 is not a perfect square.

Question 22(vi)

Find whether the trinomial is a perfect square or not :

x2 - 4x + 16

Answer

x2 - 4x + 16

Using the formula

[∵ (x - y)2 = x2- 2xy + y2 ]

Since, (x2 - 4x + 16) cannot be expressed as a2 - 2ab + b2.

Hence, x2 - 4x + 16 is not a perfect square.

Exercise 13(D)

Question 1(i)

x3 - 4x is equal to:

  1. x(x + 4) (x - 4)

  2. x(x + 2) (x - 2)

  3. (x + 4) (x - 4)

  4. (x + 2) (x - 2)

Answer

x3 - 4x

= x (x2 - 4)

= x ((x)2 - (2)2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= x (x + 2)(x - 2)

Hence, option 2 is the correct option.

Question 1(ii)

x4 - y4 + x2 - y2 is equal to:

  1. (x + y + 1) (x + y - 1) (x2 + y2)

  2. (x + y) (x - y) (x2 + y2 - 1)

  3. (x + y) (x - y) (x2 + y2 + 1)

  4. none of these

Answer

x4 - y4 + x2 - y2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (x4 - y4) + (x2 - y2)

= (x2 - y2)(x2 + y2) + (x - y)(x + y)

= (x - y)(x + y)(x2 + y2) + (x - y)(x + y)

= (x - y)(x + y)((x2 + y2) + 1)

= (x - y)(x + y)(x2 + y2 + 1)

Hence, option 3 is the correct option.

Question 1(iii)

x3 - x2 + ax + x - a - 1 is equal to:

  1. (x — 1) (x2 + a - 1)

  2. (x — 1) (x2 + a + 1)

  3. (x — 1) (x2 - a + 1)

  4. (x — 1) (x2 - a - 1)

Answer

x3 - x2 + ax + x - a - 1

= (x3 - x2) + (ax - a) + (x - 1)

= x2(x - 1) + a(x - 1) + 1(x - 1)

= (x - 1)(x2 + a + 1)

Hence, option 2 is the correct option.

Question 1(iv)

8x3 - 18x is equal to:

  1. x(2x + 3) (2x - 3)

  2. 2x(3 - 2x) (3 + 2x)

  3. 2x(2x + 3) (2x - 3)

  4. x(4x + 6y) (4x - 6y)

Answer

8x3 - 18x

= 2x (4x2 - 9)

= 2x ((2x)2 - (3)2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 2x (2x - 3)(2x + 3)

Hence, option 3 is the correct option.

Question 1(v)

x2 - (a - b) x - ab is equal to:

  1. (x - a) (x - b)

  2. (x + a) (x - b)

  3. (x - a) (x + b)

  4. (x + a) (x + b)

Answer

x2 - (a - b) x - ab

= x2 - ax + bx - ab

= (x2 - ax) + (bx - ab)

= x(x - a) + b(x - a)

= (x - a)(x + b)

Hence, option 3 is the correct option.

Question 2

8x2y - 18y3

Answer

8x2y - 18y3

= 2y (4x2 - 9y2)

= 2y ((2x)2 - (3y)2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 2y (2x - 3y)(2x + 3y)

Hence,8x2y - 18y3 = 2y (2x - 3y)(2x + 3y)

Question 3

25x3 - x

Answer

25x3 - x

= x (25x2 - 1)

= x ((5x)2 - (1)2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= x (5x + 1)(5x - 1)

Hence,25x3 - x = x (5x + 1)(5x - 1)

Question 4

16x4 - 81y4

Answer

16x4 - 81y4

= (4x2)2 - (9y2)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (4x2 + 9y2)(4x2 - 9y2)

= (4x2 + 9y2)((2x)2 - (3y)2)

= (4x2 + 9y2)(2x - 3y)(2x + 3y)

Hence,16x4 - 81y4 = (4x2 + 9y2)(2x - 3y)(2x + 3y)

Question 5

x2 - y2 - 3x - 3y

Answer

x2 - y2 - 3x - 3y

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (x2 - y2) - (3x + 3y)

= (x - y)(x + y) - 3(x + y)

= (x + y)((x - y) - 3)

= (x + y)(x - y - 3)

Hence,x2 - y2 - 3x - 3y = (x + y)(x - y - 3)

Question 6

x2 - y2 - 2x + 2y

Answer

x2 - y2 - 2x + 2y

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (x2 - y2) - (2x - 2y)

= (x - y)(x + y) - 2(x - y)

= (x - y)((x + y) - 2)

= (x - y)(x + y - 2)

Hence,x2 - y2 - 2x + 2y = (x - y)(x + y - 2)

Question 7

3x2 + 15x - 72

Answer

3x2 + 15x - 72

= 3(x2 + 5x - 24)

= 3(x2 + (8 - 3)x - 24)

= 3(x2 + 8x - 3x - 24)

= 3[(x2 + 8x) - (3x + 24)]

= 3[x(x + 8) - 3(x + 8)]

= 3(x + 8)(x - 3)

Hence,3x2 + 15x - 72 = 3(x + 8)(x - 3)

Question 8

2a2 - 8a - 64

Answer

2a2 - 8a - 64

= 2(a2 - 4a - 32)

= 2(a2 - (8 - 4)a - 32)

= 2(a2 - 8a + 4a - 32)

= 2[(a2 - 8a) + (4a - 32)]

= 2[a(a - 8) + 4(a - 8)]

= 2(a - 8)(a + 4)

Hence,2a2 - 8a - 64 = 2(a - 8)(a + 4)

Question 9

3x2y + 11xy + 6y

Answer

3x2y + 11xy + 6y

= 3x2y + (9 + 2)xy + 6y

= 3x2y + 9xy + 2xy + 6y

= (3x2y + 9xy) + (2xy + 6y)

= 3xy(x + 3) + 2y(x + 3)

= (x + 3)(3xy + 2y)

= (x + 3)y(3x + 2)

Hence,3x2y + 11xy + 6y = y(x + 3)(3x + 2)

Question 10

5ap2 + 11ap + 2a

Answer

5ap2 + 11ap + 2a

= a(5p2 + 11p + 2)

= a(5p2 + (10 + 1)p + 2)

= a(5p2 + 10p + 1p + 2)

= a[(5p2 + 10p) + (1p + 2)]

= a[5p(p + 2) + 1(p + 2)]

= a(p + 2)(5p + 1)

Hence,5ap2 + 11ap + 2a = a(p + 2)(5p + 1)

Question 11

a2 + 2ab + b2 - c2

Answer

a2 + 2ab + b2 - c2

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= (a2 + 2ab + b2) - c2

= ((a)2 + 2 ×\times a ×\times b + (b)2) - c2

= (a + b)2 - (c)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((a + b) + c)((a + b) - c)

= (a + b + c)(a + b - c)

Hence,a2 + 2ab + b2 - c2 = (a + b + c)(a + b - c)

Question 12

x2 + 6xy + 9y2 + x + 3y

Answer

x2 + 6xy + 9y2 + x + 3y

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= (x2 + 6xy + 9y2) + x + 3y

= ((x)2 + 2 ×\times x ×\times 3y + (3y)2) + x + 3y

= (x + 3y)2 + (x + 3y)

= (x + 3y)((x + 3y) + 1)

= (x + 3y)(x + 3y + 1)

Hence,x2 + 6xy + 9y2 + x + 3y = (x + 3y)(x + 3y + 1)

Question 13

4a2 - 12ab + 9b2 + 4a - 6b

Answer

4a2 - 12ab + 9b2 + 4a - 6b

Using the formula

[∵ (x - y)2 = x2 + y2 - 2xy]

= (4a2 - 12ab + 9b2) + 4a - 6b

= ((2a)2 - 2 ×\times 2a ×\times 3b + (3b)2) + 4a - 6b

= (2a - 3b)2 + (4a - 6b)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (2a - 3b)2 + 2(2a - 3b)

= (2a - 3b)((2a - 3b) + 2)

= (2a - 3b)(2a - 3b + 2)

Hence,4a2 - 12ab + 9b2 + 4a - 6b = = (2a - 3b)(2a - 3b + 2)

Question 14

2a2b2 - 98b4

Answer

2a2b2 - 98b4

= 2b2(a2 - 49b2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 2b2((a)2 - (7b)2)

= 2b2(a - 7b)(a + 7b)

Hence,2a2b2 - 98b4 = 2b2(a - 7b)(a + 7b)

Question 15

a2 - 16b2 - 2a - 8b

Answer

a2 - 16b2 - 2a - 8b

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (a2 - 16b2) - (2a + 8b)

= (a - 4b)(a + 4b) - 2(a + 4b)

= (a + 4b)((a - 4b) - 2)

= (a + 4b)(a - 4b - 2)

Hence,a2 - 16b2 - 2a - 8b = (a + 4b)(a - 4b - 2)

Test Yourself

Question 1(i)

(a + b)2 - 4ab is equal to:

  1. (a + b + 2ab) (a + b - 2ab)

  2. (a + b) (a - b)

  3. (a + b) (a + b)

  4. (a - b) (a - b)

Answer

(a + b)2 - 4ab

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= a2 + b2 + 2ab - 4ab

= a2 + b2 + (2ab - 4ab)

= a2 + b2 - 2ab

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= (a)2 + (b)2 - 2 x a x b

= (a - b)2

= (a - b)(a - b)

Hence, option 4 is the correct option.

Question 1(ii)

a4 + 4a2 - 32 is equal to:

  1. (a2 + 8) (a + 2) (a + 2)

  2. (a2 - 8) (a - 2) (a + 2)

  3. (a2 + 8) (a2 + 4)

  4. (a2 + 8) (a + 2) (a - 2)

Answer

a4 + 4a2 - 32

= a4 + (8 - 4)a2 - 32

= a4 + 8a2 - 4a2 - 32

= (a4 + 8a2) - (4a2 + 32)

= a2(a2 + 8) - 4(a2 + 8)

= (a2 + 8)(a2 - 4)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (a2 + 8)(a - 2)(a + 2)

Hence, option 4 is the correct option.

Question 1(iii)

36 - 60y + 25y2 is equal to :

  1. (3 + 5y) (3 + 5y)

  2. (3 - 5y) (6 - 5y)

  3. (3 + 4y) (3 - 4y)

  4. none of these

Answer

36 - 60y + 25y2

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= (6)2 - 2 x 6 x 5y + (5y)2

= (6 - 5y)2

= (6 - 5y)(6 - 5y)

Hence, option 4 is the correct option.

Question 1(iv)

(x - 2y)2 - 3x + 6y is equal to :

  1. (x - 3y) (x + 2y)

  2. (x - 2y) (x - 2y + 3)

  3. (x + 2y - 3) (x + 2y)

  4. (x - 2y) (x - 2y - 3)

Answer

(x - 2y)2 - 3x + 6y

= (x - 2y)2 - 3(x - 2y)

= (x - 2y)((x - 2y) - 3)

= (x - 2y)(x - 2y - 3)

Hence, option 4 is the correct option.

Question 1(v)

a(x - y)2 - by + bx is equal to :

  1. (x - y) (ax + by + b)

  2. (x - y) (ax + by - b)

  3. (x - y) (x + y + a - b)

  4. (x - y) (ax - ay + b)

Answer

a(x - y)2 - by + bx

= a(x - y)2 - (by - bx)

= a(x - y)2 - b(y - x)

= a(x - y)2 + b(x - y)

= (x - y)(a(x - y) + b)

= (x - y)(ax - ay + b)

Hence, option 4 is the correct option.

Question 1(vi)

Statement 1: The product of two binomials is a trinomial, conversely if we factorise a trinomial we always obtain two binomial factors

Statement 2: The square of the difference of two terms = the sum of the same two terms x their difference.

Which of the following options is correct?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

The product of two binomials is a trinomial.

For example; (x + 2)(x + 3) = x(x + 3) + 2(x + 3)

= x2 + 3x + 2x + 6

= x2 + 5x + 6

Here, (x + 2) and (x + 3) are two binomials and their product x2 + 5x + 6 is a trinomial.

But, (x + 2)(x - 2) = x2 - 22

= x2 - 4

Here, (x + 2) and (x - 2) are two binomials but x2 - 4 is not a trinomial.

So, we can say that the product of two binomials is not always a trinomial.

Conversely, if we factorize a trinomial, we always obtain two binomial factors.

This is not always true:

Some trinomials cannot be factorized into binomials with real coefficients.

So, statement 1 is false.

The square of the difference of two terms = the sum of the same two terms × their difference.

L.H.S. = (a - b)2

= a2 + 2ab - b2

R.H.S. = (a + b) x (a - b)

= a2 - b2

As, L.H.S. ≠ R.H.S.

So, statement 2 is false.

Hence, option 2 is the correct option.

Question 1(vii)

Assertion (A) : 25x2 - 5x + 1 is a perfect square trinomial.

Reason (R) : Any trinomial which can be expressed as x2 + y2 + 2xy or x2 + y2 - 2xy is a perfect square trinomial.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

(5x - 1)2 = (5x)2 - 2.5x.1 + 12

= 25x2 - 10x + 12

Thus, 25x2 - 5x + 1 is not a perfect square trinomial.

So, assertion (A) is false.

(x - y)2 = x2 - 2.x.y + y2

= x2 - 2xy + y2

And, (x + y)2 = x2 + 2.x.y + y2

= x2 + 2xy + y2

Thus, x2 + y2 + 2xy or x2 + y2 - 2xy is a perfect square trinomial.

So, reason (R) is true.

Hence, option 4 is the correct option.

Question 1(viii)

Assertion (A) : x2 + 7x + 12

= x2 + (4 + 3)x + 3 x 4

= x2 + 4x + 3x + 3 x 4

= (x + 4)(x + 3)

Reason (R) : To factorise a given trinomial, the product of the first and the last term of the trinomial is always the sum of the two parts when we split the middle term.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

For a quadratic trinomial of the form :

ax2 + bx + c

To factor using the middle-term splitting method;

Find two numbers that multiply to give ac and add to give b.

So, reason (R) is true.

Given; x2 + 7x + 12

= x2 + (3 + 4)x + 3 x 4

= x2 + 3x + 4x + 3 x 4

= x(x + 3) + 4(x + 3)

= (x + 3)(x + 4)

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Hence, option 1 is the correct option.

Question 1(ix)

Assertion (A) : The value of k so that the factors of (x2kx+12116)\Big(x^2 - kx + \dfrac{121}{16}\Big) are same is 112\dfrac{11}{2}.

Reason (R) : (x + a) (x + b) = x2 + (a + b)x + ab.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Given; (x2kx+12116)\Big(x^2 - kx + \dfrac{121}{16}\Big)

We know that this quadratic has equal factors if it's a perfect square trinomial, meaning :

(x2kx+12116)=(x114)2(x2kx+12116)=x22×x×114+(114)2(x2kx+12116)=x2112x+12116k=112.\Rightarrow \Big(x^2 - kx + \dfrac{121}{16}\Big) = \Big(x - \dfrac{11}{4}\Big)^2 \\[1em] \Rightarrow \Big(x^2 - kx + \dfrac{121}{16}\Big) = x^2 - 2 \times x \times \dfrac{11}{4} + \Big(\dfrac{11}{4}\Big)^2 \\[1em] \Rightarrow \Big(x^2 - kx + \dfrac{121}{16}\Big) = x^2 - \dfrac{11}{2}x + \dfrac{121}{16} \\[1em] \Rightarrow k = \dfrac{11}{2}.

So, assertion (A) is true.

Solving,

(x + a)(x + b) = x(x + b) + a(x + b)

= x2 + bx + ax + ab

= x2 + x(a + b) + ab

So, reason (R) is true but, reason (R) is not the correct explanation of assertion (A).

Hence, option 2 is the correct option.

Question 1(x)

Assertion (A) : There are two values of b so that x2 + by - 24 is factorisable.

Reason (R) : Two values have:

Product = -24 and sum = 2.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Given: x2 + bx - 24

To factor such a quadratic, we look for two numbers that :

Multiply to get −24 (the constant term)

Add to b (the coefficient of the middle term)

Values of b such that their sum is 2 and product is -24 are 6 and -4.

So, reason (R) is true.

⇒ x2 + bx - 24

⇒ x2 + 6x - 4x - 24

⇒ x(x + 6) - 4(x + 6)

⇒ (x - 4)(x + 6)

So, assertion (A) is true and reason clearly explains assertion (A).

Hence, option 1 is the correct option.

Question 2(i)

Factorise :

6x3 - 8x2

Answer

6x3 - 8x2

= 2x2 (3x - 4)

Hence,6x3 - 8x2 = 2x2 (3x - 4)

Question 2(ii)

Factorise :

36x2y2 - 30x3y3 + 48x3y2

Answer

36x2y2 - 30x3y3 + 48x3y2

= 6x2y2 (6 - 5xy + 8x)

Hence,36x2y2 - 30x3y3 + 48x3y2 = 6x2y2 (6 - 5xy + 8x)

Question 2(iii)

Factorise :

8(2a + 3b)3 - 12(2a + 3b)2

Answer

8(2a + 3b)3 - 12(2a + 3b)2

= 4(2(2a + 3b)3 - 3(2a + 3b)2)

= 4(2a + 3b)2(2(2a + 3b) - 3)

= 4(2a + 3b)2(4a + 6b - 3)

Hence,8(2a + 3b)3 - 12(2a + 3b)2 = 4(2a + 3b)2(4a + 6b - 3)

Question 2(iv)

Factorise :

9a(x - 2y)4 - 12a(x - 2y)3

Answer

9a(x - 2y)4 - 12a(x - 2y)3

= 3a(3(x - 2y)4 - 4(x - 2y)3)

= 3a(x - 2y)3(3(x - 2y) - 4)

= 3a(x - 2y)3(3x - 6y - 4)

Hence,9a(x - 2y)4 - 12a(x - 2y)3 = 3a(x - 2y)3(3x - 6y - 4)

Question 3(i)

Factorise :

a2 - ab(1 - b) - b3

Answer

a2 - ab(1 - b) - b3

= a2 - ab + ab2 - b3

= (a2 - ab) + (ab2 - b3)

= a(a - b) + b2(a - b)

= (a - b)(a + b2)

Hence,a2 - ab(1 - b) - b3 = (a - b)(a + b2)

Question 3(ii)

Factorise :

xy2 + (x - 1)y - 1

Answer

xy2 + (x - 1)y - 1

= xy2 + xy - y - 1

= (xy2 + xy) - (y + 1)

= xy(y + 1) - 1(y + 1)

= (y + 1)(xy - 1)

Hence,xy2 + (x - 1)y - 1 = (y + 1)(xy - 1)

Question 3(iii)

Factorise :

(ax + by)2 + (bx - ay)2

Answer

(ax + by)2 + (bx - ay)2

Using the formula

[∵ (x + y)2 = x2 + y2 + 2xy]

= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 - 2abxy

= (a2x2 + a2y2) + (b2y2 + b2x2) + 2abxy- 2abxy

= a2(x2 + y2) + b2(x2 + y2)

= (x2 + y2)(a2 + b2)

Hence,(ax + by)2 + (bx - ay)2 = (x2 + y2)(a2 + b2)

Question 3(iv)

Factorise :

ab(x2 + y2) - xy(a2 + b2)

Answer

ab(x2 + y2) - xy(a2 + b2)

= abx2 + aby2 - xya2 - xyb2

= (abx2 - xya2) + (aby2 - xyb2)

= (abx2 - a2xy) + (aby2 - b2xy)

= ax(bx - ay) + by(ay - bx)

= ax(bx - ay) - by(bx - ay)

= (bx - ay)(ax - by)

Hence,ab(x2 + y2) - xy(a2 + b2) = (bx - ay)(ax - by)

Question 3(v)

Factorise :

m - 1 - (m - 1)2 + am - a

Answer

m - 1 - (m - 1)2 + am - a

= (m - 1) - (m - 1)2 + (am - a)

= 1(m - 1) - (m - 1)2 + a(m - 1)

= (m - 1)(1 - (m - 1) + a)

= (m - 1)(1 - m + 1 + a)

= (m - 1)(2 - m + a)

Hence,m - 1 - (m - 1)2 + am - a = (m - 1)(2 - m + a)

Question 4(i)

Factorise :

25(2x - y)2 - 16(x - 2y)2

Answer

25(2x - y)2 - 16(x - 2y)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (5(2x - y))2 - (4(x - 2y))2

= (5(2x - y) + 4(x - 2y))(5(2x - y) - 4(x - 2y))

= (10x - 5y + 4x - 8y)(10x - 5y - 4x + 8y)

= (10x + 4x - 5y - 8y)(10x - 4x - 5y + 8y)

= (14x - 13y)(6x + 3y)

= (14x - 13y)3(2x + y)

Hence,25(2x - y)2 - 16(x - 2y)2 = 3(14x - 13y)(2x + y)

Question 4(ii)

Factorise :

16(5x + 4)2 - 9(3x - 2)2

Answer

16(5x + 4)2 - 9(3x - 2)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (4(5x + 4))2 - (3(3x - 2))2

= (4(5x + 4) + 3(3x - 2))(4(5x + 4) - 3(3x - 2))

= (20x + 16 + 9x - 6)(20x + 16 - 9x + 6)

= (20x + 9x + 16 - 6)(20x - 9x + 16 + 6)

= (29x + 10)(11x + 22y)

= (29x + 10)11(x + 2y)

Hence,16(5x + 4)2 - 9(3x - 2)2 = 11(29x + 10)(x + 2y)

Question 4(iii)

Factorise :

25(x - 2y)2 - 4

Answer

25(x - 2y)2 - 4

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (5(x - 2y))2 - (2)2

= (5(x - 2y) - 2)(5(x - 2y) + 2)

= (5x - 10y - 2)(5x - 10y + 2)

Hence,25(x - 2y)2 - 4 = (5x - 10y - 2)(5x - 10y + 2)

Question 5(i)

Factorise :

a2 - 23a + 42

Answer

a2 - 23a + 42

= a2 - (21 + 2)a + 42

= a2 - 21a - 2a + 42

= (a2 - 21a) - (2a - 42)

= a(a - 21) - 2(a - 21)

= (a - 21)(a - 2)

Hence,a2 - 23a + 42 = (a - 21)(a - 2)

Question 5(ii)

Factorise :

a2 - 23a - 108

Answer

a2 - 23a - 108

= a2 - 23a - 108

= a2 - (27 - 4)a - 108

= a2 - 27a + 4a - 108

= (a2 - 27a) + (4a - 108)

= a(a - 27) + 4(a - 27)

= (a - 27)(a + 4)

Hence,a2 - 23a - 108 = (a - 27)(a + 4)

Question 5(iii)

Factorise :

1 - 18x - 63x2

Answer

1 - 18x - 63x2

= 1 - (21 - 3)x - 63x2

= 1 - 21x + 3x - 63x2

= (1 - 21x) + (3x - 63x2)

= 1(1 - 21x) + 3x(1 - 9x)

= (1 - 21x)(1 + 3x)

Hence,1 - 18x - 63x2 = (1 - 21x)(1 + 3x)

Question 5(iv)

Factorise :

5x2 - 4xy - 12y2

Answer

5x2 - 4xy - 12y2

= 5x2 - (10 - 6)xy - 12y2

= 5x2 - 10xy + 6xy - 12y2

= (5x2 - 10xy) + (6xy - 12y2)

= 5x(x - 2y) + 6y(x - 2y)

= (x - 2y)(5x + 6y)

Hence,5x2 - 4xy - 12y2 = (x - 2y)(5x + 6y)

Question 5(v)

Factorise :

x(3x + 14) + 8

Answer

x(3x + 14) + 8

= 3x2 + 14x + 8

= 3x2 + (12 + 2)x + 8

= 3x2 + 12x + 2x + 8

= (3x2 + 12x) + (2x + 8)

= 3x(x + 4) + 2(x + 4)

= (x + 4)(3x + 2)

Hence,x(3x + 14) + 8 = (x + 4)(3x + 2)

Question 5(vi)

Factorise :

5 - 4x(1 + 3x)

Answer

5 - 4x(1 + 3x)

= 5 - 4x - 12x2

= 5 - (10 - 6)x - 12x2

= 5 - 10x + 6x - 12x2

= (5 - 10x) + (6x - 12x2)

= 5(1 - 2x) + 6x(1 - 2x)

= (1 - 2x)(5 + 6x)

Hence,5 - 4x(1 + 3x) = (1 - 2x)(5 + 6x)

Question 5(vii)

Factorise :

x2y2 - 3xy - 40

Answer

x2y2 - 3xy - 40

= x2y2 - (8 - 5)xy - 40

= x2y2 - 8xy + 5xy - 40

= (x2y2 - 8xy) + (5xy - 40)

= xy(xy - 8) + 5(xy - 8)

= (xy - 8)(xy + 5)

Hence, x2y2 - 3xy - 40 = (xy - 8)(xy + 5)

Question 5(viii)

Factorise :

(3x - 2y)2 - 5(3x - 2y) - 24

Answer

(3x - 2y)2 - 5(3x - 2y) - 24

= (3x - 2y)2 - (8 - 3)(3x - 2y) - 24

= (3x - 2y)2 - 8(3x - 2y) + 3(3x - 2y) - 24

= (3x - 2y)((3x - 2y) - 8) + 3((3x - 2y) - 8)

= (3x - 2y)(3x - 2y - 8) + 3(3x - 2y - 8)

= (3x - 2y - 8)(3x - 2y + 3)

Hence,(3x - 2y)2 - 5(3x - 2y) - 24 = (3x - 2y - 8)(3x - 2y + 3)

Question 5(ix)

Factorise :

12(a + b)2 - (a + b) - 35

Answer

12(a + b)2 - (a + b) - 35

= 12(a + b)2 - (21 - 20)(a + b) - 35

= 12(a + b)2 - 21(a + b) + 20(a + b) - 35

= 3(a + b)(4(a + b) - 7) + 5(4(a + b) - 7)

= 3(a + b)(4a + 4b - 7) + 5(4a + 4b - 7)

= (4a + 4b - 7)(3(a + b) + 5)

= (4a + 4b - 7)(3a + 3b + 5)

Hence,12(a + b)2 - (a + b) - 35 = (4a + 4b - 7)(3a + 3b + 5)

Question 6(i)

Factorise :

15(5x - 4)2 - 10(5x - 4)

Answer

15(5x - 4)2 - 10(5x - 4)

= 5(5x - 4)(3(5x - 4) - 2)

= 5(5x - 4)(15x - 12 - 2)

= 5(5x - 4)(15x - 14)

Hence,15(5x - 4)2 - 10(5x - 4) = 5(5x - 4)(15x - 14)

Question 6(ii)

Factorise :

3a2x - bx + 3a2 - b

Answer

3a2x - bx + 3a2 - b

= x(3a2 - b) + 1(3a2 - b)

= (3a2 - b)(x + 1)

Hence,3a2x - bx + 3a2 - b = (3a2 - b)(x + 1)

Question 6(iii)

Factorise :

b(c - d)2 + a(d - c) + 3(c - d)

Answer

b(c - d)2 + a(d - c) + 3(c - d)

= b(c - d)2 - a(c - d) + 3(c - d)

= (c - d)(b(c - d) - a + 3)

= (c - d)(bc - bd - a + 3)

Hence,b(c - d)2 + a(d - c) + 3(c - d) = (c - d)(bc - bd - a + 3)

Question 6(iv)

Factorise :

ax2 + b2y - ab2 - x2y

Answer

ax2 + b2y - ab2 - x2y

= (ax2 - ab2) + (b2y - x2y)

= a (x2 - b2) + y(b2 - x2)

= a (x2 - b2) - y(x2 - b2)

= (x2 - b2)(a - y)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (x - b)(x + b)(a - y)

Hence,ax2 + b2y - ab2 - x2y = (x - b)(x + b)(a - y)

Question 6(v)

Factorise :

1 - 3x - 3y - 4(x + y)2

Answer

1 - 3x - 3y - 4(x + y)2

= 1 - 3(x + y) - 4(x + y)2

= 1 - (4 - 1)(x + y) - 4(x + y)2

= 1 - 4(x + y) + 1(x + y) - 4(x + y)2

= (1 - 4(x + y)) + (1(x + y) - 4(x + y)2)

= 1(1 - 4(x + y)) + (x + y)(1 - 4(x + y))

= 1(1 - 4x - 4y) + (x + y)(1 - 4x - 4y)

= (1 - 4x - 4y)(1 + (x + y))

= (1 - 4x - 4y)(1 + x + y)

Hence,1 - 3x - 3y - 4(x + y)2 = (1 - 4x - 4y)(1 + x + y)

Question 7(i)

Factorise :

2a3 - 50a

Answer

2a3 - 50a

= 2a(a2 - 25)

= 2a(a2 - 52)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 2a(a - 5)(a + 5)

Hence,2a3 - 50a = 2a(a - 5)(a + 5)

Question 7(ii)

Factorise :

54a2b2 - 6

Answer

54a2b2 - 6

= 6(9a2b2 - 1)

= 6((3ab)2 - (1)2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 6(3ab - 1)(3ab + 1)

Hence, 54a2b2 - 6 = 6(3ab - 1)(3ab + 1)

Question 7(iii)

Factorise :

64a2b - 144b3

Answer

64a2b - 144b3

= 16b(4a2 - 9b2)

= 16b((2a)2 - (3b)2)

= 16b(2a + 3b)(2a - 3b)

Hence, 64a2b - 144b3 = 16b(2a + 3b)(2a - 3b)

Question 7(iv)

Factorise :

(2x - y)3 - (2x - y)

Answer

(2x - y)3 - (2x - y)

= (2x - y)((2x - y)2 - 1)

= (2x - y)((2x - y)2 - (1)2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (2x - y)((2x - y) - 1)((2x - y) + 1)

= (2x - y)(2x - y - 1)(2x - y + 1)

Hence, (2x - y)3 - (2x - y) = (2x - y)(2x - y - 1)(2x - y + 1)

Question 7(v)

Factorise :

x2 - 2xy + y2 - z2

Answer

x2 - 2xy + y2 - z2

Using the formula

[∵ (x - y)2 = x2 + y2 - 2xy]

= (x2 - 2xy + y2) - z2

= (x - y)2 - z2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((x - y) - z)((x - y) + z)

= (x - y - z)(x - y + z)

Hence, x2 - 2xy + y2 - z2 = (x - y - z)(x - y + z)

Question 7(vi)

Factorise :

x2 - y2 - 2yz - z2

Answer

x2 - y2 - 2yz - z2

Using the formula

[∵ (x - y)2 = x2 + y2 - 2xy]

= x2 - (y2 + 2yz + z2)

= x2 - (y + z)2

= (x)2 - (y + z)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (x - (y + z))(x + (y + z))

= (x - y - z)(x + y + z)

Hence, x2 - y2 - 2yz - z2 = (x - y - z)(x + y + z)

Question 7(vii)

Factorise :

7a5 - 567a

Answer

7a5 - 567a

= 7a(a4 - 81)

= 7a((a2)4 - (9)4)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 7a(a2 - 9)(a2 + 9)

= 7a((a)2 - (3)2)(a2 + 9)

= 7a(a - 3)(a + 3)(a2 + 9)

Hence, 7a5 - 567a = 7a(a - 3)(a + 3)(a2 + 9)

Question 7(viii)

Factorise :

5x220x495x^2 - \dfrac{20x^4}{9}

Answer

5x220x49=5x2(14x29)=5x2(12(2x3)2)5x^2 - \dfrac{20x^4}{9}\\[1em] = 5x^2\Big(1 - \dfrac{4x^2}{9}\Big)\\[1em] = 5x^2\Big(1^2 - \Big(\dfrac{2x}{3}\Big)^2\Big)\\[1em]

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

=5x2(12x3)(1+2x3)= 5x^2\Big(1 - \dfrac{2x}{3}\Big)\Big(1 + \dfrac{2x}{3}\Big)

Hence, 5x220x49=5x2(12x3)(1+2x3)5x^2 - \dfrac{20x^4}{9} = 5x^2\Big(1 - \dfrac{2x}{3}\Big)\Big(1 + \dfrac{2x}{3}\Big)

Question 8

Factorise xy2 - xz2, Hence, find the value of :

(i) 9 x 82 - 9 x 22

(ii) 40 x 5.52 - 40 x 4.52

Answer

xy2 - xz2

= x(y2 - z2)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= x(y - z)(y + z)

Hence, xy2 - xz2 = x(y - z)(y + z)

(i) 9 x 82 - 9 x 22

= 9 x (8 - 2)(8 + 2)

= 9 x 6 x 10

= 540

Hence, 9 x 82 - 9 x 22 = 540

(ii) 40 x 5.52 - 40 x 4.52

= 40 x (5.5 - 4.5) x (5.5 + 4.5)

= 40 x 1 x 10

= 400

Hence, 40 x 5.52 - 40 x 4.52 = 400

Question 9(i)

Factorise :

(a - 3b)2 - 36 b2

Answer

(a - 3b)2 - 36 b2

= (a - 3b)2 - (6 b)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= ((a - 3b) - 6b)((a - 3b) + 6b)

= (a - 3b - 6b)(a - 3b + 6b)

= (a - 9b)(a + 3b)

Hence, (a - 3b)2 - 36 b2 = (a - 9b)(a + 3b)

Question 9(ii)

Factorise :

25(a - 5b)2 - 4(a - 3b)2

Answer

25(a - 5b)2 - 4(a - 3b)2

= (5(a - 5b))2 - (2(a - 3b))2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (5(a - 5b) - 2(a - 3b))((5(a - 5b)) + 2(a - 3b))

= (5a - 25b - 2a + 6b)(5a - 25b + 2a - 6b)

= (5a - 2a - 25b + 6b)(5a + 2a - 25b - 6b)

= (3a - 19b)(7a - 31b)

Hence, 25(a - 5b)2 - 4(a - 3b)2 = (3a - 19b)(7a - 31b)

Question 9(iii)

Factorise :

a2 - 0.36 b2

Answer

a2 - 0.36 b2

= (a)2 - (0.6 b)2

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= (a + 0.6b)(a - 0.6b)

Hence, a2 - 0.36 b2 = (a + 0.6b)(a - 0.6b)

Question 9(iv)

Factorise :

x4 - 5x2 - 36

Answer

x4 - 5x2 - 36

= x4 - (9 - 4)x2 - 36

= x4 - 9x2 + 4x2 - 36

= (x4 - 9x2) + (4x2 - 36)

= x2(x2 - 9) + 4(x2 - 9)

= (x2 - 9)(x2 + 4)

= ((x)2 - (3)2)(x2 + 4)

= (x - 3)(x + 3)(x2 + 4)

Hence, x4 - 5x2 - 36 = (x - 3)(x + 3)(x2 + 4)

Question 9(v)

Factorise :

15(2x - y)2 - 16(2x - y) - 15

Answer

15(2x - y)2 - 16(2x - y) - 15

= 15(2x - y)2 - (25 - 9)(2x - y) - 15

= 15(2x - y)2 - 25(2x - y) + 9(2x - y) - 15

= (15(2x - y)2 - 25(2x - y)) + (9(2x - y) - 15)

= 5(2x - y)(3(2x - y) - 5) + 3(3(2x - y) - 5)

= 10x - 5y(6x - 3y - 5) + 3(6x - 3y - 5)

= (6x - 3y - 5)( 10x - 5y + 3)

Hence, 15(2x - y)2 - 16(2x - y) - 15 = (6x - 3y - 5)( 10x - 5y + 3)

Question 10

Evaluate (using factors) : 3012 x 300 - 3003

Answer

3012 x 300 - 3003

= 300 x (3012 - 3002)

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

= 300 x (301 - 300) x (301 + 300)

= 300 x 1 x 601

= 1,80,300

Hence, 3012 x 300 - 3003 = 1,80,300

Question 11(i)

Use factor method to evaluate:

(5z2 - 80) ÷ (z - 4)

Answer

(5z280)(z4)=5(z216)z4\dfrac{(5z^2 - 80)}{(z - 4)}\\[1em] = \dfrac{5(z^2 - 16)}{z - 4}

Using the formula

[∵ (x2 - y2) = (x + y)(x - y)]

5(z242)(z4)=5(z4)(z+4)(z4)=5(z4)(z+4)(z4)=5(z+4)\dfrac{5(z^2 - 4^2)}{(z - 4)}\\[1em] = \dfrac{5(z - 4)(z + 4)}{(z - 4)}\\[1em] = \dfrac{5\cancel{(z - 4)}(z + 4)}{\cancel{(z - 4)}}\\[1em] = 5(z + 4)

Hence, (5z2 - 80) ÷ (z - 4) = 5(z + 4)

Question 11(ii)

Use factor method to evaluate:

10y(6y + 21) ÷ (2y + 7)

Answer

10y(6y+21)(2y+7)10y×3(2y+7)(2y+7)=30y(2y+7)2y+7=30y(2y+7)(2y+7)=30y\dfrac{10y(6y + 21)}{(2y + 7)}\\[1em] \dfrac{10y \times 3(2y + 7)}{(2y + 7)}\\[1em] = \dfrac{30y(2y + 7)}{2y + 7}\\[1em] = \dfrac{30y\cancel{(2y + 7)}}{\cancel{(2y + 7)}}\\[1em] = 30y

Hence, 10y(6y + 21) ÷ (2y + 7) = 30y

Question 11(iii)

Use factor method to evaluate:

(a2 - 14a - 32) ÷ (a + 2)

Answer

(a214a32)(a+2)=(a2(162)a32)(a+2)=(a216a+2a32)(a+2)=((a216a)+(2a32))(a+2)=(a(a16)+2(a16))(a+2)=((a16)(a+2))(a+2)=(a16)(a+2)(a+2)=(a16)\dfrac{(a^2 - 14a - 32)}{(a + 2)}\\[1em] = \dfrac{(a^2 - (16 - 2)a - 32)}{(a + 2)}\\[1em] = \dfrac{(a^2 - 16a + 2a - 32)}{(a + 2)}\\[1em] = \dfrac{((a^2 - 16a) + (2a - 32))}{(a + 2)}\\[1em] = \dfrac{(a(a - 16) + 2(a - 16))}{(a + 2)}\\[1em] = \dfrac{((a - 16)(a + 2))}{(a + 2)}\\[1em] = \dfrac{(a - 16)\cancel{(a + 2)}}{\cancel{(a + 2)}}\\[1em] = (a - 16)

Hence, (a2 - 14a - 32) ÷ (a + 2) = (a - 16)

Question 11(iv)

Use factor method to evaluate:

39x3(50x2 - 98) ÷ 26x2(5x + 7)

Answer

39x3(50x298)26x2(5x+7)=3x(50x298)2(5x+7)=3x×2(25x249)2(5x+7)=6x((5x)2(7)2)2(5x+7)=3x(5x7)(5x+7)(5x+7)=3x(5x7)(5x+7)(5x+7)=3x(5x7)\dfrac{39x^3 (50x^2 - 98)}{26x^2(5x + 7)}\\[1em] = \dfrac{3x (50x^2 - 98)}{2(5x + 7)}\\[1em] = \dfrac{3x \times 2(25x^2 - 49)}{2(5x + 7)}\\[1em] = \dfrac{6x((5x)^2 - (7)^2)}{2(5x + 7)}\\[1em] = \dfrac{3x(5x - 7)(5x + 7)}{(5x + 7)}\\[1em] = \dfrac{3x(5x - 7)\cancel{(5x + 7)}}{\cancel{(5x + 7)}}\\[1em] = 3x(5x - 7)

Hence, 39x3(50x2 - 98) ÷ 26x2(5x + 7) = 3x(5x - 7)

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