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Chapter 5

Factorisation — Exercise 5(E)

Class - 9 Concise Mathematics Selina

Exercise 5(E)

Question 1(a)

x4+1x4−2x^4 + \dfrac{1}{x^4} - 2 in the form of factors is :

  1. (x−1x)2(x+1x+1)2\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} + 1\Big)^2

  2. (x−1x)2(x+1x)2\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2

  3. (x+1x)2(x−1x+1)2\Big(x + \dfrac{1}{x}\Big)^2\Big(x - \dfrac{1}{x} + 1\Big)^2

  4. (x+1x)2(x+1x−1)2\Big(x + \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} - 1\Big)^2

Answer

Given,

=x4+1x4−2=(x2)2+(1x2)2−2×x2×1x2=(x2−1x2)2=[(x−1x)(x+1x)]2=(x−1x)2(x+1x)2.\phantom{=}x^4 + \dfrac{1}{x^4} - 2 \\[1em] = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 - 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] = \Big(x^2 - \dfrac{1}{x^2}\Big)^2 \\[1em] = \Big[\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\Big]^2 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2.

Hence, Option 2 is the correct option.

Question 1(b)

x2+1x2−3x^2 + \dfrac{1}{x^2} - 3 in the form of factors is :

  1. (x−1x)(x+1x)\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big) - 1

  2. (x+1x+1)(x−1x−1)\Big(x + \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)

  3. (x−1x)(x−1x−1)\Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 1\Big)

  4. (x−1x+1)(x−1x−1)\Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)

Answer

Given,

=x2+1x2−3=x2+1x2−2−1=x2+1x2−(2×x×1x)−1=[x2+1x2−(2×x×1x)]−1=(x−1x)2−1=(x−1x)2−12=(x−1x+1)(x−1x−1).\phantom{=}x^2 + \dfrac{1}{x^2} - 3 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 1 \\[1em] = x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big) - 1 \\[1em] = \Big[x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big)\Big] - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).

Hence, Option 4 is the correct option.

Factorise :

Question 2

x2+14x2+1−7x−72xx^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x}

Answer

Given,

=x2+14x2+1−7x−72x=x2+(12x)2+1−7(x+12x)=x2+(12x)2+2×x×12x−7(x+12x)=(x+12x)2−7(x+12x)=(x+12x)(x+12x−7).\phantom{=} x^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x} \\[1em] = x^2 + \Big(\dfrac{1}{2x}\Big)^2 + 1 - 7\Big(x + \dfrac{1}{2x}\Big) \\[1em] = x^2 + \Big(\dfrac{1}{2x}\Big)^2 + 2 \times x \times \dfrac{1}{2x} - 7\Big(x + \dfrac{1}{2x}\Big) \\[1em] = \Big(x + \dfrac{1}{2x}\Big)^2 - 7\Big(x + \dfrac{1}{2x}\Big) \\[1em] = \Big(x + \dfrac{1}{2x}\Big)\Big(x + \dfrac{1}{2x} - 7\Big).

Hence, x2+14x2+1−7x−72x=(x+12x)(x+12x−7).x^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x} = \Big(x + \dfrac{1}{2x}\Big)\Big(x + \dfrac{1}{2x} - 7\Big).

Question 3

9a2+19a2−2−12a+43a9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a}

Answer

Given,

=9a2+19a2−2−12a+43a=(3a)2+(13a)2−2−4(3a−13a)=(3a−13a)2−4(3a−13a)=(3a−13a)(3a−13a−4).\phantom{=} 9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a} \\[1em] = (3a)^2 + \Big(\dfrac{1}{3a}\Big)^2 - 2 - 4\Big(3a - \dfrac{1}{3a}\Big) \\[1em] = \Big(3a - \dfrac{1}{3a}\Big)^2 - 4\Big(3a - \dfrac{1}{3a}\Big) \\[1em] = \Big(3a - \dfrac{1}{3a}\Big)\Big(3a - \dfrac{1}{3a} - 4\Big).

Hence, 9a2+19a2−2−12a+43a=(3a−13a)(3a−13a−4).9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a} = \Big(3a - \dfrac{1}{3a}\Big)\Big(3a - \dfrac{1}{3a} - 4\Big).

Question 4

x2+a2+1ax+1x^2 + \dfrac{a^2 + 1}{a}x + 1

Answer

Given,

=x2+a2+1ax+1=ax2+(a2+1)x+aa=ax2+a2x+x+aa=ax(x+a)+(x+a)a=(x+a)(ax+1)a=(x+a)ax+1a=(x+a)(x+1a).\phantom{=} x^2 + \dfrac{a^2 + 1}{a}x + 1 \\[1em] = \dfrac{ax^2 + (a^2 + 1)x + a}{a} \\[1em] = \dfrac{ax^2 + a^2x + x + a}{a} \\[1em] = \dfrac{ax(x + a) + (x + a)}{a} \\[1em] = \dfrac{(x + a)(ax + 1)}{a} \\[1em] = (x + a)\dfrac{ax + 1}{a} \\[1em] = (x + a)\Big(x + \dfrac{1}{a}\Big).

Hence, x2+a2+1ax+1=(x+a)(x+1a).x^2 + \dfrac{a^2 + 1}{a}x + 1 = (x + a)\Big(x + \dfrac{1}{a}\Big).

Question 5

x4 + y4 - 27x2y2

Answer

Given,

    x4 + y4 - 27x2y2

= x4 + y4 - 2x2y2 - 25x2y2

= (x2 - y2)2 - (5xy)2

= (x2 - y2 + 5xy)(x2 - y2 - 5xy).

Hence, x4 + y4 - 27x2y2 = (x2 - y2 + 5xy)(x2 - y2 - 5xy).

Question 6

4x4 + 9y4 + 11x2y2

Answer

Given,

    4x4 + 9y4 + 11x2y2

= (2x2)2 + (3y2)2 + 12x2y2 - x2y2

= (2x2)2 + (3y2)2 + 2 × 2x2 × 3y2 - x2y2

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

= (2x2 + 3y2 + xy)(2x2 + 3y2 - xy).

Hence, 4x4 + 9y4 + 11x2y2 = (2x2 + 3y2 + xy)(2x2 + 3y2 - xy).

Question 7

x2+1x2−3x^2 + \dfrac{1}{x^2} - 3

Answer

Given,

=x2+1x2−3=x2+1x2−2−1=(x−1x)2−12=(x−1x+1)(x−1x−1).\phantom{=}x^2 + \dfrac{1}{x^2} - 3 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).

Hence, x2+1x2−3=(x−1x+1)(x−1x−1).x^2 + \dfrac{1}{x^2} - 3 = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).

Question 8

a - b - 4a2 + 4b2

Answer

Given,

    a - b - 4a2 + 4b2

= a - b - 4(a2 - b2)

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

= (a - b)[1 - 4(a + b)]

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

Hence, a - b - 4a2 + 4b2 = (a - b)(1 - 4a - 4b).

Question 9

(2a - 3)2 - 2(2a - 3)(a - 1) + (a - 1)2

Answer

Given,

    (2a - 3)2 - 2(2a - 3)(a - 1) + (a - 1)2

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

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

= (a - 2)2.

Hence, (2a - 3)2 - 2(2a - 3)(a - 1) + (a - 1)2 = (a - 2)2.

Question 10

(a2 - 3a)(a2 - 3a + 7) + 10

Answer

Given,

    (a2 - 3a)(a2 - 3a + 7) + 10

Substituting a2 - 3a = x, we get :

⇒ x(x + 7) + 10

= x2 + 7x + 10

= x2 + 2x + 5x + 10

= x(x + 2) + 5(x + 2)

= (x + 5)(x + 2)

= (a2 - 3a + 5)(a2 - 3a + 2)

= (a2 - 3a + 5)(a2 - 2a - a + 2)

= (a2 - 3a + 5)[a(a - 2) -1(a - 2)]

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

Hence, (a2 - 3a)(a2 - 3a + 7) + 10 = (a2 - 3a + 5)(a - 1)(a - 2).

Question 11

(a2 - a)(4a2 - 4a - 5) - 6

Answer

Given,

    (a2 - a)(4a2 - 4a - 5) - 6

= (a2 - a)[4(a2 - a) - 5] - 6

Substituting a2 - a = x, we get :

= x(4x - 5) - 6

= 4x2 - 5x - 6

= 4x2 - 8x + 3x - 6

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

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

= (a2 - a - 2)[4(a2 - a) + 3]

= (a2 - a - 2)(4a2 - 4a + 3)

= (a2 - 2a + a - 2)(4a2 - 4a + 3)

= [a(a - 2) + 1(a - 2)](4a2 - 4a + 3)

= (a - 2)(a + 1)(4a2 - 4a + 3).

Hence, (a2 - a)(4a2 - 4a - 5) - 6 = (a - 2)(a + 1)(4a2 - 4a + 3).

Question 12

x4 + y4 - 3x2y2

Answer

Given,

    x4 + y4 - 3x2y2

= (x2)2 + (y2)2 - 2x2y2 - x2y2

= (x2 - y2)2 - (xy)2

= (x2 - y2 + xy)(x2 - y2 - xy).

Hence, x4 + y4 - 3x2y2 = (x2 - y2 + xy)(x2 - y2 - xy).

Question 13

5a2 - b2 - 4ab + 7a - 7b

Answer

Given,

    5a2 - b2 - 4ab + 7a - 7b

= 5a2 - 4ab - b2 + 7a - 7b

= 5a2 - 5ab + ab - b2 + 7a - 7b

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

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

Hence, 5a2 - b2 - 4ab + 7a - 7b = (a - b)(5a + b + 7).

Question 14

12(3x - 2y)2 - 3x + 2y - 1

Answer

Given,

    12(3x - 2y)2 - 3x + 2y - 1

= 12(3x - 2y)2 - (3x - 2y) - 1

Substituting (3x - 2y) = a, we get :

= 12a2 - a - 1

= 12a2 - 4a + 3a - 1

= 4a(3a - 1) + 1(3a - 1)

= (4a + 1)(3a - 1)

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

= (12x - 8y + 1)(9x - 6y - 1).

Hence, 12(3x - 2y)2 - 3x + 2y - 1 = (12x - 8y + 1)(9x - 6y - 1).

Question 15

4(2x - 3y)2 - 8x + 12y - 3

Answer

Given,

    4(2x - 3y)2 - 8x + 12y - 3

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

Substituting (2x - 3y) = a, we get :

= 4a2 - 4a - 3

= 4a2 - 6a + 2a - 3

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

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

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

= (4x - 6y + 1)(4x - 6y - 3).

Hence, 4(2x - 3y)2 - 8x + 12y - 3 = (4x - 6y + 1)(4x - 6y - 3).

Question 16

3 - 5x + 5y - 12(x - y)2

Answer

Given,

    3 - 5x + 5y - 12(x - y)2

= 3 - 5(x - y) - 12(x - y)2

Substituting x - y = a, we get :

= 3 - 5a - 12a2

= 3 - 9a + 4a - 12a2

= 3(1 - 3a) + 4a(1 - 3a)

= (1 - 3a)(3 + 4a)

= [1 - 3(x - y)][3 + 4(x - y)]

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

Hence, 3 - 5x + 5y - 12(x - y)2 = (1 - 3x + 3y)(3 + 4x - 4y).

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