nrop: RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3A

EXERCISE 3A

PAGE NO 99

Question 1:

Factorize:
9x² + 12xy

Answer 1:

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

Question 2:

Factorize:
18x²y − 24xyz

Answer 2:

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

Question 3:

Factorize:
27a³b³ − 45a⁴b²

Answer 3:

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

Question 4:

Factorize:
2a(x + y) − 3b(x + y)

Answer 4:

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

Question 5:

Factorize:
2x(p² + q²) + 4y(p² + q²)

Answer 5:

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

Question 6:

Factorize:
x(a − 5) + y(5 − a)

Answer 6:

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

Question 7:

Factorize:
4(a + b) − 6(a + b)²

Answer 7:

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

Question 8:

Factorize:
8(3a − 2b)² − 10(3a − 2b)

Answer 8:

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

Question 9:

Factorize:
x(x + y)³ − 3x²y(x + y)

Answer 9:

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

Question 10:

Factorize:
x³ + 2x² + 5x + 10

Answer 10:

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

Question 11:

Factorize:
x² + xy − 2xz − 2yz

Answer 11:

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

Question 12:

Factorize:
a³b − a²b + 5ab − 5b

Answer 12:

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

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

Question 13:

Factorize:
8 − 4a − 2a³ + a⁴

Answer 13:

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

Question 14:

Factorize:
x³ − 2x²y + 3xy² − 6y³

Answer 14:

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

Question 15:

Factorize:
px − 5q + pq − 5x

Answer 15:

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

Question 16:

Factorize:
x² + y − xy − x

Answer 16:

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

Question 17:

Factorize:
(3a − 1)² − 6a + 2

Answer 17:

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

Question 18:

Factorize:
(2x − 3)² − 8x + 12

Answer 18:

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

Question 19:

Factorize:
a³ + a − 3a² − 3

Answer 19:

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

Question 20:

Factorize:
3ax − 6ay − 8by + 4bx

Answer 20:

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

Question 21:

Factorize:
abx² + a²x + b²x + ab

Answer 21:

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

Question 22:

Factorize:
x³ − x² + ax + x − a − 1

Answer 22:

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 23:

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

Question 24:

Factorize:
ab(x² + y²) − xy(a² + b²)

Answer 24:

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

Question 25:

Factorize:
a² + ab(b + 1) + b³

Answer 25:

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

Question 26:

Factorize:
a³ + ab(1 − 2a) − 2b²

Answer 26:

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

Question 27:

Factorize:
2a² + bc − 2ab − ac2

Answer 27:

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

Question 28:

Factorize:
(ax + by)² + (bx − ay)²

Answer 28:

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

Question 29:

Factorize:
a(a + b − c) − bc

Answer 29:

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

Question 30:

Factorize:
a(a − 2b − c) + 2bc

Answer 30:

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

Question 31:

Factorize:
a²x² + (ax² + 1)x + a

Answer 31:

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

Question 32:

Factorize:
ab(x² + 1) + x(a² + b²)

Answer 32:

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

Question 33:

Factorize:
x² − (a + b)x + ab

Answer 33:

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

Question 34:

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

Answer 34:

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

RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3A