## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 4 Factorisation Chapter Test

Factorize the following (1 to 12)

Question 1.

15(2x – 3)^{3} – 10(2x – 3)

Answer:

(i) 15(2x – 3)^{3} – 10(2x – 3)

= 5(2x – 3) [3(2x – 3)^{2} – 2]

= 5(2x – 3)[3(2x – 3)^{2} – 2]

(ii) a(b – c) (b + c) – d (c – b)

Answer:

a(b – c) (b + c) – d (c – b)

= a(b – c) (b + c) + d (b – c)

= (b – c) [a (b + c) + d]

= (b – c) (ab + ac + d)

Question 2.

(i) 2a^{2}x – bx + 2a^{2} – b

Answer:

(i) 2a^{2}x – bx + 2a^{2} – b

= 2a^{2}x + 2a^{2} – bx – b

= 2a^{2} (x + 1) – b (x + 1)

= (x + 1) (2a^{2} – b)

(ii) p^{2} – (a + 2b) p + 2ab

Answer:

p^{2} – (a + 2b) p + 2ab

= p^{2} – ap – 2bp + 2ab

= p(p – a) – 2b (p – a)

= (p – a) (p – 2b)

Question 3.

(i) (x^{2} – y^{2}) z + (y^{2} – z^{2}) x

Answer:

(i) (x^{2} – y^{2}) z + (y^{2} – z^{2}) x

= x^{2}z – y^{2}z + xy^{2} – z^{2} x

= x^{2}z – z^{2}x + xy^{2} – y^{2}z

= xz (x – z) + y^{2} (x – z)

= (x – 2) (xz + y^{2})

(ii) 5a^{4} – 5a^{3} + 30a^{2} – 30a

Answer:

5a^{4} – 5a^{3} + 30a^{2} – 30a

= 5a [a^{3} – a^{2} + 6a – 6]

= 5a [a^{2} (a – 1) + 6 (a – 1)]

= 5a (a – 1) (a^{2} + 6)

Question 4.

(i) b(c – d)^{2} + a(d – c) + 3c – 3d

Answer:

b(c – d)^{2} + a(d – c) + 3c – 3d

= b(c – d)^{2} – a(c – d) + 3c – 3d

= b(c – d)^{2} – a (c – d) + 3 (c – d)

= (c – d) [b (c – d) – a + 3]

= (c – d) (bc – bd – a + 3)

(ii) x^{3} – x^{2} – xy + x + y – 1

Answer:

x^{3} – x^{2} – xy + x + y – 1

= x^{3} – x^{2} – xy + y + x – 1.

= x^{2} (x – 1) – y (x – 1) + 1(x – 1)

= (x – 1) (x^{2} – y + 1)

Question 5.

(i) x(x + z) – y (y + z)

Answer:

x (x + z) – y(y + z)

= x^{2} + xz – y^{2} – yz

= x^{2} – y^{2} + xz – yz

= (x + y)(x – y) + z(x – y) {∵ x^{2} – y^{2} = (x + y)(x – y)}

= (x – y)(x + y + z)

(ii) a^{12}x^{4} – a^{4}x^{12}

Answer:

a^{12}x^{4} – a^{4}x^{12} = a^{4}x^{4} (a^{8} – x^{8})

= a^{4}x^{4} {(a^{4})^{2} – (x^{4})^{2}}

= a^{4}x^{4} (a^{4} + x^{4}) (a^{4} – x^{4}) (∵ a^{2} – b^{2} = (a + b) (a – b)}

= a^{4}x^{4} (a^{4} + x^{4}) {(a^{2})^{2} – (x^{2})^{2}}

= a^{4}x^{4} (a^{4} + x^{4}) (a^{2} + x^{2}) (a^{2} – x^{2})

= a^{4}x^{4} (a^{4} + x^{4}) (a^{2} + x^{2}) (a + x) (a – x)

Question 6.

(i) 9x^{2} + 12x + 4 – 16y^{2}

Answer:

9x^{2} + 12x + 4 – 16y^{2}

⇒ (3x)^{2} + 2 × 3x × 2 + (2)^{2} – 16y^{2}

⇒ (3x + 2)^{2} + (4y)^{2}

⇒ (3x + 2 + 4y) (3x + 2 – 4y)

(ii) x^{4} + 3x^{2} + 4

Answer:

x^{4} + 3x^{2} + 4

= (x^{2})^{3} + 3(x^{2}) + 4

= (x^{2})^{2} + (2)^{2} + 4x^{2} – x^{2}

= (x^{2} + 2)^{2} – (x)^{2} {∵ a^{2} – b^{2} = (a + b) (a – b)}

= (x^{2} + 2 + x) (x^{2} + 2 – x)

= (x^{2} + x + 2) (x^{2} – x + 2)

Question 7.

(i) 21x^{2} – 59xy + 40y^{2}

Answer:

(i) 21x^{2} – 59xy + 40y^{2}

∵ 21 × 40 = 840

∴ 840 = (- 35) (- 29)

and – 59 = – 35 – 24

= 7x (3x – 5y) – 8y (3x – 5y)

= (3x – 5y) (7x – 8y)

(ii) 4x^{3}y – 44x^{2}y + 112xy

Answer:

4x^{3}y – 44x^{2}y + 112xy

= 4xy (x^{2} – 11x + 28)

= 4xy [x^{2} – 7x – 4x + 28]

∵ – 11 = -7 – 4

28 = (-7) (-4)

= 4xy [x (x – 7) – 4 (x – 7)]

= 4xy(x – 7) (x – 4)

Question 8.

(i) x^{2}y^{2} – xy – 72

Answer:

x^{2}y^{2} – xy – 72

= x^{2}y^{2} – 9xy + 8xy – 72 { ∵ – 72 = -9 × 8l

-1 = -9 + 8 }

= xy(xy – 9) + 8 (xy – 9)

= (xy – 9) (xy + 8)

(ii) 9x^{3}y + 41x^{2}y^{2} + 20xy^{3}

Answer:

9x^{3}y + 41x^{2}y^{2} + 20xy^{3}

= xy (9x^{2} + 41xy + 20y^{2})

= xy {9x^{2} + 36xy + 5xy + 2oy^{2}}

{∵ 9 × 20 = 180

180 = 36 × 5

41 = 36 + 5}

= xy {9x (x + 4y) + 5y (x + 4y)}

= xy (x + 4y) (9x + 5y)

Question 9.

(i) (3a – 2b)^{2} + 3 (3a – 2b) – 10

Answer:

(3a – 2b)^{2} + 3 (3a – 2b) – 10

= x^{2} + 3x – 10 where x = (3a- 2b)

= (x^{2} + 5x) – (2x +10)

= x (x + 5) – 2 (x + 5)

= (x + 5) (x – 2)

= (3a – 2b + 5) (3a – 2b – 2)

(ii) (x^{2} – 3x) (x^{2} – 3x + 7) + 10

Answer:

(x^{2} – 3x) (x^{2} – 3x + 7) + 10

Put x^{2} – 3x = 3

then, y(y + 7) + 10

= y^{2} + 7y + 10

= y^{2} + 5y + 2y + 10

= y(y + 5) + 2(y + 5)

= (y + 5) (y + 2)

= (x^{2} – 3x + 5) (x^{2} – 3x + 2)

Question 10.

(i) (x^{2} – x) (4x^{2} – 4x – 5) – 6

Answer:

(x^{2} – x) (4x^{2} – 4x – 5) – 6

= (x^{2} -x) [4 (x^{2} – x) – 5] – 6

= y(4y – 5) – 6 where y = x^{2} – x

= 4y^{2} – 5y – 6

= 4y – 8y + 3y – 6

= 4y (y – 2) + 3 (y – 2)

= (4y + 3)(y – 2)

= (4x^{2} – 4x + 3) (x^{2} – x – 2)

= (x^{2} – x – 2) (4x^{2} – 4x + 3)

= (x^{2} – 2x + x – 2) (4x^{2} – 4x + 3)

= [x(x – 2) + 1 (x – 2)] (4x^{2} – 4x + 3)

= (x – 2) (x +1) (4x^{2} – 4x + 3)

(ii) x^{4} + 9x^{2}y^{2} + 81y^{4}

Answer:

x^{4} + 9x^{2}y^{2} + 81y^{4}

= (x^{4} + 18x^{2}y^{2} + 81y^{4}) – 9x^{2}y^{2}

= (x^{2} + 9y^{2})^{2} – (3xy)^{2}

= (x^{2} + 9y^{2} + 3xy) (x^{2} + 9y^{2} – 3xy)

Question 11.

(i) \(\frac{8}{27} x^{3}-\frac{1}{8} y^{3}\)

Answer:

(ii) x^{6} + 63x^{3} – 64

Answer:

x^{6} + 63x^{3} – 64 = x^{6} + 64x^{3} – x – 64

= x^{3} (x^{3} + 64) – 1 (x^{3} + 64)

= (x^{3} + 64) (x^{3} – 1)

= [(x)^{3} + (4)^{3}][(x)^{3} – (1)^{3}]

= (x + 4) [(x)^{2} – (4) (x) + (4)^{2}] (x – 1) [(x)^{2} + (1)(x) + (1)^{2}]

= (x + 4) (x^{2} – 4x + 16) (x – 1) (x^{2} + x + 1)

Question 12.

(i) \(x^{3}+x^{2}-\frac{1}{x^{2}}+\frac{1}{x^{3}}\)

Answer:

(ii) (x + 1)^{6} – (x – 1)^{6}

Answer:

(x + 1)^{6} – (x – 1)^{6}

[(X + 1)^{3}]^{2} – [(x – 1)^{3}]^{2} [(a^{2} – b^{2} = (a + b) (a – b)

= [(x + 1)^{3} + (x – 1)^{3}] [(x + 1)^{3} – (x – 1)^{3}]

= [(x + 1) + (x – 1)]

[(x+ 1)^{2} – (x – 1)(x- 1) + (x- 1)^{2}] [(x + 1) – (x – 1)] [(x + 1)^{2} + (x + 1) (x – 1) + (x – 1)^{2}]

= (x + 1 + x – 1) [x^{2} + 2x + 1 – x^{2} + 1 + x^{2} + 1 – 2x (x + 1) – x + 1) [x^{2} + 2x + 1 + x^{2} – 1 + x^{2} – 2x + 1 ]

= 2x (x^{2} + 3) (2) (3x^{2} + 1)

= 4x (x^{2} + 3) (3x^{2} + 1)

Question 13.

Show that (97)^{3} + (14)^{3} is divisible by 111.

Answer:

(97)^{3} + (14)^{3}

= (97 + 14) [(97)^{2} – (97) (14) + (14)^{2}]

= 111 × [(97)^{2} – (97)(14) + (14)^{2}]

Clearly given expression is divisible by 111.

Question 14.

If a + b = 8 and ab = 15, find the value of a^{4} + a^{2}b^{2} + b^{4}

Solution:

a^{4} + a^{2}b^{2} + b^{4}

= a^{4} + 2a^{2}b^{2} + b^{4} – a^{2}b^{2}

= (a^{2})^{2} + 2a^{2}b^{2} + (b^{2})^{2} – (ab)^{2}

= (a^{2} + b^{2})^{2} – (ab)^{2}

= (a^{2} + b^{2} + ab) (a^{2} + b – ab)

But a + b = 8, ab = 15

∴ (a + b)2^{2} = 8^{2}

⇒ a^{2} + b^{2} + 2ab = 64

⇒ a^{2} + b^{2} + 2 × 15 = 64

⇒ a^{2} + b^{2} + 30 = 64

a^{2} + b^{2} = 64 – 30 = 34

Now a^{4} + a^{2}b^{2} + b^{4}

= (a^{2} + b^{2} + ab) (a^{2} + b^{2} – ab)

= (34 + 15) (34 – 15)

= 49 × 19 = 931