Rs Aggarwal 2019 for Class 9 Math Chapter 3 - Factorisation Of Polynomials

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Rs Aggarwal 2019 Solutions for Class 9 Math Chapter 3 Factorisation Of Polynomials are provided here with simple step-by-step explanations. These solutions for Factorisation Of Polynomials are extremely popular among Class 9 students for Math Factorisation Of Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 99:

Question 1:

Factorize:
9x² + 12xy

Answer:

We have:
$9 x^{2} + 12 x y = 3 x (3 x + 4 y)$

Page No 99:

Question 2:

Factorize:
18x²y − 24xyz

Answer:

We have:
$18 x^{2} y - 24 x y z = 6 x y (3 y - 4 z)$

Page No 99:

Question 3:

Factorize:
27a³b³ − 45a⁴b²

Answer:

We have:
$27 a^{3} b^{3} - 45 a^{4} b^{2} = 9 a^{3} b^{2} (3 b - 5 a)$

Page No 99:

Question 4:

Factorize:
2a(x + y) − 3b(x + y)

Answer:

We have:
$2 a (x + y) - 3 b (x + y) = (x + y) (2 a - 3 b)$

Page No 99:

Question 5:

Factorize:
2x(p² + q²) + 4y(p² + q²)

Answer:

We have:
$2 x (p^{2} + q^{2}) + 4 y (p^{2} + q^{2}) = 2 [x (p^{2} + q^{2}) + 2 y (p^{2} + q^{2})] = 2 (p^{2} + q^{2}) (x + 2 y)$

Page No 99:

Question 6:

Factorize:
x(a − 5) + y(5 − a)

Answer:

We have:
$x (a - 5) + y (5 - a) = x (a - 5) - y (a - 5)$
$= (a - 5) (x - y)$

Page No 99:

Question 7:

Factorize:
4(a + b) − 6(a + b)²

Answer:

We have:
$4 (a + b) - 6 {(a + b)}^{2} = 2 (a + b) [2 - 3 (a + b)]$
$= 2 (a + b) (2 - 3 a - 3 b)$

Page No 99:

Question 8:

Factorize:
8(3a − 2b)² − 10(3a − 2b)

Answer:

We have:
$8 {(3 a - 2 b)}^{2} - 10 (3 a - 2 b) = 2 (3 a - 2 b) [4 (3 a - 2 b) - 5]$
$= 2 (3 a - 2 b) (12 a - 8 b - 5)$

Page No 99:

Question 9:

Factorize:
x(x + y)³ − 3x²y(x + y)

Answer:

We have:
$x {(x + y)}^{3} - 3 x^{2} y (x + y) = x (x + y) [{(x + y)}^{2} - 3 x y]$
$= x (x + y) [x^{2} + y^{2} + 2 x y - 3 x y] = x (x + y) (x^{2} + y^{2} - x y)$

Page No 99:

Question 10:

Factorize:
x³ + 2x² + 5x + 10

Answer:

We have:
$x^{3} + 2 x^{2} + 5 x + 10 = (x^{3} + 2 x^{2}) + (5 x + 10)$
$= x^{2} (x + 2) + 5 (x + 2) = (x + 2) (x^{2} + 5)$

Page No 99:

Question 11:

Factorize:
x² + xy − 2xz − 2yz

Answer:

We have:
$x^{2} + x y - 2 x z - 2 y z = (x^{2} + x y) - (2 x z + 2 y z) = x (x + y) - 2 z (x + y) = (x + y) (x - 2 z)$

Page No 99:

Question 12:

Factorize:
a³b − a²b + 5ab − 5b

Answer:

We have:
$a^{3} b - a^{2} b + 5 a b - 5 b = b (a^{3} - a^{2} + 5 a - 5) = b [(a^{3} - a^{2}) + (5 a - 5)]$

$= b [a^{2} (a - 1) + 5 (a - 1)] = b (a - 1) (a^{2} + 5)$

Page No 99:

Question 13:

Factorize:
8 − 4a − 2a³ + a⁴

Answer:

We have:
$8 - 4 a - 2 a^{3} + a^{4} = (8 - 4 a) - (2 a^{3} - a^{4}) = 4 (2 - a) - a^{3} (2 - a) = (2 - a) (4 - a^{3})$

Page No 99:

Question 14:

Factorize:
x³ − 2x²y + 3xy² − 6y³

Answer:

We have:
$x^{3} - 2 x^{2} y + 3 x y^{2} - 6 y^{3} = (x^{3} - 2 x^{2} y) + (3 x y^{2} - 6 y^{3})$
$= x^{2} (x - 2 y) + 3 y^{2} (x - 2 y) = (x - 2 y) (x^{2} + 3 y^{2})$

Page No 99:

Question 15:

Factorize:
px − 5q + pq − 5x

Answer:

We have:
$p x - 5 q + p q - 5 x = (p x - 5 x) + (p q - 5 q)$
$= x (p - 5) + q (p - 5) = (p - 5) (x + q)$

Page No 99:

Question 16:

Factorize:
x² + y − xy − x

Answer:

We have:
$x^{2} + y - x y - x = (x^{2} - x y) - (x - y)$
$= x (x - y) - 1 (x - y) = (x - y) (x - 1)$

Page No 99:

Question 17:

Factorize:
(3a − 1)² − 6a + 2

Answer:

We have:
${(3 a - 1)}^{2} - 6 a + 2 = {(3 a - 1)}^{2} - 2 (3 a - 1)$
$= (3 a - 1) [(3 a - 1) - 2] = (3 a - 1) (3 a - 1 - 2) = (3 a - 1) (3 a - 3) = 3 (3 a - 1) (a - 1)$

Page No 99:

Question 18:

Factorize:
(2x − 3)² − 8x + 12

Answer:

We have:
${(2 x - 3)}^{2} - 8 x + 12 = {(2 x - 3)}^{2} - 4 (2 x - 3)$
$= (2 x - 3) [(2 x - 3) - 4] = (2 x - 3) (2 x - 3 - 4) = (2 x - 3) (2 x - 7)$

Page No 99:

Question 19:

Factorize:
a³ + a − 3a² − 3

Answer:

We have:
$a^{3} + a - 3 a^{2} - 3 = (a^{3} - 3 a^{2}) + (a - 3)$
$= a^{2} (a - 3) + 1 (a - 3) = (a - 3) (a^{2} + 1)$

Page No 99:

Question 20:

Factorize:
3ax − 6ay − 8by + 4bx

Answer:

We have:
$3 a x - 6 a y - 8 b y + 4 b x = (3 a x - 6 a y) + (4 b x - 8 b y)$
$= 3 a (x - 2 y) + 4 b (x - 2 y) = (x - 2 y) (3 a + 4 b)$

Page No 99:

Question 21:

Factorize:
abx² + a²x + b²x + ab

Answer:

We have:
$a b x^{2} + a^{2} x + b^{2} x + a b = (a b x^{2} + b^{2} x) + (a^{2} x + a b)$
$= b x (a x + b) + a (a x + b) = (a x + b) (b x + a)$

Page No 99:

Question 22:

Factorize:
x³ − x² + ax + x − a − 1

Answer:

We have:
$x^{3} - x^{2} + a x + x - a - 1 = (x^{3} - x^{2}) + (a x - a) + (x - 1)$
$= x^{2} (x - 1) + a (x - 1) + 1 (x - 1) = (x - 1) (x^{2} + a + 1)$

Page No 100:

Question 23:

Factorize:
2x + 4y − 8xy − 1

Answer:

We have:
$2 x + 4 y - 8 x y - 1 = (2 x - 8 x y) - (1 - 4 y)$
$= 2 x (1 - 4 y) - 1 (1 - 4 y) = (1 - 4 y) (2 x - 1)$

Page No 100:

Question 24:

Factorize:
ab(x² + y²) − xy(a² + b²)

Answer:

We have:
$a b (x^{2} + y^{2}) - x y (a^{2} + b^{2}) = a b x^{2} + a b y^{2} - a^{2} x y - b^{2} x y$
$= (a b x^{2} - a^{2} x y) - (b^{2} x y - a b y^{2}) = a x (b x - a y) - b y (b x - a y) = (b x - a y) (a x - b y)$

Page No 100:

Question 25:

Factorize:
a² + ab(b + 1) + b³

Answer:

We have:
$a^{2} + a b (b + 1) + b^{3} = a^{2} + a b^{2} + a b + b^{3}$
$= (a^{2} + a b^{2}) + (a b + b^{3}) = a (a + b^{2}) + b (a + b^{2}) = (a + b^{2}) (a + b)$

Page No 100:

Question 26:

Factorize:
a³ + ab(1 − 2a) − 2b²

Answer:

We have:
$a^{3} + a b (1 - 2 a) - 2 b^{2} = a^{3} + a b - 2 a^{2} b - 2 b^{2}$
$= (a^{3} - 2 a^{2} b) + (a b - 2 b^{2}) = a^{2} (a - 2 b) + b (a - 2 b) = (a - 2 b) (a^{2} + b)$

Page No 100:

Question 27:

Factorize:
2a² + bc − 2ab − ac2

Answer:

We have:
$2 a^{2} + b c - 2 a b - a c = (2 a^{2} - 2 a b) - (a c - b c)$
$= 2 a (a - b) - c (a - b) = (a - b) (2 a - c)$

Page No 100:

Question 28:

Factorize:
(ax + by)² + (bx − ay)²

Answer:

We have:
${(a x + b y)}^{2} + {(b x - a y)}^{2} = [{(a x)}^{2} + 2 \times a x \times b y + {(b y)}^{2}] + [{(b x)}^{2} - 2 \times b x \times a y + {(a y)}^{2}]$
$= a^{2} x^{2} + 2 a b x y + b^{2} y^{2} + b^{2} x^{2} - 2 a b x y + a^{2} y^{2} = a^{2} x^{2} + b^{2} y^{2} + b^{2} x^{2} + a^{2} y^{2} = (a^{2} x^{2} + b^{2} x^{2}) + (a^{2} y^{2} + b^{2} y^{2}) = x^{2} (a^{2} + b^{2}) + y^{2} (a^{2} + b^{2}) = (a^{2} + b^{2}) (x^{2} + y^{2})$

Page No 100:

Question 29:

Factorize:
a(a + b − c) − bc

Answer:

We have:
$a (a + b - c) - b c = a^{2} + a b - a c - b c$
$= (a^{2} - a c) + (a b - b c) = a (a - c) + b (a - c) = (a - c) (a + b)$

Page No 100:

Question 30:

Factorize:
a(a − 2b − c) + 2bc

Answer:

We have:
$a (a - 2 b - c) + 2 b c = a^{2} - 2 a b - a c + 2 b c$
$= (a^{2} - 2 a b) - (a c - 2 b c) = a (a - 2 b) - c (a - 2 b) = (a - 2 b) (a - c)$

Page No 100:

Question 31:

Factorize:
a²x² + (ax² + 1)x + a

Answer:

We have:
$a^{2} x^{2} + (a x^{2} + 1) x + a = (a x^{2} + 1) x + (a^{2} x^{2} + a)$
$= x (a x^{2} + 1) + a (a x^{2} + 1) = (a x^{2} + 1) (x + a)$

Page No 100:

Question 32:

Factorize:
ab(x² + 1) + x(a² + b²)

Answer:

We have:
$a b (x^{2} + 1) + x (a^{2} + b^{2}) = a b x^{2} + a b + a^{2} x + b^{2} x$
$= (a b x^{2} + a^{2} x) + (b^{2} x + a b) = a x (b x + a) + b (b x + a) = (b x + a) (a x + b)$

Page No 100:

Question 33:

Factorize:
x² − (a + b)x + ab

Answer:

We have:
$x^{2} - (a + b) x + a b = x^{2} - a x - b x + a b$
$= (x^{2} - a x) - (b x - a b) = x (x - a) - b (x - a) = (x - a) (x - b)$

Page No 100:

Question 34:

Factorize:
$x^{2} + \frac{1}{x^{2}} - 2 - 3 x + \frac{3}{x}$

Answer:

$We have : x^{2} + \frac{1}{x^{2}} - 2 - 3 x + \frac{3}{x} = x^{2} - 2 + \frac{1}{x^{2}} - 3 x + \frac{3}{x} = {(x)}^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} - 3 (x - \frac{1}{x}) = {(x - \frac{1}{x})}^{2} - 3 (x - \frac{1}{x}) = (x - \frac{1}{x}) (x - \frac{1}{x} - 3)$

Page No 105:

Question 1:

Factorise:
9x² – 16y²

Answer:

$9 x^{2} - 16 y^{2} = {(3 x)}^{2} - {(4 y)}^{2} = (3 x + 4 y) (3 x - 3 y) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 2:

Factorise:
$(\frac{25}{4} x^{2} - \frac{1}{9} y^{2})$

Answer:

$(\frac{25}{4} x^{2} - \frac{1}{9} y^{2}) = {(\frac{5}{2} x)}^{2} - {(\frac{1}{3} y)}^{2} = (\frac{5}{2} x + \frac{1}{3} y) (\frac{5}{2} x - \frac{1}{3} y) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 3:

Factorise:
81 – 16x²

Answer:

$81 - 16 x^{2} = 9^{2} - {(4 x)}^{2} = (9 + 4 x) (9 - 4 x) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 4:

Factorise:
5 – 20x²

Answer:

$5 - 20 x^{2} = 5 (1 - 4 x^{2}) = 5 [1^{2} - {(2 x)}^{2}] = 5 (1 + 2 x) (1 - 2 x) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 5:

Factorise:
2x⁴ – 32

Answer:

$2 x^{4} - 32 = 2 (x^{4} - 16) = 2 [{(x^{2})}^{2} - 4^{2}] = 2 (x^{2} + 4) (x^{2} - 4) [a^{2} - b^{2} = (a + b) (a - b)]$

$= 2 (x^{2} + 4) (x^{2} - 2^{2}) = 2 (x^{2} + 4) (x + 2) (x - 2) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 6:

Factorize:
3a³b − 243ab³

Answer:

$3 a^{3} b - 243 a b^{3} = 3 a b (a^{2} - 81 b^{2}) = 3 a b [a^{2} - {(9 b)}^{2}] = 3 a b (a - 9 b) (a + 9 b)$

Page No 105:

Question 7:

Factorize:
3x³ − 48x

Answer:

$3 x^{3} - 48 x = 3 x (x^{2} - 16) = 3 x (x^{2} - 4^{2}) = 3 x (x - 4) (x + 4)$

Page No 105:

Question 8:

Factorize:
27a² − 48b²

Answer:

$27 a^{2} - 48 b^{2} = 3 (9 a^{2} - 16 b^{2}) = 3 [{(3 a)}^{2} - {(4 b)}^{2}] = 3 (3 a - 4 b) (3 a + 4 b)$

Page No 105:

Question 9:

Factorize:
x − 64x³

Answer:

$x - 64 x^{3} = x (1 - 64 x^{2}) = x [1 - {(8 x)}^{2}] = x (1 - 8 x) (1 + 8 x)$

Page No 105:

Question 10:

Factorize:
8ab² − 18a³

Answer:

$8 a b^{2} - 18 a^{3} = 2 a (4 b^{2} - 9 a^{2}) = 2 a [{(2 b)}^{2} - {(3 a)}^{2}] = 2 a (2 b - 3 a) (2 b + 3 a)$

Page No 105:

Question 11:

Factorize:
150 − 6x²

Answer:

$150 - 6 x^{2} = 6 (25 - x^{2}) = 6 (5^{2} - x^{2}) = 6 (5 - x) (5 + x)$

Page No 105:

Question 12:

Factorise:
2 – 50x²

Answer:

$2 - 50 x^{2} = 2 (1 - 25 x^{2}) = 2 [1^{2} - {(5 x)}^{2}] = 2 (1 + 5 x) (1 - 5 x) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 13:

Factorise:
20x²– 45

Answer:

$20 x^{2} - 45 = 5 (4 x^{2} - 9) = 5 [{(2 x)}^{2} - 3^{2}] = 5 (2 x + 3) (2 x - 3) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 14:

Factorise:
(3a + 5b)² – 4c²

Answer:

${(3 a + 5 b)}^{2} - 4 c^{2} = {(3 a + 5 b)}^{2} - {(2 c)}^{2} = (3 a + 5 b + 2 c) (3 a + 5 b - 2 c) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 15:

Factorise:
a²– b² – a – b

Answer:

$a^{2} - b^{2} - a - b = (a + b) (a - b) - 1 (a + b) [a^{2} - b^{2} = (a + b) (a - b)] = (a + b) [(a - b) - 1] = (a + b) (a - b - 1)$

Page No 105:

Question 16:

Factorise:
4a²– 9b²– 2a – 3b

Answer:

$4 a^{2} - 9 b^{2} - 2 a - 3 b = {(2 a)}^{2} - {(3 b)}^{2} - 1 (2 a + 3 b) = (2 a + 3 b) (2 a - 3 b) - 1 (2 a + 3 b) [a^{2} - b^{2} = (a + b) (a - b)] = (2 a + 3 b) [(2 a - 3 b) - 1] = (2 a + 3 b) (2 a - 3 b - 1)$

Page No 105:

Question 17:

Factorise:
a²– b²+ 2bc – c²

Answer:

$a^{2} - b^{2} + 2 b c - c^{2} = a^{2} - (b^{2} - 2 b c + c^{2}) = a^{2} - {(b - c)}^{2} [a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = [a + (b - c)] [a - (b - c)] [a^{2} - b^{2} = (a + b) (a - b)] = (a + b - c) (a - b + c)$

Page No 105:

Question 18:

Factorise:
4a²– 4b²+ 4a + 1

Answer:

$4 a^{2} - 4 b^{2} + 4 a + 1 = (4 a^{2} + 4 a + 1) - 4 b^{2} = [{(2 a)}^{2} + 2 \times 2 a \times 1 + 1^{2}] - 4 b^{2} = {(2 a + 1)}^{2} - {(2 b)}^{2} [a^{2} + 2 a b + b^{2} = {(a + b)}^{2}]$

$= [(2 a + 1) + 2 b] [(2 a + 1) - 2 b] [a^{2} - b^{2} = (a + b) (a - b)] = (2 a + 1 + 2 b) (2 a + 1 - 2 b) = (2 a + 2 b + 1) (2 a - 2 b + 1)$

Page No 105:

Question 19:

Factorize:
a² + 2ab + b² − 9c²

Answer:

$a^{2} + 2 a b + b^{2} - 9 c^{2} = {(a + b)}^{2} - {(3 c)}^{2} = (a + b - 3 c) (a + b + 3 c)$

Page No 105:

Question 20:

Factorize:
108a² − 3(b − c)²

Answer:

$108 a^{2} - 3 {(b - c)}^{2} = 3 [36 a^{2} - {(b - c)}^{2}] = 3 [{(6 a)}^{2} - {(b - c)}^{2}] = 3 (6 a - b + c) (6 a + b - c)$

Page No 105:

Question 21:

Factorize:
(a + b)³ − a − b

Answer:

${(a + b)}^{3} - a - b = {(a + b)}^{3} - (a + b) = (a + b) [{(a + b)}^{2} - 1] = (a + b) [{(a + b)}^{2} - 1^{2}] = (a + b) (a + b - 1) (a + b + 1)$

Page No 105:

Question 22:

Factorise:
x²+ y² – z²– 2xy

Answer:

$x^{2} + y^{2} - z^{2} - 2 x y = (x^{2} + y^{2} - 2 x y) - z^{2} = {(x - y)}^{2} - z^{2} [a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = (x - y + z) (x - y - z) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 23:

Factorise:
x²+ 2xy + y²– a² + 2ab – b²

Answer:

$x^{2} + 2 x y + y^{2} - a^{2} + 2 a b - b^{2} = (x^{2} + 2 x y + y^{2}) - (a^{2} - 2 a b + b^{2}) = {(x + y)}^{2} - {(a - b)}^{2} [a^{2} + 2 a b + b^{2} = {(a + b)}^{2} and a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = [(x + y) + (a - b)] [(x + y) - (a - b)] [a^{2} - b^{2} = (a + b) (a - b)] = (x + y + a - b) (x + y - a + b)$

Page No 105:

Question 24:

Factorise:
25x²– 10x + 1 – 36y²

Answer:

$25 x^{2} - 10 x + 1 - 36 y^{2} = [{(5 x)}^{2} - 2 \times 5 x \times 1 + 1^{2}] - {(6 y)}^{2} = {(5 x - 1)}^{2} - {(6 y)}^{2} [a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = [(5 x - 1 + 6 y)] [(5 x - 1 - 6 y)] [a^{2} - b^{2} = (a + b) (a - b)] = (5 x + 6 y - 1) (5 x - 6 y - 1)$

Page No 105:

Question 25:

Factorize:
a − b − a² + b²

Answer:

$a - b - a^{2} + b^{2} = (a - b) - (a^{2} - b^{2}) = (a - b) - (a - b) (a + b) = (a - b) [1 - (a + b)] = (a - b) (1 - a - b)$

Page No 105:

Question 26:

Factorize:
a² − b² − 4ac + 4c²

Answer:

$a^{2} - b^{2} - 4 a c + 4 c^{2} = (a^{2} - 4 a c + 4 c^{2}) - b^{2} = a^{2} - 2 \times 2 a \times c + {(2 c)}^{2} - b^{2} = {(a - 2 c)}^{2} - b^{2} = (a - 2 c + b) (a - 2 c - b)$

Page No 105:

Question 27:

Factorize:
9 − a² + 2ab − b²

Answer:

$9 - a^{2} + 2 a b - b^{2} = 9 - (a^{2} - 2 a b + b^{2}) = 3^{2} - {(a - b)}^{2} = [3 - (a - b)] [3 + (a - b)] = (3 - a + b) (3 + a - b)$

Page No 105:

Question 28:

Factorize:
x³ − 5x² − x + 5

Answer:

$x^{3} - 5 x^{2} - x + 5 = x^{2} (x - 5) - 1 (x - 5) = (x - 5) (x^{2} - 1) = (x - 5) (x^{2} - 1^{2}) = (x - 5) (x - 1) (x + 1)$

Page No 105:

Question 29:

Factorise:
1 + 2ab – (a²+ b²)

Answer:

$1 + 2 a b - (a^{2} + b^{2}) = 1 + 2 a b - a^{2} - b^{2} = 1 - a^{2} + 2 a b - b^{2} = 1^{2} - (a^{2} - 2 a b + b^{2})$

$= 1^{2} - {(a - b)}^{2} [a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = [1 + (a - b)] [1 - (a - b)] [a^{2} - b^{2} = (a + b) (a - b)] = (1 + a - b) (1 - a + b)$

Page No 105:

Question 30:

Factorise:
9a² + 6a + 1 – 36b²

Answer:

$9 a^{2} + 6 a + 1 - 36 b^{2} = [{(3 a)}^{2} + 2 \times 3 a \times 1 + 1^{2}] - {(6 b)}^{2} = {(3 a + 1)}^{2} - {(6 b)}^{2} [a^{2} + 2 a b + b^{2} = {(a + b)}^{2}] = (3 a + 1 - 6 b) (3 a + 1 + 6 b) [a^{2} - b^{2} = (a - b) (a + b)] = (3 a - 6 b + 1) (3 a + 6 b + 1)$

Page No 105:

Question 31:

Factorize:
x² − y² + 6y − 9

Answer:

$x^{2} - y^{2} + 6 y - 9 = x^{2} - (y^{2} - 6 y + 9) = x^{2} - (y^{2} - 2 \times y \times 3 + 3^{2}) = x^{2} - {(y - 3)}^{2} = [x + (y - 3)] [x - (y - 3)] = (x + y - 3) (x - y + 3)$

Page No 105:

Question 32:

Factorize:
4x² − 9y² − 2x − 3y

Answer:

$4 x^{2} - 9 y^{2} - 2 x - 3 y = (4 x^{2} - 9 y^{2}) - (2 x + 3 y) = [{(2 x)}^{2} - {(3 y)}^{2}] - (2 x + 3 y) = (2 x - 3 y) (2 x + 3 y) - 1 (2 x + 3 y) = (2 x + 3 y) (2 x - 3 y - 1)$

Page No 105:

Question 33:

Factorize:
9a² + 3a − 8b − 64b²

Answer:

$9 a^{2} + 3 a - 8 b - 64 b^{2} = 9 a^{2} - 64 b^{2} + 3 a - 8 b = {(3 a)}^{2} - {(8 b)}^{2} + (3 a - 8 b) = (3 a - 8 b) (3 a + 8 b) + 1 (3 a - 8 b) = (3 a - 8 b) (3 a + 8 b + 1)$

Page No 105:

Question 34:

Factorise:
$x^{2} + \frac{1}{x^{2}} - 3$

Answer:

$x^{2} + \frac{1}{x^{2}} - 3 = x^{2} + \frac{1}{x^{2}} - 2 - 1 = [x^{2} + {(\frac{1}{x})}^{2} - 2 \times x \times \frac{1}{x}] - 1 = {(x - \frac{1}{x})}^{2} - 1^{2} [a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = (x - \frac{1}{x} + 1) (x - \frac{1}{x} - 1) [a^{2} - b^{2} = (a - b) (a + b)]$

Page No 105:

Question 35:

Factorise:
$x^{2} - 2 + \frac{1}{x^{2}} y^{2}$

Answer:

$x^{2} - 2 + \frac{1}{x^{2}} - y^{2} = [x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2}] - y^{2} = {(x - \frac{1}{x})}^{2} - y^{2} [a^{2} - 2 a b + b^{2} = {(a - b)}^{2}] = (x - \frac{1}{x} + y) (x - \frac{1}{x} - y) [a^{2} - b^{2} = (a - b) (a + b)]$

Disclaimer: The expression of the question should be $x^{2} - 2 + \frac{1}{x^{2}} - y^{2}$ . The same has been done before solving the question.

Page No 105:

Question 36:

Factorise:
$x^{4} + \frac{4}{x^{4}}$

Answer:

$x^{4} + \frac{4}{x^{4}} = x^{4} + \frac{4}{x^{4}} + 4 - 4 = [{(x^{2})}^{2} + {(\frac{2}{x^{2}})}^{2} + 2 \times (x^{2}) \times (\frac{2}{x^{2}})] - 2^{2} = {(x^{2} + \frac{2}{x^{2}})}^{2} - 2^{2} [a^{2} + 2 a b + b^{2} = {(a + b)}^{2}] = (x^{2} + \frac{2}{x^{2}} + 2) (x^{2} + \frac{2}{x^{2}} - 2) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 37:

Factorise:
x⁸ – 1

Answer:

$x^{8} - 1 = {(x^{4})}^{2} - 1^{2} = (x^{4} + 1) (x^{4} - 1) [a^{2} - b^{2} = (a + b) (a - b)] = (x^{4} + 1) [{(x^{2})}^{2} - 1^{2}]$

$= (x^{4} + 1) (x^{2} + 1) (x^{2} - 1) [a^{2} - b^{2} = (a + b) (a - b)] = (x^{4} + 1) (x^{2} + 1) {(x^{2} - 1)}^{2} = (x^{4} + 1) (x^{2} + 1) (x + 1) (x - 1) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 38:

Factorise:
16x⁴ – 1

Answer:

$16 x^{4} - 1 = {(4 x^{2})}^{2} - 1^{2} = (4 x^{2} + 1) (4 x^{2} - 1) [a^{2} - b^{2} = (a + b) (a - b)] = (4 x^{2} + 1) [{(2 x)}^{2} - 1^{2}] = (4 x^{2} + 1) (2 x + 1) (2 x - 1) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 39:

81x⁴ – y⁴

Answer:

$81 x^{4} - y^{4} = {(9 x^{2})}^{2} - {(y^{2})}^{2} = (9 x^{2} + y^{2}) (9 x^{2} - y^{2}) [a^{2} - b^{2} = (a + b) (a - b)] = (9 x^{2} + y^{2}) [{(3 x)}^{2} - y^{2}] = (9 x^{2} + y^{2}) (3 x + y) (3 x - y) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 105:

Question 40:

x⁴ – 625

Answer:

$x^{4} - 625 = {(x^{2})}^{2} - 25^{2} = (x^{2} + 25) (x^{2} - 25) [a^{2} - b^{2} = (a + b) (a - b)] = (x^{2} + 25) (x^{2} - 5^{2}) = (x^{2} + 25) (x + 5) (x - 5) [a^{2} - b^{2} = (a + b) (a - b)]$

Page No 114:

Question 1:

Factorize:
x² + 11x + 30

Answer:

We have:
$x^{2} + 11 x + 30$
We have to split 11 into two numbers such that their sum of is 11 and their product is 30.
Clearly, $5 + 6 = 11 and 5 \times 6 = 30$ .

$∴ x^{2} + 11 x + 30 = x^{2} + 5 x + 6 x + 30 = x (x + 5) + 6 (x + 5) = (x + 5) (x + 6)$

Page No 114:

Question 2:

Factorize:
x² + 18x + 32

Answer:

We have:
$x^{2} + 18 x + 32$
We have to split 18 into two numbers such that their sum is 18 and their product is 32.
Clearly, $16 + 2 = 18 and 16 \times 2 = 32$ .

$∴ x^{2} + 18 x + 32 = x^{2} + 16 x + 2 x + 32 = x (x + 16) + 2 (x + 16) = (x + 16) (x + 2)$

Page No 114:

Question 3:

Factorise:
x² + 20x – 69

Answer:

$x^{2} + 20 x - 69 = x^{2} + 23 x - 3 x - 69 = x (x + 23) - 3 (x + 23) = (x + 23) (x - 3)$

Page No 114:

Question 4:

x² + 19x – 150

Answer:

$x^{2} + 19 x - 150 = x^{2} + 25 x - 6 x - 150 = x (x + 25) - 6 (x + 25) = (x + 25) (x - 6)$

Page No 114:

Question 5:

Factorise:
x² + 7x – 98

Answer:

$x^{2} + 7 x - 98 = x^{2} + 14 x - 7 x - 98 = x (x + 14) - 7 (x + 14) = (x + 14) (x - 7)$

Page No 114:

Question 6:

Factorise:
$x^{2} + 2 \sqrt{3} x - 24$

Answer:

$x^{2} + 2 \sqrt{3} x - 24 = x^{2} + 4 \sqrt{3} x - 2 \sqrt{3} x - 24 = x (x + 4 \sqrt{3}) - 2 \sqrt{3} (x + 4 \sqrt{3}) = (x + 4 \sqrt{3}) (x - 2 \sqrt{3})$

Page No 114:

Question 7:

Factorise:
x² – 21x + 90

Answer:

$x^{2} - 21 x + 90 = x^{2} - 15 x - 6 x + 90 = x (x - 15) - 6 (x - 15) = (x - 6) (x - 15)$

Page No 114:

Question 8:

Factorise:
x² – 22x + 120

Answer:

$x^{2} - 22 x + 120 = x^{2} - 12 x - 10 x + 120 = x (x - 12) - 10 (x - 12) = (x - 10) (x - 12)$

Page No 114:

Question 9:

Factorise:
x² – 4x + 3

Answer:

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

Page No 114:

Question 10:

Factorise:
$x^{2} + 7 \sqrt{6} x + 60$

Answer:

$x^{2} + 7 \sqrt{6} x + 60 = x^{2} + 5 \sqrt{6} x + 2 \sqrt{6} x + 60 = x (x + 5 \sqrt{6}) + 2 \sqrt{6} (x + 5 \sqrt{6}) = (x + 5 \sqrt{6}) (x + 2 \sqrt{6})$

Page No 114:

Question 11:

Factorise:
$x^{2} + 3 \sqrt{3} x + 6$

Answer:

$x^{2} + 3 \sqrt{3} x + 6 = x^{2} + 2 \sqrt{3} x + \sqrt{3} x + 6 = x (x + 2 \sqrt{3}) + \sqrt{3} (x + 2 \sqrt{3}) = (x + 2 \sqrt{3}) (x + \sqrt{3})$

Page No 114:

Question 12:

Factorise:
$x^{2} + 6 \sqrt{6} x + 48$

Answer:

$x^{2} + 6 \sqrt{6} x + 48 = x^{2} + 4 \sqrt{6} x + 2 \sqrt{6} x + 48 = x (x + 4 \sqrt{6}) + 2 \sqrt{6} (x + 4 \sqrt{6}) = (x + 4 \sqrt{6}) (x + 2 \sqrt{6})$

Page No 114:

Question 13:

Factorise:
$x^{2} + 5 \sqrt{5} x + 30$

Answer:

$x^{2} + 5 \sqrt{5} x + 30 = x^{2} + 3 \sqrt{5} x + 2 \sqrt{5} x + 30 = x (x + 3 \sqrt{5}) + 2 \sqrt{5} (x + 3 \sqrt{5}) = (x + 3 \sqrt{5}) (x + 2 \sqrt{5})$

Page No 114:

Question 14:

Factorise:
$x^{2} - 24 x - 180$

Answer:

$x^{2} - 24 x - 180 = x^{2} - 30 x + 6 x - 180 = x (x - 30) + 6 (x - 30) = (x - 30) (x + 6)$

Page No 114:

Question 15:

Factorise:
x² – 32x – 105

Answer:

$x^{2} - 32 x - 105 = x^{2} - 35 x + 3 x - 105 = x (x - 35) + 3 (x - 35) = (x - 35) (x + 3)$

Page No 114:

Question 16:

Factorise:
x² – 11x – 80

Answer:

$x^{2} - 11 x - 80 = x^{2} - 16 x + 5 x - 80 = x (x - 16) + 5 (x - 16) = (x - 16) (x + 5)$

Page No 114:

Question 17:

Factorise:
6 – x – x²

Answer:

$- x^{2} - x + 6 = - x^{2} - 3 x + 2 x + 6 = - x (x + 3) + 2 (x + 3) = (x + 3) (- x + 2) = (x + 3) (2 - x)$

Page No 114:

Question 18:

Factorise:
$x^{2} - \sqrt{3} x - 6$

Answer:

$x^{2} - \sqrt{3} x - 6 = x^{2} - 2 \sqrt{3} x + \sqrt{3} x - 6 = x (x - 2 \sqrt{3}) + \sqrt{3} (x - 2 \sqrt{3}) = (x - 2 \sqrt{3}) (x + \sqrt{3})$

Page No 114:

Question 19:

Factorise:
40 + 3x – x²

Answer:

$- x^{2} + 3 x + 40 = - x^{2} + 8 x - 5 x + 40 = - x (x - 8) - 5 (x - 8) = (x - 8) (- x - 5) = (8 - x) (x + 5)$

Page No 114:

Question 20:

Factorise:
x² – 26x + 133

Answer:

$x^{2} - 26 x + 133 = x^{2} - 19 x - 7 x + 133 = x (x - 19) - 7 (x - 19) = (x - 19) (x - 7)$

Page No 114:

Question 21:

Factorise:
$x^{2} - 2 \sqrt{3} x - 24$

Answer:

$x^{2} - 2 \sqrt{3} x - 24 = x^{2} - 4 \sqrt{3} x + 2 \sqrt{3} x - 24 = x (x - 4 \sqrt{3}) + 2 \sqrt{3} (x - 4 \sqrt{3}) = (x - 4 \sqrt{3}) (x + 2 \sqrt{3})$

Page No 114:

Question 22:

Factorise:
$x^{2} - 3 \sqrt{5} x - 20$

Answer:

$x^{2} - 3 \sqrt{5} x - 20 = x^{2} - 4 \sqrt{5} x + \sqrt{5} x - 20 = x (x - 4 \sqrt{5}) + \sqrt{5} (x - 4 \sqrt{5}) = (x - 4 \sqrt{5}) (x + \sqrt{5})$

Page No 114:

Question 23:

Factorise:
$x^{2} + \sqrt{2} x - 24$

Answer:

$x^{2} + \sqrt{2} x - 24 = x^{2} + 4 \sqrt{2} x - 3 \sqrt{2} x - 24 = x (x + 4 \sqrt{2}) - 3 \sqrt{2} (x + 4 \sqrt{2}) = (x + 4 \sqrt{2}) (x - 3 \sqrt{2})$

Page No 114:

Question 24:

Factorise:
$x^{2} - 2 \sqrt{2} x - 30$

Answer:

$x^{2} - 2 \sqrt{2} x - 30 = x^{2} - 5 \sqrt{2} x + 3 \sqrt{2} x - 30 = x (x - 5 \sqrt{2}) + 3 \sqrt{2} (x - 5 \sqrt{2}) = (x - 5 \sqrt{2}) (x + 3 \sqrt{2})$

Page No 114:

Question 25:

Factorize:
x² − x − 156

Answer:

We have:
$x^{2} - x - 156$
We have to split ( $-$ 1) into two numbers such that their sum is ( $-$ 1) and their product is ( $-$ 156).
Clearly, $- 13 + 12 = - 1 and - 13 \times 12 = - 156$ .

$∴ x^{2} - x - 156 = x^{2} - 13 x + 12 x - 156 = x (x - 13) + 12 (x - 13) = (x - 13) (x + 12)$

Page No 114:

Question 26:

Factorise:
x² – 32x – 105

Answer:

$x^{2} - 32 x - 105 = x^{2} - 35 x + 3 x - 105 = x (x - 35) + 3 (x - 35) = (x - 35) (x + 3)$

Page No 114:

Question 27:

Factorise:
9x²+ 18x + 8

Answer:

$9 x^{2} + 18 x + 8 = 9 x^{2} + 12 x + 6 x + 8 = 3 x (3 x + 4) + 2 (3 x + 4) = (3 x + 4) (3 x + 2)$

Page No 114:

Question 28:

Factorise:
6x²+ 17x + 12

Answer:

$6 x^{2} + 17 x + 12 = 6 x^{2} + 9 x + 8 x + 12 = 3 x (2 x + 3) + 4 (2 x + 3) = (2 x + 3) (3 x + 4)$

Page No 114:

Question 29:

Factorize:
18x² + 3x − 10

Answer:

We have:
$18 x^{2} + 3 x - 10$
We have to split 3 into two numbers such that their sum is 3 and their product is ( $-$ 180), i.e., $18 \times (- 10)$ .
Clearly, $15 + (- 12) = 3 and 15 \times (- 12) = - 180$ .

$∴ 18 x^{2} + 3 x - 10 = 18 x^{2} + 15 x - 12 x - 10 = 3 x (6 x + 5) - 2 (6 x + 5) = (6 x + 5) (3 x - 2)$

Page No 114:

Question 30:

Factorize:
2x² + 11x − 21

Answer:

We have:
$2 x^{2} + 11 x - 21$
We have to split 11 into two numbers such that their sum is 11 and their product is ( $-$ 42), i.e., $2 \times (- 21)$ .
Clearly, $14 + (- 3) = 11 and 14 \times (- 3) = - 42$ .

$∴ 2 x^{2} + 11 x - 21 = 2 x^{2} + 14 x - 3 x - 21 = 2 x (x + 7) - 3 (x + 7) = (x + 7) (2 x - 3)$

Page No 114:

Question 31:

Factorize:
15x² + 2x − 8

Answer:

We have:
$15 x^{2} + 2 x - 8$
We have to split 2 into two numbers such that their sum is 2 and their product is ( $-$ 120), i.e., $15 \times (- 8)$ .
Clearly, $12 + (- 10) = 2 and 12 \times (- 10) = - 120$ .

$∴ 15 x^{2} + 2 x - 8 = 15 x^{2} + 12 x - 10 x - 8 = 3 x (5 x + 4) - 2 (5 x + 4) = (5 x + 4) (3 x - 2)$

Page No 114:

Question 32:

Factorise:
21x²+ 5x – 6

Answer:

$21 x^{2} + 5 x - 6 = 21 x^{2} + 14 x - 9 x - 6 = 7 x (3 x + 2) - 3 (3 x + 2) = (3 x + 2) (7 x - 3)$

Page No 114:

Question 33:

Factorize:
24x² − 41x + 12

Answer:

We have:
$24 x^{2} - 41 x + 12$
We have to split ( $-$ 41) into two numbers such that their sum is ( $-$ 41) and their product is 288, i.e., $24 \times 12$ .
Clearly, $(- 32) + (- 9) = - 41 and (- 32) \times (- 9) = 288$ .

$∴ 24 x^{2} - 41 x + 12 = 24 x^{2} - 32 x - 9 x + 12 = 8 x (3 x - 4) - 3 (3 x - 4) = (3 x - 4) (8 x - 3)$

Page No 114:

Question 34:

Factorise:
3x²– 14x + 8

Answer:

$3 x^{2} - 14 x + 8 = 3 x^{2} - 12 x - 2 x + 8 = 3 x (x - 4) - 2 (x - 4) = (x - 4) (3 x - 2)$

Hence, factorisation of 3x²– 14x + 8 is $(x - 4) (3 x - 2)$ .

Page No 114:

Question 35:

Factorize:
2x² + 3x − 90

Answer:

We have:
$2 x^{2} + 3 x - 90$
We have to split 3 into two numbers such that their sum is 3 and their product is ( $-$ 180), i.e., $2 \times (- 90)$ .
Clearly, $- 12 + 15 = 3 and - 12 \times 15 = - 180$ .

$∴ 2 x^{2} + 3 x - 90 = 2 x^{2} - 12 x + 15 x - 90 = 2 x (x - 6) + 15 (x - 6) = (x - 6) (2 x + 15)$

Page No 114:

Question 36:

Factorize:
$\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$

Answer:

We have:
$\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$
We have to split 2 into two numbers such that their sum is 2 and product is ( $-$ 15), i.e., $\sqrt{5} \times (- 3 \sqrt{5})$ .
Clearly, $5 + (- 3) = 2 and 5 \times (- 3) = - 15$ .

$∴ \sqrt{5} x^{2} + 2 x - 3 \sqrt{5} = \sqrt{5} x^{2} + 5 x - 3 x - 3 \sqrt{5} = \sqrt{5} x (x + \sqrt{5}) - 3 (x + \sqrt{5}) = (x + \sqrt{5}) (\sqrt{5} x - 3)$

Page No 114:

Question 37:

Factorize:
$2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$

Answer:

We have:
$2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$
We have to split 1 into two numbers such that their sum is 1 and product is 30, i.e., $2 \sqrt{3} \times (- 5 \sqrt{3})$ .
Clearly, $6 + (- 5) = 1 and 6 \times (- 5) = - 30$ .

$∴ 2 \sqrt{3} x^{2} + x - 5 \sqrt{3} = 2 \sqrt{3} x^{2} + 6 x - 5 x - 5 \sqrt{3} = 2 \sqrt{3} x (x + \sqrt{3}) - 5 (x + \sqrt{3}) = (x + \sqrt{3}) (2 \sqrt{3} x - 5)$

Page No 114:

Question 38:

Factorize:
$7 x^{2} + 2 \sqrt{14} x + 2$

Answer:

We have:
$7 x^{2} + 2 \sqrt{14} x + 2$
We have to split $2 \sqrt{14}$ into two numbers such that their sum is $2 \sqrt{14}$ and product is 14.
Clearly, $\sqrt{14} + \sqrt{14} = 2 \sqrt{14} and \sqrt{14} \times \sqrt{14} = 14$ .
$∴ 7 x^{2} + 2 \sqrt{14} x + 2 = 7 x^{2} + \sqrt{14} x + \sqrt{14} x + 2 = \sqrt{7} x (\sqrt{7} x + \sqrt{2}) + \sqrt{2} (\sqrt{7} x + \sqrt{2}) = (\sqrt{7} x + \sqrt{2}) (\sqrt{7} x + \sqrt{2}) = {(\sqrt{7} x + \sqrt{2})}^{2}$

Page No 114:

Question 39:

Factorize:
$6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$

Answer:

We have:
$6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$
Now, we have to split ( $-$ 47) into two numbers such that their sum is ( $-$ 47) and their product is 90.
Clearly, $(- 45) + (- 2) = - 47 and (- 45) \times (- 2) = 90$ .

$∴ 6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3} = 6 \sqrt{3} x^{2} - 2 x - 45 x + 5 \sqrt{3} = 2 x (3 \sqrt{3} x - 1) - 5 \sqrt{3} (3 \sqrt{3} x - 1) = (3 \sqrt{3} x - 1) (2 x - 5 \sqrt{3})$

Page No 114:

Question 40:

Factorize:
$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

Answer:

We have:
$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$
We have to split 20 into two numbers such that their sum is 20 and their product is 75.
Clearly,
$15 + 5 = 20 and 15 \times 5 = 75$

$∴ 5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5} = 5 \sqrt{5} x^{2} + 15 x + 5 x + 3 \sqrt{5} = 5 x (\sqrt{5} x + 3) + \sqrt{5} (\sqrt{5} x + 3) = (\sqrt{5} x + 3) (5 x + \sqrt{5})$

Page No 114:

Question 41:

Factorise:
$\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$

Answer:

$\sqrt{3} x^{2} + 10 x + 8 \sqrt{3} = \sqrt{3} x^{2} + 6 x + 4 x + 8 \sqrt{3} = \sqrt{3} x (x + 2 \sqrt{3}) + 4 (x + 2 \sqrt{3}) = (x + 2 \sqrt{3}) (\sqrt{3} x + 4)$

Hence, factorisation of $\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$ is $(x + 2 \sqrt{3}) (\sqrt{3} x + 4)$ .

Page No 114:

Question 42:

Factorize:
$\sqrt{2} x^{2} + 3 x + \sqrt{2}$

Answer:

We have:
$\sqrt{2} x^{2} + 3 x + \sqrt{2}$
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., $\sqrt{2} \times \sqrt{2}$ .
Clearly, $2 + 1 = 3 and 2 \times 1 = 2$ .

$∴ \sqrt{2} x^{2} + 3 x + \sqrt{2} = \sqrt{2} x^{2} + 2 x + x + \sqrt{2} = \sqrt{2} x (x + \sqrt{2}) + 1 (x + \sqrt{2}) = (x + \sqrt{2}) (\sqrt{2} x + 1)$

Page No 114:

Question 43:

Factorize:
$2 x^{2} + 3 \sqrt{3} x + 3$

Answer:

We have:
$2 x^{2} + 3 \sqrt{3} x + 3$
We have to split $3 \sqrt{3}$ into two numbers such that their sum is $3 \sqrt{3}$ and their product is 6, i.e., $2 \times 3$ .
Clearly, $2 \sqrt{3} + \sqrt{3} = 3 \sqrt{3} and 2 \sqrt{3} \times \sqrt{3} = 6$ .

$∴ 2 x^{2} + 3 \sqrt{3} x + 3 = 2 x^{2} + 2 \sqrt{3} x + \sqrt{3} x + 3 = 2 x (x + \sqrt{3}) + \sqrt{3} (x + \sqrt{3}) = (x + \sqrt{3}) (2 x + \sqrt{3})$

Page No 114:

Question 44:

Factorize:
15x² − x − 128

Answer:

We have:
$15 x^{2} - x - 28$
We have to split ( $-$ 1) into two numbers such that their sum is ( $-$ 1) and their product is ( $-$ 420), i.e., $15 \times (- 28)$ .
Clearly, $(- 21) + 20 = - 1 and (- 21) \times 20 = - 420$ .

$∴ 15 x^{2} - x - 28 = 15 x^{2} - 21 x + 20 x - 28 = 3 x (5 x - 7) + 4 (5 x - 7) = (5 x - 7) (3 x + 4)$

Page No 114:

Question 45:

Factorize:
6x² − 5x − 21

Answer:

We have:
$6 x^{2} - 5 x - 21$
We have to split ( $-$ 5) into two numbers such that their sum is ( $-$ 5) and their product is ( $-$ 126), i.e., $6 \times (- 21)$ .
Clearly, $9 + (- 14) = - 5 and 9 \times (- 14) = - 126$ .

$∴ 6 x^{2} - 5 x - 21 = 6 x^{2} + 9 x - 14 x - 21 = 3 x (2 x + 3) - 7 (2 x + 3) = (2 x + 3) (3 x - 7)$

Page No 114:

Question 46:

Factorize:
2x² − 7x − 15

Answer:

We have:
$2 x^{2} - 7 x - 15$
We have to split ( $-$ 7) into two numbers such that their sum is ( $-$ 7) and their product is ( $-$ 30), i.e., $2 \times (- 15)$ .
Clearly, $(- 10) + 3 = - 7 and (- 10) \times 3 = - 30$ .

$∴ 2 x^{2} - 7 x - 15 = 2 x^{2} - 10 x + 3 x - 15 = 2 x (x - 5) + 3 (x - 5) = (x - 5) (2 x + 3)$

Page No 114:

Question 47:

Factorize:
5x² − 16x − 21

Answer:

We have:
$5 x^{2} - 16 x - 21$
We have to split ( $-$ 16) into two numbers such that their sum is ( $-$ 16) and their product is ( $-$ 105), i.e., $5 \times (- 21)$ .
Clearly, $(- 21) + 5 = - 16 and (- 21) \times 5 = - 105$ .

$∴ 5 x^{2} - 16 x - 21 = 5 x^{2} + 5 x - 21 x - 21 = 5 x (x + 1) - 21 (x + 1) = (x + 1) (5 x - 21)$

Page No 114:

Question 48:

Factorise:
6x² – 11x – 35

Answer:

$6 x^{2} - 11 x - 35 = 6 x^{2} - 21 x + 10 x - 35 = 3 x (2 x - 7) + 5 (2 x - 7) = (2 x - 7) (3 x + 5)$

Hence, factorisation of 6x² – 11x – 35 is $(2 x - 7) (3 x + 5)$ .

Page No 114:

Question 49:

Factorise:
9x² – 3x – 20

Answer:

$9 x^{2} - 3 x - 20 = 9 x^{2} - 15 x + 12 x - 20 = 3 x (3 x - 5) + 4 (3 x - 5) = (3 x - 5) (3 x + 4)$

Hence, factorisation of 9x² – 3x – 20 is $(3 x - 5) (3 x + 4)$ .

Page No 114:

Question 50:

Factorize:
10x² − 9x − 7

Answer:

We have:
$10 x^{2} - 9 x - 7$

We have to split ( $-$ 9) into two numbers such that their sum is ( $-$ 9) and their product is ( $-$ 70), i.e., $10 \times (- 7)$ .
Clearly, $(- 14) + 5 = - 9 and (- 14) \times 5 = - 70$ .

$∴ 10 x^{2} - 9 x - 7 = 10 x^{2} + 5 x - 14 x - 7 = 5 x (2 x + 1) - 7 (2 x + 1) = (2 x + 1) (5 x - 7)$

Page No 114:

Question 51:

Factorize:
$x^{2} - 2 x + \frac{7}{16}$

Answer:

$We have : x^{2} - 2 x + \frac{7}{16} = \frac{16 x^{2} - 32 x + 7}{16} = \frac{1}{16} (16 x^{2} - 32 x + 7)$

Now, we have to split ( $-$ 32) into two numbers such that their sum is ( $-$ 32) and their product is 112, i.e., $16 \times 7$ .
Clearly, $(- 4) + (- 28) = - 32 and (- 4) \times (- 28) = 112$ .

$∴ x^{2} - 2 x + \frac{7}{16} = \frac{1}{16} (16 x^{2} - 32 x + 7) = \frac{1}{16} (16 x^{2} - 4 x - 28 x + 7) = \frac{1}{16} [4 x (4 x - 1) - 7 (4 x - 1)] = \frac{1}{16} (4 x - 1) (4 x - 7)$

Page No 114:

Question 52:

Factorise:
$\frac{1}{3} x^{2} - 2 x - 9$

Answer:

$\frac{1}{3} x^{2} - 2 x - 9 = \frac{x^{2} - 6 x - 27}{3} = \frac{x^{2} - 9 x + 3 x - 27}{3} = \frac{x (x - 9) + 3 (x - 9)}{3} = \frac{(x - 9) (x + 3)}{3} = \frac{(x - 9)}{3} \times \frac{(x + 3)}{1} = (\frac{1}{3} x - 3) (x + 3)$

Hence, factorisation of $\frac{1}{3} x^{2} - 2 x - 9$ is $(\frac{1}{3} x - 3) (x + 3)$ .

Page No 114:

Question 53:

Factorise:
$x^{2} + \frac{12}{35} x + \frac{1}{35}$

Answer:

$x^{2} + \frac{12}{35} x + \frac{1}{35} = \frac{35 x^{2} + 12 x + 1}{35} = \frac{35 x^{2} + 7 x + 5 x + 1}{35} = \frac{7 x (5 x + 1) + 1 (5 x + 1)}{35} = \frac{(5 x + 1) (7 x + 1)}{35} = \frac{(5 x + 1) (7 x + 1)}{5 \times 7} = \frac{(5 x + 1)}{5} \times \frac{(7 x + 1)}{7} = (x + \frac{1}{5}) (x + \frac{1}{7})$

Hence, factorisation of $x^{2} + \frac{12}{35} x + \frac{1}{35}$ is $(x + \frac{1}{5}) (x + \frac{1}{7})$ .

Page No 114:

Question 54:

Factorise:
$21 x^{2} - 2 x + \frac{1}{21}$

Answer:

$21 x^{2} - 2 x + \frac{1}{21} = 21 x^{2} - x - x + \frac{1}{21} = 21 x (x - \frac{1}{21}) - 1 (x - \frac{1}{21}) = (x - \frac{1}{21}) (21 x - 1)$

Hence, factorisation of $21 x^{2} - 2 x + \frac{1}{21}$ is $(x - \frac{1}{21}) (21 x - 1)$ .

Page No 114:

Question 55:

Factorise:
$\frac{3}{2} x^{2} + 16 x + 10$

Answer:

$\frac{3}{2} x^{2} + 16 x + 10 = \frac{3}{2} x^{2} + 15 x + x + 10 = 3 x (\frac{1}{2} x + 5) + 1 (x + 10) = \frac{3}{2} x (x + 10) + 1 (x + 10) = (x + 10) (\frac{3}{2} x + 1)$

Hence, factorisation of $\frac{3}{2} x^{2} + 16 x + 10$ is $(x + 10) (\frac{3}{2} x + 1)$ .

Page No 114:

Question 56:

Factorise:
$\frac{2}{3} x^{2} - \frac{17}{3} x - 28$

Answer:

$\frac{2}{3} x^{2} - \frac{17}{3} x - 28 = \frac{2}{3} x^{2} - 8 x + \frac{7}{3} x - 28 = 2 x (\frac{1}{3} x - 4) + 7 (\frac{1}{3} x - 4) = (\frac{1}{3} x - 4) (2 x + 7)$

Hence, factorisation of $\frac{2}{3} x^{2} - \frac{17}{3} x - 28$ is $(\frac{1}{3} x - 4) (2 x + 7)$ .

Page No 114:

Question 57:

Factorise:
$\frac{3}{5} x^{2} - \frac{19}{5} x + 4$

Answer:

$\frac{3}{5} x^{2} - \frac{19}{5} x + 4 = \frac{3}{5} x^{2} - 3 x - \frac{4}{5} x + 4 = 3 x (\frac{1}{5} x - 1) - 4 (\frac{1}{5} x - 1) = (\frac{1}{5} x - 1) (3 x - 4)$

Hence, factorisation of $\frac{3}{5} x^{2} - \frac{19}{5} x + 4$ is $(\frac{1}{5} x - 1) (3 x - 4)$ .

Page No 114:

Question 58:

Factorise:
$2 x^{2} - x + \frac{1}{8}$

Answer:

$2 x^{2} - x + \frac{1}{8} = 2 x^{2} - \frac{1}{2} x - \frac{1}{2} x + \frac{1}{8} = 2 x (x - \frac{1}{4}) - \frac{1}{2} (x - \frac{1}{4}) = (x - \frac{1}{4}) (2 x - \frac{1}{2})$

Hence, factorisation of $2 x^{2} - x + \frac{1}{8}$ is $(x - \frac{1}{4}) (2 x - \frac{1}{2})$ .

Page No 114:

Question 59:

Factorize:
2(x + y)² − 9(x + y) − 5

Answer:

We have:
$2 {(x + y)}^{2} - 9 (x + y) - 5 Let : (x + y) = u$
Thus, the given expression becomes
$2 u^{2} - 9 u - 5$
Now, we have to split ( $-$ 9) into two numbers such that their sum is ( $-$ 9) and their product is ( $-$ 10).
Clearly, $- 10 + 1 = - 9 and - 10 \times 1 = - 10$ .

$∴ 2 u^{2} - 9 u - 5 = 2 u^{2} - 10 u + u - 5 = 2 u (u - 5) + 1 (u - 5) = (u - 5) (2 u + 1)$
Putting $u = (x + y)$ , we get:
$2 {(x + y)}^{2} - 9 (x + y) - 5 = (x + y - 5) [2 (x + y) + 1] = (x + y - 5) (2 x + 2 y + 1)$

Page No 114:

Question 60:

Factorize:
9(2a − b)² − 4(2a − b) − 13

Answer:

We have:
$9 (2 a - b)^{2} - 4 (2 a - b) - 13 Let : (2 a - b) = p$
Thus, the given expression becomes
$9 p^{2} - 4 p - 13$
Now, we must split ( $-$ 4) into two numbers such that their sum is ( $-$ 4) and their product is ( $-$ 117).
Clearly, $- 13 + 9 = - 4 and - 13 \times 9 = - 117$ .
$∴ 9 p^{2} - 4 p - 13 = 9 p^{2} + 9 p - 13 p - 13 = 9 p (p + 1) - 13 (p + 1) = (p + 1) (9 p - 13)$
Putting $p = (2 a - b)$ , we get:
$9 {(2 a - b)}^{2} - 4 (2 a - b) - 13 = [(2 a - b) + 1] [9 (2 a - b) - 13] = (2 a - b + 1) [18 a - 9 b - 13]$

Page No 115:

Question 61:

Factorise:
$7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$

Answer:

$7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12 = 7 {(x - 2 y)}^{2} - 21 (x - 2 y) - 4 (x - 2 y) + 12 = [7 (x - 2 y)] (x - 2 y - 3) - 4 (x - 2 y - 3) = [7 (x - 2 y) - 4] (x - 2 y - 3) = (7 x - 14 y - 4) (x - 2 y - 3)$

Hence, factorisation of $7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$ is $(7 x - 14 y - 4) (x - 2 y - 3)$ .

Page No 115:

Question 62:

Factorise:
$10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$

Answer:

$10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3 = 10 {(3 x + \frac{1}{x})}^{2} - 6 (3 x + \frac{1}{x}) + 5 (3 x + \frac{1}{x}) - 3 = [2 (3 x + \frac{1}{x})] [5 (3 x + \frac{1}{x}) - 3] + 1 [5 (3 x + \frac{1}{x}) - 3] = [5 (3 x + \frac{1}{x}) - 3] [2 (3 x + \frac{1}{x}) + 1] = (15 x + \frac{5}{x} - 3) (6 x + \frac{2}{x} + 1)$

Hence, factorisation of $10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$ is $(15 x + \frac{5}{x} - 3) (6 x + \frac{2}{x} + 1)$ .

Page No 115:

Question 63:

Factorise:
$6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20$

Answer:

$6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20 = 6 {(2 x - \frac{3}{x})}^{2} + 15 (2 x - \frac{3}{x}) - 8 (2 x - \frac{3}{x}) - 20 = [3 (2 x - \frac{3}{x})] [2 (2 x - \frac{3}{x}) + 5] - 4 [2 (2 x - \frac{3}{x}) + 5] = [2 (2 x - \frac{3}{x}) + 5] [3 (2 x - \frac{3}{x}) - 4] = (4 x - \frac{6}{x} + 5) (6 x - \frac{9}{x} - 4)$

Hence, factorisation of $6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20$ is $(4 x - \frac{6}{x} + 5) (6 x - \frac{9}{x} - 4)$ .

Page No 115:

Question 64:

Factorise:
${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$

Answer:

${(a + 2 b)}^{2} + 101 (a + 2 b) + 100 = {(a + 2 b)}^{2} + 100 (a + 2 b) + 1 (a + 2 b) + 100 = (a + 2 b) [(a + 2 b) + 100] + 1 [(a + 2 b) + 100] = [(a + 2 b) + 1] [(a + 2 b) + 100] = (a + 2 b + 1) (a + 2 b + 100)$

Hence, factorisation of ${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$ is $(a + 2 b + 1) (a + 2 b + 100)$ .

Page No 115:

Question 65:

Factorise:
4x⁴ + 7x² – 2

Answer:

$4 x^{4} + 7 x^{2} - 2 = 4 x^{4} + 8 x^{2} - x^{2} - 2 = 4 x^{2} (x^{2} + 2) - 1 (x^{2} + 2) = (4 x^{2} - 1) (x^{2} + 2)$

Hence, factorisation of 4x⁴ + 7x² – 2 is $(4 x^{2} - 1) (x^{2} + 2)$ .

Page No 115:

Question 66:

Evaluate {(999)2 – 1}.

Answer:

$\{{(999)}^{2} - 1\} = \{{(999)}^{2} - 1^{2}\} = (999 - 1) (999 + 1) = (998) (1000) = 998000$

Hence, {(999)² – 1} = 998000.

Page No 119:

Question 1:

Expand:
(i) (a + 2b + 5c)²
(ii) (2b − b + c)²
(iii) (a − 2b − 3c)²

Answer:

$(i) {(a + 2 b + 5 c)}^{2} = {(a)}^{2} + {(2 b)}^{2} + {(5 c)}^{2} + 2 (a) (2 b) + 2 (2 b) (5 c) + 2 (5 c) (a) = a^{2} + 4 b^{2} + 25 c^{2} + 4 a b + 20 b c + 10 a c$

$(ii) {(2 a - b + c)}^{2} = {[(2 a) + (- b) + (c)]}^{2} = {(2 a)}^{2} + {(- b)}^{2} + {(c)}^{2} + 2 (2 a) (- b) + 2 (- b) (c) + 4 (a) (c) = 4 a^{2} + b^{2} + c^{2} - 4 a b - 2 b c + 4 a c$

$(iii) {(a - 2 b - 3 c)}^{2} = {[a + (- 2 b) + (- 3 c)]}^{2} = {(a)}^{2} + {(- 2 b)}^{2} + {(- 3 c)}^{2} + 2 (a) (- 2 b) + 2 (- 2 b) (- 3 c) + 2 (a) (- 3 c) = a^{2} + 4 b^{2} + 9 c^{2} - 4 a b + 12 b c - 6 a c$

Page No 119:

Question 2:

Expand:
(i) (2a − 5b − 7c)²
(ii) (−3a + 4b − 5c)²
(iii) ${(\frac{1}{2} a - \frac{1}{4} a + 2)}^{2}$

Answer:

$(i) {(2 a - 5 b - 7 c)}^{2} = {[(2 a) + (- 5 b) + (- 7 c)]}^{2} = {(2 a)}^{2} + {(- 5 b)}^{2} + {(- 7 c)}^{2} + 2 (2 a) (- 5 b) + 2 (- 5 b) (- 7 c) + 2 (2 a) (- 7 c) = 4 a^{2} + 25 b^{2} + 49 c^{2} - 20 a b + 70 b c - 28 a c$

$(ii) {(- 3 a + 4 b - 5 c)}^{2} = {[(- 3 a) + (4 b) + (- 5 c)]}^{2} = {(- 3 a)}^{2} + {(4 b)}^{2} + {(- 5 c)}^{2} + 2 (- 3 a) (4 b) + 2 (4 b) (- 5 c) + 2 (- 3 a) (- 5 c) = 9 a^{2} + 16 b^{2} + 25 c^{2} - 24 a b - 40 b c + 30 a c$

$(iii) {(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2} = {[(\frac{a}{2}) + (- \frac{b}{4}) + (2)]}^{2} = {(\frac{a}{2})}^{2} + {(- \frac{b}{4})}^{2} + {(2)}^{2} + 2 (\frac{a}{2}) (\frac{- b}{4}) + 2 (- \frac{b}{4}) (2) + 2 (\frac{a}{2}) (2) = \frac{a^{2}}{4} + \frac{b^{2}}{16} + 4 - \frac{a b}{4} - b + 2 a$

Page No 119:

Question 3:

Factorize: 4x² + 9y² + 16z² + 12xy − 24yz − 16xz.

Answer:

$We have : 4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 x z = {(2 x)}^{2} + {(3 y)}^{2} + {(- 4 z)}^{2} + 2 (2 x) (3 y) + 2 (3 y) (- 4 z) + 2 (- 4 z) (2 x) = {[(2 x) + (3 y) + (- 4 z)]}^{2} = {(2 x + 3 y - 4 z)}^{2}$

Page No 119:

Question 4:

Factorize: 9x² + 16y² + 4z² − 24xy + 16yz − 12xz

Answer:

$We have : 9 x^{2} + 16 y^{2} + 4 z^{2} - 24 x y + 16 y z - 12 x z = {(- 3 x)}^{2} + {(4 y)}^{2} + {(2 z)}^{2} + 2 (- 3 x) (4 y) + 2 (4 y) (2 z) + 2 (2 z) (- 3 x) = {[(- 3 x) + (4 y) + (2 z)]}^{2} = {(- 3 x + 4 y + 2 z)}^{2}$

Page No 119:

Question 5:

Factorize: 25x²+ 4y² + 9z² − 20xy − 12yz + 30xz.

Answer:

$We have : 25 x^{2} + 4 y^{2} + 9 z^{2} - 20 x y - 12 y z + 30 x z = {(5 x)}^{2} + {(- 2 y)}^{2} + {(3 z)}^{2} + 2 (5 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (3 z) (5 x) = {[(5 x) + (- 2 y) + (3 z)]}^{2} = {(5 x - 2 y + 3 z)}^{2}$

Page No 119:

Question 6:

16x² + 4y² + 9z² – 16xy – 12yz + 24xz

Answer:

$16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z = {(4 x)}^{2} + {(- 2 y)}^{2} + {(3 z)}^{2} + 2 (4 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (3 z) (4 x) = {(4 x - 2 y + 3 z)}^{2} [using a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = {(a + b + c)}^{2}]$

Hence, 16x² + 4y² + 9z² – 16xy – 12yz + 24xz = ${(4 x - 2 y + 3 z)}^{2}$ .

Page No 119:

Question 7:

Evaluate
(i) (99)²
(ii) (995)²
(iii) (107)²

Answer:

$(i) {(99)}^{2} = {(100 - 1)}^{2} = {[(100) + (- 1)]}^{2} = {(100)}^{2} + 2 \times (100) \times (- 1) + {(- 1)}^{2} = 10000 - 200 + 1 = 9801$

$(ii) {(995)}^{2} = {(1000 - 5)}^{2} = {[(1000) + (- 5)]}^{2} = {(1000)}^{2} + 2 \times (1000) \times (- 5) + {(- 5)}^{2} = 1000000 - 10000 + 25 = 990025$

$(iii) {(107)}^{2} = {(100 + 7)}^{2} = {(100)}^{2} + 2 \times (100) \times (7) + {(7)}^{2} = 10000 + 1400 + 49 = 11449$

Page No 123:

Question 1:

Expand
(i) (3x + 2)³
(ii) ${(3 a + \frac{1}{4 b})}^{3}$
(iii) ${(1 + \frac{2}{3} a)}^{3}$

Answer:

$(i) {(3 x + 2)}^{3} = {(3 x)}^{3} + 3 \times {(3 x)}^{2} x 2 + 3 \times 3 x \times {(2)}^{2} + {(2)}^{3} = 27 x^{3} + 54 x^{2} + 36 x + 8$
$(ii) {(3 a + \frac{1}{4 b})}^{3} = {(3 a)}^{3} + {(\frac{1}{4 b})}^{3} + 3 {(3 a)}^{2} (\frac{1}{4 b}) + 3 (3 a) {(\frac{1}{4 b})}^{2} = 27 a^{3} + \frac{1}{64 b^{3}} + \frac{27 a^{2}}{4 b} + \frac{9 a}{16 b^{2}}$
$(iii) {(1 + \frac{2}{3} a)}^{3} = {(\frac{2}{3} a)}^{3} + 3 \times {(\frac{2}{3} a)}^{2} \times 1 + 3 a \frac{2}{3} a \times {(1)}^{2} + {(1)}^{3} = \frac{8}{27} a^{3} + \frac{4}{3} a^{2} + 2 a + 1$

Page No 123:

Question 2:

Expand
(i) (5a – 3b)³
(ii) ${(3 x - \frac{5}{x})}^{3}$
(iii) ${(\frac{4}{5} a - 2)}^{3}$

Answer:

$(i) {(5 a - 3 b)}^{3} = {(5 a)}^{3} - {(3 b)}^{3} - 3 {(5 a)}^{2} (3 b) + 3 (5 a) {(3 b)}^{2} = 125 a^{3} - 27 b^{3} - 225 a^{2} b + 135 a b^{2}$

$(ii) {(3 x - \frac{5}{x})}^{3} = {(3 x)}^{3} - {(\frac{5}{x})}^{3} - 3 {(3 x)}^{2} (\frac{5}{x}) + 3 (3 x) {(\frac{5}{x})}^{2} = 27 x^{3} - \frac{125}{x^{3}} - 135 x + \frac{225}{x}$

$(iii) {(\frac{4}{5} a - 2)}^{3} = {(\frac{4}{5} a)}^{3} - {(2)}^{3} - 3 {(\frac{4}{5} a)}^{2} (2) + 3 (\frac{4}{5} a) {(2)}^{2} = \frac{64}{125} a^{3} - 8 - \frac{96}{25} a^{2} + \frac{48}{5} a$

Page No 123:

Question 3:

Factorise
$8 a^{3} + 27 b^{3} + 36 a^{2} b + 54 a b^{2}$

Answer:

$8 a^{3} + 27 b^{3} + 36 a^{2} b + 54 a b^{2} = {(2 a)}^{3} + {(3 b)}^{3} + 3 {(2 a)}^{2} (3 b) + 3 (2 a) {(3 b)}^{2} = {(2 a + 3 b)}^{3}$

Hence, factorisation of $8 a^{3} + 27 b^{3} + 36 a^{2} b + 54 a b^{2}$ is ${(2 a + 3 b)}^{3}$ .

Page No 123:

Question 4:

Factorise
$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

Answer:

$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2} = {(4 a)}^{3} - {(3 b)}^{3} - 3 {(4 a)}^{2} (3 b) + 3 (4 a) {(3 b)}^{2} = {(4 a - 3 b)}^{3}$

Hence, factorisation of $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$ is ${(4 a - 3 b)}^{3}$ .

Page No 123:

Question 5:

Factorise
$1 + \frac{27}{125} a^{3} + \frac{9 a}{5} + \frac{27 a^{2}}{24}$

Answer:

$1 + \frac{27}{125} a^{3} + \frac{9 a}{5} + \frac{27 a^{2}}{25} = {(1)}^{3} + {(\frac{3}{5} a)}^{3} + 3 {(1)}^{2} (\frac{3}{5} a) + 3 (1) {(\frac{3}{5} a)}^{2} = {(1 + \frac{3}{5} a)}^{3}$

Hence, factorisation of $1 + \frac{27}{125} a^{3} + \frac{9 a}{5} + \frac{27 a^{2}}{25}$ is ${(1 + \frac{3}{5} a)}^{3}$ .

Page No 123:

Question 6:

Factorise
$125 x^{3} - 27 y^{3} - 225 x^{2} y + 135 x y^{2}$

Answer:

$125 x^{3} - 27 y^{3} - 225 x^{2} y + 135 x y^{2} = {(5 x)}^{3} - {(3 y)}^{3} - 3 {(5 x)}^{2} (3 y) + 3 (5 x) {(3 y)}^{2} = {(5 x - 3 y)}^{3}$

Hence, factorisation of $125 x^{3} - 27 y^{3} - 225 x^{2} y + 135 x y^{2}$ is ${(5 x - 3 y)}^{3}$ .

Page No 123:

Question 7:

Factorise
$a^{3} x^{3} - 3 a^{2} b x^{2} + 3 a b^{2} x - b^{3}$

Answer:

$a^{3} x^{3} - 3 a^{2} b x^{2} + 3 a b^{2} x - b^{3} = {(a x)}^{3} - {(b)}^{3} - 3 {(a x)}^{2} (b) + 3 (a x) {(b)}^{2} = {(a x - b)}^{3}$

Hence, factorisation of $a^{3} x^{3} - 3 a^{2} b x^{2} + 3 a b^{2} x - b^{3}$ is ${(a x - b)}^{3}$ .

Page No 123:

Question 8:

Factorise
$\frac{64}{125} a^{3} - \frac{96}{25} a^{2} + \frac{48}{5} a - 8$

Answer:

$\frac{64}{125} a^{3} - \frac{96}{25} a^{2} + \frac{48}{5} a - 8 = {(\frac{4}{5} a)}^{3} - {(2)}^{3} - 3 {(\frac{4}{5} a)}^{2} (2) + 3 (\frac{4}{5} a) {(2)}^{2} = {(\frac{4}{5} a - 2)}^{3}$

Hence, factorisation of $\frac{64}{125} a^{3} - \frac{96}{25} a^{2} + \frac{48}{5} a - 8$ is ${(\frac{4}{5} a - 2)}^{3}$ .

Page No 123:

Question 9:

Factorise
a³ – 12a(a – 4) – 64

Answer:

$a^{3} - 12 a (a - 4) - 64 = a^{3} - 12 a^{2} + 48 a - 64 = {(a)}^{3} - {(4)}^{3} - 3 {(a)}^{2} (4) + 3 (a) {(4)}^{2} = {(a - 4)}^{3}$

Hence, factorisation of a³ – 12a(a – 4) – 64 is ${(a - 4)}^{3}$ .

Page No 123:

Question 10:

Evaluate
(i) (103)³
(ii) (99)³

Answer:

$(i) {(103)}^{3} = {(100 + 3)}^{3} = {(100)}^{3} + {(3)}^{3} + 3 {(100)}^{2} (3) + 3 (100) {(3)}^{2} = 1000000 + 27 + 90000 + 2700 = 1092727 (ii) {(99)}^{3} = {(100 - 1)}^{3} = {(100)}^{3} - {(1)}^{3} - 3 {(100)}^{2} (1) + 3 (100) {(1)}^{2} = 1000000 - 1 - 30000 + 300 = 1000300 - 30001 = 970299$

Page No 129:

Question 1:

Factorize:
x³ + 27

Answer:

$x^{3} + 27 = {(x)}^{3} + {(3)}^{3} = (x + 3) (x^{2} - 3 x + 3^{2}) = (x + 3) (x^{2} - 3 x + 9)$

Page No 129:

Question 2:

Factorise
27a³ + 64b³

Answer:

We know that
$x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$
Given: 27a³ + 64b³
x = 3a, y = 4b
$27 a^{3} + 64 b^{3} = (3 a + 4 b) (9 a^{2} + 16 b^{2} - 12 a b)$

Page No 129:

Question 3:

Factorize:
$125 a^{3} + \frac{1}{8}$

Answer:

$125 a^{3} + \frac{1}{8} = {(5 a)}^{3} + {(\frac{1}{2})}^{3} = (5 a + \frac{1}{2}) [{(5 a)}^{2} - 5 a \times \frac{1}{2} + {(\frac{1}{2})}^{2}] = (5 a + \frac{1}{2}) (25 a^{2} - \frac{5 a}{2} + \frac{1}{4})$

Page No 129:

Question 4:

Factorize:
$216 x^{3} + \frac{1}{125}$

Answer:

$216 x^{3} + \frac{1}{125} = {(6 x)}^{3} + {(\frac{1}{5})}^{3} = (6 x + \frac{1}{5}) [{(6 x)}^{2} - 6 x \times \frac{1}{5} + {(\frac{1}{5})}^{2}] = (6 x + \frac{1}{5}) (36 x^{2} - \frac{6 x}{5} + \frac{1}{25})$

Page No 129:

Question 5:

Factorize:
16x⁴ + 54x

Answer:

$16 x^{4} + 54 x = 2 x (8 x^{3} + 27) = 2 x [{(2 x)}^{3} + {(3)}^{3}] = 2 x (2 x + 3) [{(2 x)}^{2} - 2 x \times 3 + 3^{2}] = 2 x (2 x + 3) (4 x^{2} - 6 x + 9)$

Page No 129:

Question 6:

Factorize:
7a³ + 56b³

Answer:

$7 a^{3} + 56 b^{3} = 7 (a^{3} + 8 b^{3}) = 7 [{(a)}^{3} + {(2 b)}^{3}] = 7 (a + 2 b) [a^{2} - a \times 2 b + {(2 b)}^{2}] = 7 (a + 2 b) (a^{2} - 2 a b + 4 b^{2})$

Page No 129:

Question 7:

Factorize:
x⁵ + x²

Answer:

$x^{5} + x^{2} = x^{2} (x^{3} + 1) = x^{2} (x^{3} + 1^{3}) = x^{2} (x + 1) (x^{2} - x \times 1 + 1^{2}) = x^{2} (x + 1) (x^{2} - x + 1)$

Page No 129:

Question 8:

Factorize:
a³ + 0.008

Answer:

$a^{3} + 0.008 = a^{3} + {(0.2)}^{3} = (a + 0.2) [a^{2} - a \times (0.2) + {(0.2)}^{2}] = (a + 0.2) (a^{2} - 0.2 a + 0.04)$

Page No 129:

Question 9:

Factorise
1 – 27a³

Answer:

$1 - 27 a^{3} = 1^{3} - {(3 a)}^{3} = (1 - 3 a) [1^{2} + 1 \times 3 x + {(3 a)}^{2}] = (1 - 3 a) (1 + 3 a + 9 a^{2})$

Page No 129:

Question 10:

Factorize:
64a³ − 343

Answer:

$64 a^{3} - 343 = {(4 a)}^{3} - {(7)}^{3} = (4 a - 7) (16 a^{2} + 4 a \times 7 + 7^{2}) = (4 a - 7) (16 a^{2} + 28 a + 49)$

Page No 129:

Question 11:

Factorize:
x³ − 512

Answer:

$x^{3} - 512 = x^{3} - 8^{3} = (x - 8) (x^{2} + 8 x + 8^{2}) = (x - 8) (x^{2} + 8 x + 64)$

Page No 129:

Question 12:

Factorize:
a³ − 0.064

Answer:

$a^{3} - 0.064 = {(a)}^{3} - {(0.4)}^{3} = (a - 0.4) [a^{2} + a \times (0.4) + {(0.4)}^{2}] = (a - 0.4) (a^{2} + 0.4 a + 0.16)$

Page No 129:

Question 13:

Factorize:
$8 x^{3} - \frac{1}{27 y^{3}}$

Answer:

$8 x^{3} - \frac{1}{27 y^{3}} = {(2 x)}^{3} - {(\frac{1}{3 y})}^{3} = (2 x - \frac{1}{3 y}) [{(2 x)}^{2} + 2 x \times \frac{1}{3 y} + {(\frac{1}{3 y})}^{2}] = (2 x - \frac{1}{3 y}) (4 x^{2} + \frac{2 x}{3 y} + \frac{1}{9 y^{2}})$

Page No 129:

Question 14:

Factorise
$\frac{x^{3}}{216} - 8 y^{3}$

Answer:

We know
$a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$
We have,
$\frac{x^{3}}{216} - 8 y^{3} = {(\frac{x}{6})}^{3} - {(2 y)}^{3}$
So, $a = \frac{x}{6}, b = 2 y$
$\frac{x^{3}}{216} - 8 y^{3} = (\frac{x}{6} - 2 y) ({(\frac{x}{6})}^{2} + \frac{x}{6} \times 2 y + {(2 y)}^{2}) = (\frac{x}{6} - 2 y) (\frac{x^{2}}{36} + \frac{x y}{3} + 4 y^{2})$

Page No 129:

Question 15:

Factorize:
x − 8xy³

Answer:

$x - 8 x y^{3} = x (1 - 8 y^{3}) = x [1^{3} - {(2 y)}^{3}] = x (1 - 2 y) (1^{2} + 1 \times 2 y + {(2 y)}^{2}) = x (1 - 2 y) (1 + 2 y + 4 y^{2})$

Page No 129:

Question 16:

Factorise
32x⁴ – 500x

Answer:

$32 x^{4} - 500 x = 4 x (8 x^{3} - 125) = 4 x ({(2 x)}^{3} - 5^{3}) we know a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b) a = 2 x, b = 5 32 x^{4} - 500 x = 4 x ({(2 x)}^{3} - 5^{3}) = 4 x (2 x - 5) (4 x^{2} + 25 + 10 x)$

Page No 129:

Question 17:

Factorize:
3a⁷b − 81a⁴b⁴

Answer:

$3 a^{7} b - 81 a^{4} b^{4} = 3 a^{4} b (a^{3} - 27 b^{3}) = 3 a^{4} b [a^{3} - {(3 b)}^{3}] = 3 a^{4} b (a - 3 b) [a^{2} + a \times 3 b + {(3 b)}^{2}] = 3 a^{4} b (a - 3 b) (a^{2} + 3 a b + 9 b^{2})$

Page No 129:

Question 18:

Factorise
x⁴ y⁴ – xy

Answer:

Using the identity
$a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$
$x^{4} y^{4} - x y = x y (x^{3} y^{3} - 1) = x y (x y - 1) (x^{2} y^{2} + 1 + x y)$

Page No 129:

Question 19:

Factorise
8x² y³ – x⁵

Answer:

$8 x^{2} y^{3} - x^{5} = x^{2} (8 y^{3} - x^{3}) = x^{2} (2 y - x) (4 y^{2} + x^{2} + 2 x y)$

Page No 129:

Question 20:

Factorise
1029 – 3x³

Answer:

$1029 - 3 x^{3} = 3 (343 - x^{3}) = 3 (7^{3} - x^{3}) = 3 (7 - x) (49 + x^{2} + 7 x)$

Page No 129:

Question 21:

Factorize:
x⁶ − 729

Answer:

$x^{6} - 729 = {(x^{2})}^{3} - {(9)}^{3} = [x^{2} - 9] [{(x^{2})}^{2} + x^{2} \times 9 + 9^{2}] = [x^{2} - 3^{2}] (x^{4} + 9 x^{2} + 81) = (x + 3) (x - 3) (x^{4} + 18 x^{2} + 81 - 9 x^{2}) = (x + 3) (x - 3) [{(x^{2})}^{2} + 2 \times x^{2} \times 9 + 9^{2} - 9 x^{2}] = (x + 3) (x - 3) [{(x^{2} + 9)}^{2} - {(3 x)}^{2}] = (x + 3) (x - 3) (x^{2} + 9 + 3 x) (x^{2} + 9 - 3 x) = (x + 3) (x - 3) (x^{2} + 3 x + 9) (x^{2} - 3 x + 9)$

Page No 129:

Question 22:

Factorise
x⁹ – y⁹

Answer:

$x^{9} - y^{9} = {(x^{3})}^{3} - {(y^{3})}^{3} we know a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b) a = x^{3}, b = y^{3} So, x^{9} - y^{9} = {(x^{3})}^{3} - {(y^{3})}^{3} = (x^{3} - y^{3}) (x^{6} + y^{6} + x^{3} y^{3}) = (x - y) (x^{2} + y^{2} + x y) (x^{6} + y^{6} + x^{3} y^{3})$

Page No 129:

Question 23:

Factorize:
(a + b)³ − (a − b)³

Answer:

${(a + b)}^{3} - {(a - b)}^{3} = [(a + b) - (a - b)] [{(a + b)}^{2} + (a + b) (a - b) + {(a - b)}^{2}] = (a + b - a + b) [a^{2} + 2 a b + b^{2} + a^{2} - b^{2} + a^{2} - 2 a b + b^{2}] = 2 b (3 a^{2} + b^{2})$

Page No 129:

Question 24:

Factorize:
8a³ − b³ − 4ax + 2bx

Answer:

$8 a^{3} - b^{3} - 4 a x + 2 b x = [{(2 a)}^{3} - {(b)}^{3}] - 2 x (2 a - b) = (2 a - b) [{(2 a)}^{2} + 2 a b + b^{2}] - 2 x (2 a - b) = (2 a - b) (4 a^{2} + 2 a b + b^{2}) - 2 x (2 a - b) = (2 a - b) (4 a^{2} + 2 a b + b^{2} - 2 x)$

Page No 129:

Question 25:

Factorize:
a³ + 3a²b + 3ab² + b³ − 8

Answer:

$a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 8 = (a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}) - 8 = [a^{3} + b^{3} + 3 a b (a + b)] - 8 = {(a + b)}^{3} - 2^{3} = (a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 2^{2}] = (a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 4]$

Page No 129:

Question 26:

Factorize:
$a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a}$

Answer:

$a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a} = (a^{3} - \frac{1}{a^{3}}) - 2 (a - \frac{1}{a}) = [{(a)}^{3} - {(\frac{1}{a})}^{3}] - 2 (a - \frac{1}{a}) = (a - \frac{1}{a}) [a^{2} + a \times \frac{1}{a} + {(\frac{1}{a})}^{2}] - 2 (a - \frac{1}{a}) = (a - \frac{1}{a}) (a^{2} + 1 + \frac{1}{a^{2}}) - 2 (a - \frac{1}{a}) = (a - \frac{1}{a}) (a^{2} + 1 + \frac{1}{a^{2}} - 2) = (a - \frac{1}{a}) (a^{2} - 1 + \frac{1}{a^{2}})$

Page No 129:

Question 27:

Factorize:
2a³ + 16b³ − 5a − 10b

Answer:

$2 a^{3} + 16 b^{3} - 5 a - 10 b = 2 [a^{3} + 8 b^{3}] - 5 (a + 2 b) = 2 [a^{3} + {(2 b)}^{3}] - 5 (a + 2 b) = 2 (a + 2 b) [a^{2} - a \times 2 b + {(2 b)}^{2}] - 5 (a + 2 b) = 2 (a + 2 b) (a^{2} - 2 a b + 4 b^{2}) - 5 (a + 2 b) = (a + 2 b) [2 (a^{2} - 2 a b + 4 b^{2}) - 5]$

Page No 129:

Question 28:

Factorise
a⁶ + b⁶

Answer:

$a^{6} + b^{6} = {(a^{2})}^{3} + {(b^{2})}^{3} = (a^{2} + b^{2}) [{(a^{2})}^{2} - a^{2} b^{2} + {(b^{2})}^{2}] = (a^{2} + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$

Page No 129:

Question 29:

Factorise
a¹² – b¹²

Answer:

a¹² – b^{12

$= (a^{6} + b^{6}) (a^{6} - b^{6}) = [{(a^{2})}^{3} + {(b^{2})}^{3}] [{(a^{3})}^{2} - {(b^{3})}^{2}] = [(a^{2} + b^{2}) (a^{4} + b^{4} - a^{2} b^{2})] [(a^{3} - b^{3}) (a^{3} + b^{3})] = [(a^{2} + b^{2}) (a^{4} + b^{4} - a^{2} b^{2})] [(a - b) (a^{2} + b^{2} + a b) (a + b) (a^{2} + b^{2} - a b)] = (a - b) (a^{2} + b^{2} + a b) (a + b) (a^{2} + b^{2} - a b) (a^{2} + b^{2}) (a^{4} + b^{4} - a^{2} b^{2})$}

Page No 129:

Question 30:

Factorise
x⁶ – 7x³– 8

Answer:

Let $x^{3} = y$
So, the equation becomes
$y^{2} - 7 y - 8 = y^{2} - 8 y + y - 8 = y (y - 8) + (y - 8) = (y - 8) (y + 1) = (x^{3} - 8) (x^{3} + 1) = (x - 2) (x^{2} + 4 + 2 x) (x + 1) (x^{2} + 1 - x)$

Page No 129:

Question 31:

Factorise
x³ – 3x²+ 3x + 7

Answer:

x³ – 3x²+ 3x + 7
$= x^{3} - 3 x^{2} + 3 x + 7 = x^{3} - 3 x^{2} + 3 x + 8 - 1 = x^{3} - 3 x^{2} + 3 x - 1 + 8 = (x^{3} - 3 x^{2} + 3 x - 1) + 8 = {(x - 1)}^{3} + 2^{3} = (x - 1 + 2) [{(x - 1)}^{2} + 4 - 2 (x - 1)] = (x + 1) [x^{2} + 1 - 2 x + 4 - 2 x + 2] = (x + 1) (x^{2} - 4 x + 7)$

Page No 129:

Question 32:

Factorise
(x +1)³ + (x– 1)³

Answer:

(x +1)³ + (x– 1)^{3

$= (x + 1 + x - 1) [{(x + 1)}^{2} + {(x - 1)}^{2} - (x - 1) (x + 1)] = (2 x) [{(x + 1)}^{2} + {(x - 1)}^{2} - (x^{2} - 1)] = 2 x (x^{2} + 1 + 2 x + x^{2} + 1 - 2 x - x^{2} + 1) = 2 x (x^{2} + 3)$}

Page No 129:

Question 33:

Factorise
(2a +1)³ + (a– 1)³

Answer:

(2a +1)³ + (a– 1)^{3

$= (2 a + 1 + a - 1) [{(2 a + 1)}^{2} + {(a - 1)}^{2} - (2 a + 1) (a - 1)] = (3 a) [4 a^{2} + 1 + 4 a + a^{2} + 1 - 2 a - 2 a^{2} + 2 a - a + 1] = 3 a [3 a^{2} + 3 a + 3] = 9 a (a^{2} + a + 1)$}

Page No 129:

Question 34:

Factorise
8(x +y)³ – 27(x– y)³

Answer:

8(x +y)³ – 27(x– y)^{3

$= {[2 (x + y)]}^{3} - {[3 (x - y)]}^{3} = (2 x + 2 y - 3 x + 3 y) [4 {(x + y)}^{2} + 9 {(x - y)}^{2} + 6 (x^{2} - y^{2})] = (- x + 5 y) [4 (x^{2} + y^{2} + 2 x y) + 9 (x^{2} + y^{2} - 2 x y) + 6 (x^{2} - y^{2})] = (- x + 5 y) [4 x^{2} + 4 y^{2} + 8 x y + 9 x^{2} + 9 y^{2} - 18 x y + 6 x^{2} - 6 y^{2}] = (- x + 5 y) (19 x^{2} + 7 y^{2} - 10 x y)$}

Page No 129:

Question 35:

Factorise
(x +2)³ + (x– 2)³

Answer:

(x +2)³ + (x– 2)^{3

$= (x + 2 + x - 2) [{(x + 2)}^{2} + {(x - 2)}^{2} - (x^{2} - 4)] = 2 x (x^{2} + 4 + 4 x + x^{2} + 4 - 4 x - x^{2} + 4) = 2 x (x^{2} + 12)$}

Page No 129:

Question 36:

Factorise
(x + 2)³ – (x– 2)³

Answer:

(x + 2)³ – (x– 2)^{3

$= (x + 2 - x + 2) [{(x + 2)}^{2} + {(x - 2)}^{2} + (x^{2} - 4)] = 4 [x^{2} + 4 + 4 x + x^{2} + 4 - 4 x + x^{2} - 4] = 4 (3 x^{2} + 4)$}

Page No 129:

Question 37:

Prove that $\frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15} = 1$ .

Answer:

$LHS : \frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15} = \frac{{(0.85)}^{3} + {(0.15)}^{3}}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}} We know a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) Here a = 0.85, b = 0.15$
$\frac{{(0.85)}^{3} + {(0.15)}^{3}}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}} = \frac{(0.85 + 0.15) ({(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2})}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}} = 0.85 + 0.15 = 1 : RHS$
Thus, LHS = RHS

Page No 129:

Question 38:

Prove that $\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9} = 50$ .

Answer:

$\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9} = \frac{{(59)}^{3} - 9^{3}}{59^{2} + 59 \times 9 + 9^{2}}$
$We know a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) Here a = 59, b = 9 So, \frac{(59 - 9) (59^{2} + 9^{2} + 59 \times 9)}{59^{2} + 9^{2} + 59 \times 9} = 59 - 9 = 50 : RHS$
Thus, LHS=RHS

Page No 136:

Question 1:

Find the product:
(x + y − z) (x² + y² + z² − xy + yz + zx)

Answer:

$(x + y - z) (x^{2} + y^{2} + z^{2} - x y + y z + z x) = [x + y + (- z)] [x^{2} + y^{2} + {(- z)}^{2} - x y - y \times (- z) - [- z] \times x] = x^{3} + y^{3} + {(- z)}^{3} - 3 x \times y \times (- z) = x^{3} + y^{3} - z^{3} + 3 x y z$

Page No 136:

Question 2:

Find the product:
(x – y − z) (x² + y² + z² + xy – yz + xz)

Answer:

(x – y − z) (x² + y² + z² + xy – yz + xz)
$= (x + (- y) + (- z)) (x^{2} + y^{2} + z^{2} + x y - y z + x z) We know (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = a^{3} + b^{3} + c^{3} - 3 a b c Here, a = x, b = - y, c = - z (x + (- y) + (- z)) (x^{2} + y^{2} + z^{2} + x y - y z + x z) = x^{3} - y^{3} - z^{3} - 3 x y z$

Page No 136:

Question 3:

Find the product:
(x − 2y + 3) (x² + 4y² + 2xy − 3x + 6y + 9)

Answer:

$(x − 2y + 3) (x^{2} + 4 y^{2} + 2 x y - 3 x + 6 y + 9) = (x − 2y + 3) (x^{2} + 4 y^{2} + 9 + 2 x y + 6 y - 3 x) = [x + (- 2 y) + 3] [x^{2} + {(- 2 y)}^{2} + {(3)}^{2} - x \times (- 2 y) - (- 2 y) \times 3 - 3 \times x] = {(x)}^{3} + {(- 2 y)}^{3} + {(3)}^{3} - 3 (x) (- 2 y) (3) = x^{3} - 8 y^{3} + 27 + 18 x y$

Page No 136:

Question 4:

Find the product:
(3x – 5y + 4) (9x² + 25y² + 15xy − 20y + 12x + 16)

Answer:

$(3 x - 5 y + 4) (9 x^{2} + 25 y^{2} + 15 x y - 20 y + 12 x + 16) = (3 x + (- 5 y) + 4) (9 x^{2} + 25 y^{2} + 16 + 15 x y - 20 y + 12 x)$
$(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = a^{3} + b^{3} + c^{3} - 3 a b c Here, a = 3 x, b = - 5 y, c = 4$
$(3 x + (- 5 y) + 4) (9 x^{2} + 25 y^{2} + 16 + 15 x y - 20 y + 12 x) = {(3 x)}^{3} + {(- 5 y)}^{3} + 4^{3} - 3 \times 3 x (- 5 y) (4) = 27 x^{3} - 125 y^{3} + 64 + 180 x y$

Page No 136:

Question 5:

Factorize:
125a³ + b³ + 64c³ − 60abc

Answer:

$125 a^{3} + b^{3} + 64 c^{3} - 60 a b c = {(5 a)}^{3} + {(b)}^{3} + {(4 c)}^{3} - 3 \times 5 a \times b \times 4 c = (5 a + b + 4 c) [{(5 a)}^{2} + {(b)}^{2} + {(4 c)}^{2} - 5 a \times b - b \times 4 c - 5 a \times 4 c] = (5 a + b + 4 c) (25 a^{2} + b^{2} + 16 c^{2} - 5 a b - 4 b c - 20 a c)$

Page No 136:

Question 6:

Factorize:
a³ + 8b³ + 64c³ − 24abc

Answer:

$a^{3} + 8 b^{3} + 64 c^{3} - 24 a b c = a^{3} + {(2 b)}^{3} + {(4 c)}^{3} - 3 \times a \times 2 b \times 4 c = (a + 2 b + 4 c) [a^{2} + {(2 b)}^{2} + {(4 c)}^{2} - a \times 2 b - 2 b \times 4 c - 4 c \times a] = (a + 2 b + 4 c) (a^{2} + 4 b^{2} + 16 c^{2} - 2 a b - 8 b c - 4 c a)$

Page No 136:

Question 7:

Factorize:
1 + b³ + 8c³ − 6bc

Answer:

$1 + b^{3} + 8 c^{3} - 6 b c = {(1)}^{3} + {(b)}^{3} + {(2 c)}^{3} - 3 \times 1 \times b \times 2 c = (1 + b + 2 c) [1^{2} + b^{2} + {(2 c)}^{2} - 1 \times b - b \times 2 c - 1 \times 2 c] = (1 + b + 2 c) (1 + b^{2} + 4 c^{2} - b - 2 b c - 2 c)$

Page No 136:

Question 8:

Factorize:
216 + 27b³ + 8c³ − 108abc

Answer:

$216 + 27 b^{3} + 8 c^{3} - 108 a b c = {(6)}^{3} + {(3 b)}^{3} + {(2 c)}^{3} - 3 \times 6 \times 3 b \times 2 c = (6 + 3 b + 2 c) [6^{2} + {(3 b)}^{2} + {(2 c)}^{2} - 6 \times 3 b - 3 b \times 2 c - 2 c \times 6] = (6 + 3 b + 2 c) (36 + 9 b^{2} + 4 c^{2} - 18 b - 6 b c - 12 c)$

Page No 136:

Question 9:

Factorize:
27a³ − b³ + 8c³ + 18abc

Answer:

$27 a^{3} - b^{3} + 8 c^{3} + 18 a b c = {(3 a)}^{3} + {(- b)}^{3} + {(2 c)}^{3} - 3 \times (3 a) \times (- b) \times (2 c) = [3 a + (- b) + 2 c] [{(3 a)}^{2} + {(- b)}^{2} + {(2 c)}^{2} - 3 a (- b) - (- b) 2 c - 3 a \times 2 c] = (3 a - b + 2 c) (9 a^{2} + b^{2} + 4 c^{2} + 3 a b + 2 b c - 6 a c)$

Page No 136:

Question 10:

Factorize:
8a³ + 125b³ − 64c³ + 120abc

Answer:

$8 a^{3} + 125 b^{3} - 64 c^{3} + 120 a b c = {(2 a)}^{3} + {(5 b)}^{3} + {(- 4 c)}^{3} - 3 \times (2 a) \times (5 b) \times (- 4 c) = (2 a + 5 b - 4 c) [{(2 a)}^{2} + {(5 b)}^{2} + {(- 4 c)}^{2} - (2 a) (5 b) - (5 b) (- 4 c) - (2 a) \times (- 4 c)] = (2 a + 5 b - 4 c) (4 a^{2} + 25 b^{2} + 16 c^{2} - 10 a b + 20 b c + 8 a c)$

Page No 136:

Question 11:

Factorize:
8 − 27b³ − 343c³ − 126bc

Answer:

$8 - 27 b^{3} - 343 c^{3} - 126 b c = {(2)}^{3} + {(- 3 b)}^{3} + {(- 7 c)}^{3} - 3 \times (2) \times (- 3 b) \times (- 7 c) = [2 + (- 3 b) + (- 7 c)] [{(2)}^{2} + {(- 3 b)}^{2} + {(- 7 c)}^{2} - (2) (- 3 b) - (- 3 b) (- 7 c) - (2) (- 7 c)] = (2 - 3 b - 7 c) (4 + 9 b^{2} + 49 c^{2} + 6 b - 21 b c + 14 c)$

Page No 136:

Question 12:

Factorize:
125 − 8x³ − 27y³ − 90xy

Answer:

$125 - 8 x^{3} - 27 y^{3} - 90 x y = 5^{3} + {(- 2 x)}^{3} + {(- 3 y)}^{3} - 3 \times 5 \times (- 2 x) \times (- 3 y) = [5 + (- 2 x) + (- 3 y)] [5^{2} + {(- 2 x)}^{2} + {(- 3 y)}^{2} - 5 \times (- 2 x) - (- 2 x) (- 3 y) - 5 \times (- 3 y)] = (5 - 2 x - 3 y) (25 + 4 x^{2} + 9 y^{2} + 10 x - 6 x y + 15 y)$

Page No 137:

Question 13:

Factorize:
$2 \sqrt{2} a^{3} + 16 \sqrt{2} b^{3} + c^{3} - 12 a b c$

Answer:

$2 \sqrt{2} a^{3} + 16 \sqrt{2} b^{3} + c^{3} - 12 a b c = {(\sqrt{2} a)}^{3} + {(2 \sqrt{2} b)}^{3} + c^{3} - 3 \times (\sqrt{2} a) \times (2 \sqrt{2} b) \times (c) = (\sqrt{2} a + 2 \sqrt{2} b + c) [{(\sqrt{2} a)}^{2} + {(2 \sqrt{2} b)}^{2} + c^{2} - (\sqrt{2} a) \times (2 \sqrt{2} b) - (2 \sqrt{2} b) \times (c) - (\sqrt{2} a) \times (c)] = (\sqrt{2} a + 2 \sqrt{2} b + c) (2 a^{2} + 8 b^{2} + c^{2} - 4 a b - 2 \sqrt{2} b c - \sqrt{2} a c)$

Page No 137:

Question 14:

Factorise:
27x³ – y³ – z³ – 9xyz

Answer:

$27 x^{3} - y^{3} - z^{3} - 9 x y z = {(3 x)}^{3} - y^{3} - z^{3} - 3 \times (3 x) \times (- y) \times (- z) W e k n o w, a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) a = 3 x, b = - y, c = - z {(3 x)}^{3} - y^{3} - z^{3} - 3 \times (3 x) \times (- y) \times (- z) = (3 x - y - z) (9 x^{2} + y^{2} + z^{2} + 3 x y - y z + 3 x z)$

Page No 137:

Question 15:

Factorise:
$2 \sqrt{2} a^{3} + 3 \sqrt{3} b^{3} + c^{3} - 3 \sqrt{6} a b c$

Answer:

$2 \sqrt{2} a^{3} + 3 \sqrt{3} b^{3} + c^{3} - 3 \sqrt{6} a b c = {(\sqrt{2} a)}^{3} + {(\sqrt{3} b)}^{3} + c^{3} - 3 (\sqrt{2} a) (\sqrt{3} b) c We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x = \sqrt{2} a, y = \sqrt{3} b, z = c {(\sqrt{2} a)}^{3} + {(\sqrt{3} b)}^{3} + c^{3} - 3 (\sqrt{2} a) (\sqrt{3} b) c = (\sqrt{2} a + \sqrt{3} b + c) (2 a^{2} + 3 b^{2} + c^{2} - \sqrt{6 a b} - \sqrt{3 b c} - \sqrt{2} a c)$

Page No 137:

Question 16:

Factorise:
$3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c$

Answer:

$3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c = {(\sqrt{3} a)}^{3} + {(- b)}^{3} + {(- \sqrt{5} c)}^{3} - 3 (\sqrt{3} a) (- b) (- \sqrt{5} c) W e k n o w x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) H e r e, x = (\sqrt{3} a), y = (- b), z = (- \sqrt{5} c) 3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c = {(\sqrt{3} a)}^{3} + {(- b)}^{3} + {(- \sqrt{5} c)}^{3} - 3 (\sqrt{3} a) (- b) (- \sqrt{5} c) = (\sqrt{3} a - b - \sqrt{5} c) (3 a^{2} + b^{2} + 5 c^{2} + \sqrt{3} a b - \sqrt{5} b c + \sqrt{15} c)$

Page No 137:

Question 17:

Factorize:
(a − b)³ + (b − c)³ + (c − a)³

Answer:

${(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}$
$Putting (a - b) = x, (b - c) = y and (c - a) = z, we get : {(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3} = x^{3} + y^{3} + z^{3} [Where (x + y + z) = (a - b) + (b - c) + (c - a) = 0] = 3 x y z [(x + y + z) = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z] = 3 (a - b) (b - c) (c - a)$

Page No 137:

Question 18:

Factorise:
${(a - 3 b)}^{3} + {(3 b - c)}^{3} + {(c - a)}^{3}$

Answer:

$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z H e r e, x = (a - 3 b), y = (3 b - c), z = (c - a)$
${(a - 3 b)}^{3} + {(3 b - c)}^{3} + {(c - a)}^{3} = (a - 3 b + 3 b - c + c - a) [{(a - 3 b)}^{2} + {(3 b - c)}^{2} + {(c - a)}^{2} - (a - 3 b) (3 b - c) - (3 b - c) (c - a) - (c - a) (a - 3 b)] + 3 (a - 3 b) (3 b - c) (c - a) = 0 + 3 (a - 3 b) (3 b - c) (c - a) = 3 (a - 3 b) (3 b - c) (c - a)$

Page No 137:

Question 19:

Factorize:
(3a − 2b)³ + (2b − 5c)³ + (5c − 3a)³

Answer:

$Put (3 a - 2 b) = x, (2 b - 5 c) = y and (5 c - 3 a) = z . W e have : x + y + z = 3 a - 2 b + 2 b - 5 c + 5 c - 3 a = 0 Now, {(3 a - 2 b)}^{3} + {(2 b - 5 c)}^{3} + {(5 c - 3 a)}^{3} = x^{3} + y^{3} + z^{3} = 3 x y z [Here, x + y + z = 0 . So, x^{3} + y^{3} + z^{3} = 3 x y z] = 3 (3 a - 2 b) (2 b - 5 c) (5 c - 3 a)$

Page No 137:

Question 20:

Factorize:
(5a − 7b)³ + (9c − 5a)³ + (7b − 9c)³

Answer:

$Put (5 a - 7 b) = x, (9 c - 5 a) = z and (7 b - 9 c) = y . Here, x + y + z = 5 a - 7 b + 9 c - 5 a + 7 b - 9 c = 0 Thus, we have : {(5 a - 7 b)}^{3} + {(9 c - 5 a)}^{3} + {(7 b - 9 c)}^{3} = x^{3} + z^{3} + y^{3} = 3 x z y [When x + y + z = 0, x^{3} + y^{3} + z^{3} = 3 x y z .] = 3 (5 a - 7 b) (9 c - 5 a) (7 b - 9 c)$

Page No 137:

Question 21:

Factorize:
a³(b − c)³ + b³(c − a)³ + c³(a − b)³

Answer:

$We have : a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} = {[a (b - c)]}^{3} + {[b (c - a)]}^{3} + {[c (a - b)]}^{3} Put a (b - c) = x b (c - a) = y c (a - b) = z Here, x + y + z = a (b - c) + b (c - a) + c (a - b) = a b - a c + b c - a b + a c - b c = 0 Thus, we have : a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} = x^{3} + y^{3} + z^{3} = 3 x y z [When x + y + z = 0, x^{3} + y^{3} + z^{3} = 3 x y z .] = 3 a (b - c) b (c - a) c (a - b) = 3 a b c (a - b) (b - c) (c - a)$

Page No 137:

Question 22:

Evaluate
(i) (–12)³+ 7³+ 5³
(ii) (28)³ + (–15)³ + (–13)³

Answer:

(i) (–12)³+ 7³+ 5³
$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z Here, x = (- 12), y = 7, z = 5$
${(- 12)}^{3} + 7^{3} + 5^{3} = (- 12 + 7 + 5) [{(- 12)}^{2} + 7^{2} + 5^{2} - 7 (- 12) - 35 + 60] + 3 (- 12) \times 35 = 0 - 1260 = - 1260$

(ii) (28)³ + (–15)³ + (–13)³

$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z Here, x = (- 28), y = - 15, z = - 13$
${(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3} = (28 - 15 - 13) [{(28)}^{2} + {(- 15)}^{2} + {(- 13)}^{2} - 28 (- 15) - (- 15) (- 13) - 28 (- 13)] + 3 \times 28 (- 15) (- 13) = 0 + 16380 = 16380$

Page No 137:

Question 23:

Prove that ${(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)$

Answer:

${(a + b + c)}^{3} = {[(a + b) + c]}^{3} = {(a + b)}^{3} + c^{3} + 3 (a + b) c (a + b + c) \Rightarrow {(a + b + c)}^{3} = a^{3} + b^{3} + 3 a b (a + b) + c^{3} + 3 (a + b) c (a + b + c) \Rightarrow {(a + b + c)}^{3} - a^{3} + b^{3} - c^{3} = 3 a b (a + b) + 3 (a + b) c (a + b + c) \Rightarrow {(a + b + c)}^{3} - a^{3} + b^{3} - c^{3} = 3 (a + b) [a b + c a + c b + c^{2}] \Rightarrow {(a + b + c)}^{3} - a^{3} + b^{3} - c^{3} = 3 (a + b) [a (b + c) + c (b + c)] \Rightarrow {(a + b + c)}^{3} - a^{3} + b^{3} - c^{3} = 3 (a + b) (b + c) (a + c)$

Page No 137:

Question 24:

If a, b, c are all nonzero and a + b + c = 0, prove that $\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3$ .

Answer:

$a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$

Thus, we have:
$(\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b}) = \frac{a^{3} + b^{3} + c^{3}}{a b c}$
$= \frac{3 a b c}{a b c} = 3$

Page No 137:

Question 25:

If a + b + c = 9 and a² + b² + c² = 35, find the value of (a³+ b³ + c³ – 3abc).

Answer:

a + b + c = 9
$\Rightarrow {(a + b + c)}^{2} = 9^{2} = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 81 \Rightarrow 35 + 2 (a b + b c + c a) = 81 \Rightarrow (a b + b c + c a) = 23$
We know,
(a³+ b³ + c³ – 3abc) = $(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
$= (9) (35 - 23) = 108$

Page No 138:

Question 1:

If (x + 1) is factor of the polynomial (2x² + kx) then the value of k is
(a) –2
(b) –3
(c) 2
(d) 3

Answer:

(c) 2

$(x + 1) is a factor of 2 x^{2} + k x . So, - 1 is a zero of 2 x^{2} + k x . Thus, we have : 2 \times {(- 1)}^{2} + k \times (- 1) = 0 \Rightarrow 2 - k = 0 \Rightarrow k = 2$

Page No 138:

Question 2:

The value of (249)² – (248)² is
(a) 1²
(b) 477
(c) 487
(d) 497

Answer:

(249)² – (248)²
We know
^{$a^{2} - b^{2} = (a + b) (a - b) S o, {(249)}^{2} - {(248)}^{2} = (249 - 248) (249 + 248) = 497$}
Hence, the correct answer is option (d).

Page No 138:

Question 3:

If $\frac{x}{y} + \frac{y}{x} = - 1$ , where x ≠ 0 and y ≠ 0, then the value of (x³ − y³) is
(a) 1
(b) −1
(c) 0
(d) $\frac{1}{2}$

Answer:

(c) 0

$\frac{x}{y} + \frac{y}{x} = - 1 \Rightarrow \frac{x^{2} + y^{2}}{x y} = - 1$
$\Rightarrow x$ ² + y² = $-$ xy
⇒ x²+ y² + xy = 0

Thus, we have:
$(x^{3} - y^{3}) = (x - y) (x^{2} + y^{2} + x y)$
$= (x - y) \times 0 = 0$

Page No 138:

Question 4:

If a + b + c = 0, then a³ + b³ + c³ = ?
(a) 0
(b) abc
(c) 2abc
(d) 3abc

Answer:

(d) 3abc

$a + b + c = 0 \Rightarrow a + b = - c$

$\Rightarrow {(a + b)}^{3} = {(- c)}^{3} \Rightarrow a^{3} + b^{3} + 3 a b (a + b) = - c^{3} \Rightarrow a^{3} + b^{3} + 3 a b (- c) = - c^{3} \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$

Page No 139:

Question 5:

If $(3 x + \frac{1}{2}) (3 x - \frac{1}{2}) = 9 x^{2} - p$ then the value of p is
(a) ⁰
(b) $- \frac{1}{4}$
(c) $\frac{1}{4}$
(d) $\frac{1}{2}$

Answer:

$(3 x + \frac{1}{2}) (3 x - \frac{1}{2}) = 9 x^{2} - p$
$9 x^{2} - \frac{1}{4} = 9 x^{2} - p (∵ (a^{2} - b^{2}) = (a + b) (a - b)) \Rightarrow p = \frac{1}{4}$
Hence, the correct answer is option (c).

Page No 139:

Question 6:

The coefficient of x in the expansion of (x + 3)³ is
(a) 1
(b) 9
(c) 18
(d) 27

Answer:

(x + 3)³
$= x^{3} + 3^{3} + 9 x (x + 3) = x^{3} + 27 + 9 x^{2} + 27 x$
So, the coefficient of x in (x + 3)³is 27.
Hence, the correct answer is option (d).

Page No 139:

Question 7:

Which of the following is a factor of (x + y)³ – (x³+ y³)?
(a) x²+ y² + 2xy
(b) x²+ y² – xy
(c) xy²
(d) 3xy

Answer:

(x + y)³ – (x³+ y³)
$= x^{3} + y^{3} + 3 x y (x + y) - (x^{3} + y^{3}) = 3 x y (x + y)$
Thus, the factors of (x + y)³ – (x³+ y³) are 3xy and (x + y).
Hence, the correct answer is option (d).

Page No 139:

Question 8:

One of the factors of $(25 x^{2} - 1) + {(1 + 5 x)}^{2}$ is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x

Answer:

$(25 x^{2} - 1) + {(1 + 5 x)}^{2} = (5 x - 1) (5 x + 1) + {(1 + 5 x)}^{2} = (5 x + 1) [(5 x - 1) + (1 + 5 x)] = (5 x + 1) (10 x)$
So, the factors of $(25 x^{2} - 1) + {(1 + 5 x)}^{2}$ are (5x + 1) and 10x
Hence, the correct answer is option (d).

Page No 139:

Question 9:

If (x + 5) is a factor of p(x) = x³ − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3

Answer:

(b) 5

$(x + 5) is a factor of p (x) = x^{3} - 20 x + 5 k . ∴ p (- 5) = 0 \Rightarrow {(- 5)}^{3} - 20 \times (- 5) + 5 k = 0 \Rightarrow - 125 + 100 + 5 k = 0 \Rightarrow 5 k = 25 \Rightarrow k = 5$

Page No 139:

Question 10:

If (x + 2) and (x − 1) are factors of (x³ + 10x² + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19

Answer:

(b) m = 7, n = −18

Let:
$p (x) = x^{3} + 10 x^{2} + m x + n$
Now,
$x + 2 = 0 \Rightarrow x = - 2$
(x + 2) is a factor of p(x).
So, we have p( $-$ 2)=0
$\Rightarrow {(- 2)}^{3} + 10 \times {(- 2)}^{2} + m \times (- 2) + n = 0 \Rightarrow - 8 + 40 - 2 m + n = 0 \Rightarrow 32 - 2 m + n = 0 \Rightarrow 2 m - n = 32 . . . . . (i)$
Now,
$x - 1 = 0 \Rightarrow x = 1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0
$\Rightarrow 1^{3} + 10 \times 1^{2} + m \times 1 + n = 0 \Rightarrow 1 + 10 + m + n = 0 \Rightarrow 11 + m + n = 0 \Rightarrow m + n = - 11 . . . . . (ii) From (i) and (ii), we get : 3 m = 21 \Rightarrow m = 7$
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

Page No 139:

Question 11:

104 × 96 = ?
(a) 9864
(b) 9984
(c) 9684
(d) 9884

Answer:

(b) 9984

$104 \times 96 = (100 + 4) (100 - 4)$
$= 100^{2} - 4^{2} = (10000 - 16) = 9984$

Page No 139:

Question 12:

305 × 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d) 94840

Answer:

(c) 93940

$305 \times 308 = (300 + 5) \times (300 + 8) = {(300)}^{2} + 300 \times (5 + 8) + 5 \times 8 = 90000 + 3900 + 40 = 93940$

Page No 139:

Question 13:

207 × 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d) 38951

Answer:

(b) 39951

$207 \times 193 = (200 + 7) (200 - 7) = {(200)}^{2} - {(7)}^{2} = 40000 - 49 = 39951$

Page No 139:

Question 14:

4a² + b² + 4ab + 8a + 4b + 4 = ?
(a) (2a + b + 2)²
(b) (2a − b + 2)²
(c) (a + 2b + 2)²
(d) none of these

Answer:

(a) (2a + b + 2)²

$4 a^{2} + b^{2} + 4 a b + 8 a + 4 b + 4 = 4 a^{2} + b^{2} + 4 + 4 a b + 4 b + 8 a = {(2 a)}^{2} + b^{2} + 2^{2} + 2 \times 2 a \times b + 2 \times b \times 2 + 2 \times 2 a \times 2 = {(2 a + b + 2)}^{2}$

Page No 139:

Question 15:

(x2 − 4x − 21) = ?
(a) (x − 7)(x − 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x + 3)
(d) none of these

Answer:

Page No 139:

Question 16:

(4x² + 4x − 3) = ?
(a) (2x − 1) (2x − 3)
(b) (2x + 1) (2x − 3)
(c) (2x + 3) (2x − 1)
(d) none of these

Answer:

(c) (2x + 3) (2x − 1)

$4 x^{2} + 4 x - 3 = 4 x^{2} + 6 x - 2 x - 3$
$= 2 x (2 x + 3) - 1 (2 x + 3) = (2 x + 3) (2 x - 1)$

Page No 139:

Question 17:

6x² + 17x + 5 = ?
(a) (2x + 1)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d) none of these

Answer:

(b) (2x + 5)(3x + 1)

$6 x^{2} + 17 x + 5 = 6 x^{2} + 15 x + 2 x + 5$
$= 3 x (2 x + 5) + 1 (2 x + 5) = (2 x + 5) (3 x + 1)$

Page No 139:

Question 18:

(x + 1) is a factor of the polynomial
(a) x³ − 2x² + x + 2
(b) x³+ 2x² + x − 2
(c) x³ − 2x² − x − 2
(d) x³ − 2x² − x + 2

Answer:

(c) x³ − 2x² − x − 2

Let:
$f (x) = x^{3} - 2 x^{2} + x + 2$
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:
$f (- 1) = {(- 1)}^{3} - 2 \times {(- 1)}^{2} + (- 1) + 2 = - 1 - 2 - 1 + 2 = - 2 \neq 0$
Hence, (x + 1) is not a factor of $f (x) = x^{3} - 2 x^{2} + x + 2$ .

Now,
Let:
$f (x) = x^{3} + 2 x^{2} + x - 2$
By the factor theorem, (x + 1) will be a factor of f (x) if f ( $-$ 1) = 0.
We have:
$f (- 1) = {(- 1)}^{3} + 2 \times {(- 1)}^{2} + (- 1) - 2 = - 1 + 2 - 1 - 2 = - 2 \neq 0$
Hence, (x + 1) is not a factor of $f (x) = x^{3} + 2 x^{2} + x - 2$ .

Now,
Let:
$f (x) = x^{3} + 2 x^{2} - x - 2$
By the factor theorem, (x + 1) will be a factor of f (x) if f ( $-$ 1) = 0.
We have:
$f (- 1) = {(- 1)}^{3} + 2 \times {(- 1)}^{2} - (- 1) - 2 = - 1 + 2 + 1 - 2 = 0$
Hence, (x + 1) is a factor of $f (x) = x^{3} + 2 x^{2} - x - 2$ .

Page No 140:

Question 19:

3x³ + 2x² + 3x + 2 = ?
(a) (3x − 2)(x² − 1)
(b) (3x − 2)(x² + 1)
(c) (3x + 2)(x² − 1)
(d) (3x + 2)(x² + 1)

Answer:

(d) (3x + 2)(x² + 1)

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

Page No 140:

Question 20:

If a + b + c = 0, then $(\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b})$ =?

Answer:

(d) 3

$a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$

Thus, we have:
$(\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b}) = \frac{a^{3} + b^{3} + c^{3}}{a b c}$
$= \frac{3 a b c}{a b c} = 3$

Page No 140:

Question 21:

If x + y + z = 9 and xy + yz + zx = 23, then the value of (x³ + y³ + z³ − 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d) 729

Answer:

(a) 108

$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
$= (x + y + z) [{(x + y + z)}^{2} - 3 (x y + y z + z x)] = 9 \times (81 - 3 \times 23) = 9 \times 12 = 108$

Page No 140:

Question 22:

If $\frac{a}{b} + \frac{b}{a} = - 1$ then (a³ − b³) = ?
(a) −3
(b) −2
(c) −1
(d) 0

Answer:

$\frac{a}{b} + \frac{b}{a} = - 1 \Rightarrow \frac{a^{2} + b^{2}}{a b} = - 1$
$\Rightarrow a$ ² + b² = $-$ ab
$\Rightarrow a$ ² + b² + ab = 0

Thus, we have:
$(a^{3} - b^{3}) = (a - b) (a^{2} + b^{2} + a b)$
$= (a - b) \times 0 = 0$

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