RS Aggarwal Solutions Class 9 Ex 2B

Question 1:

Without actual division, show that (x3 – 3x2 – 13x + 15) is exactly divisible by (x2 + 2x – 3).

Let f(x) = x3 – 3x2 – 13x + 15

Now, x2 + 2x – 3 = x2 + 3x – x – 3

= x (x + 3) – 1 (x + 3)

= (x + 3) (x – 1)

Thus, f(x) will be exactly divisible by x2 + 2x – 3 = (x + 3) (x – 1)

if (x + 3) and (x – 1) are both factors of f(x), so by factor theorem,

we should have f(-3) = 0 and f(1) = 0.

Now, f(-3) = (-3)3 – 3 (-3)2 – 13 (-3) + 15

= -27 – 3 x 9 + 39 + 15

= -27 – 27 + 39 + 15

= -54 + 54 = 0

And, f(1) = 13 – 3 X 12 – 13 x 1 + 15

= 1 – 3 – 13 + 15

= 16 – 16 = 0

f(-3) = 0 and f(1) = 0

So, x2 + 2x – 3 divides f(x) exactly.

 

Question 2:

If (x3 + ax2 + bx + 6) has (x-2) as a factor and leaves a remainder 3 when divided by (x-3), find the values of a and b.

Let f(x) = (x3 + ax2 + bx + 6)

Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).

So, f(3) = 33 + a x 22 + b x 3 + 6 = 3

= 27 + 9a + 3b + 6 = 3

= 9a + 3b + 33 = 3

= 9a + 3b = 3-33

= 9a + 3b = -30

= 3a + b = -10 ….(i)

Given that (x – 2) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0

. f(2) = 23 + a x 22 + b x 2 + 6 = 0

= 8 + 4a+ 2b + 6 = 0

= 4a + 2b = -14

= 2a + b = -7 ….(ii)

Subtracting (ii) from (i), we get,

a = -3

Substituting the value of a = -3 in (i), we get,

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

a = -3 and b = -1.

 

Question 3:

9x2 + 12xy  

9x2 + 12xy = 3x (3x + 4y)

Question 4:

18 x2y – 24xyz

18 x2y – 24xyz = 6xy (3x – 4z)

Question 5:

27 a3 b3 – 45 a4 b2

27 a3 b3 – 45 a4 b2 = 9a3 b2 (3b – 5a)

Question 6:

2a (x + y) – 3b (x + y)  

2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)

Question 7:

2x (p2 + q2) +4y (p2 + q2)

2x (p2 + q2) +4y (p2 + q2)

= (2x + 4y) (p2 + q2)

= 2(x+ 2y) (p2 + q2)

 

Question 8:

x (a – 5) + y (5 – a)

x (a – 5) + y (5 – a)

= x (a – 5) + y (-1) (a – 5)

= (x – y) (a – 5)

Question 9:

4 (a + b) – 6 (a + b)2

4 (a + b) – 6 (a + b)2

= (a + b) [4 – 6 (a + b)]

= 2 (a + b) (2 – 3a – 3b)

= 2 (a + b) (2 – 3a – 3b)

Question 10:

8 (3a – 2b)2 – 10 (3a – 2b)  

8 (3a – 2b)2 – 10 (3a – 2b)

= (3a – 2b) [8(3a – 2b) – 10]

= (3a – 2b) 2[4 (3a – 2b) – 5]

= 2 (3a – 2b) (12 a – 8b – 5)

Question 11:

x (x + y)3 – 3 x2 y (x + y)

x (x + y)3 – 3 x2 y (x + y)

= x (x + y) [(x + y)2 – 3xy]

= x (x + y) (x2 + y2 + 2xy – 3xy)

= x (x + y) (x2 + y2 – xy)

Question 12:

X3 + 2x2 + 5x + 10

X3 + 2x2 + 5x + 10

= x2 (X + 2) + 5 (x + 2)

= (x2 + 5) (x + 2)

Question 13:

X2 + xy – 2xz – 2yz

X2 + xy – 2xz – 2yz

= x (x + y) – 2z (x + y)

= (x+ y) (x – 2z)

 

Question 14:

a3 b – a2 b + 5ab – 5b

a3 b – a2 b + 5ab – 5b

= a2 b (a – 1) + 5b (a – 1)

= (a – 1) (a2 b + 5b)

= (a – 1) b (a2 + 5)

= b (a – 1) (a2 + 5)

Question 15:

8 – 4a – 2a3 + a4

8 – 4a – 2a3 + a4

= 4(2 – a) – a3 (2 – a)

= (2 – a) (4 – a3)

Question 16:

X3 – 2x2 y + 3xy2 – 6y3

X3 – 2x2 y + 3xy2 – 6y3

= x2 (x – 2y) + 3y2 (x – 2y)

= (x – 2y) (x2 + 3y2)

Question 17:

px + pq – 5q – 5x

px + pq – 5q – 5x

= p(x + q) – 5 (q + x)

= (x + q) (p – 5)

 

Question 18:

X2 – xy + y – x

X2 – xy + y – x

= x (x – y) – 1 (x – y)

= (x – y) (x – 1)

Question 19:

(3a – 1)2 – 6a + 2

(3a – 1)2 – 6a + 2

= (3a – 1)2 – 2 (3a – 1)

= (3a – 1) [(3a – 1) – 2]

= (3a – 1) (3a – 3)

= 3(3a – 1) (a – 1)

Question 20:

(2x – 3)2 – 8x + 12

(2x – 3)2 – 8x + 12

= (2x – 3)2 – 4 (2x – 3)

= (2x – 3) (2x – 3 – 4)

= (2x – 3) (2x – 7)

 

Question 21:

a3 + a – 3a2 – 3

a3 + a – 3a2 – 3

= a(a2 + 1) – 3 (a2 + 1)

= (a – 3) (a2 + 1)

Question 22:

3ax – 6ay – 8by + 4bx

3ax – 6ay – 8by + 4bx

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)

Question 23:

abx2 + a2 x + b2 x +ab

abx2 + a2 x + b2 x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

Question 24:

X3 – x2 + ax + x – a – 1

X3 – x2 + ax + x – a – 1

= x3 – x2 + ax – a + x – 1

= x2(x – 1) + a (x – 1) + 1 (x – 1)

= (x – 1) (x2 + a + 1)

 

Question 25:

2x + 4y – 8xy – 1

2x + 4y – 8xy – 1

= 2x – 1 – 8xy + 4y

= (2x – 1) – 4y (2x – 1)

= (2x – 1) (1 – 4y)

Question 26:

ab (x2 + y2) – xy (a2 + b2)

ab (x2 + y2) – xy (a2 + b2)

= abx2 + aby2 – a2 xy – b2 xy

= abx2 – a2 xy + aby2 – b2 xy

= ax (bx – ay) + by(ay – bx)

= (bx – ay) (ax – by)

 

Question 27:

a2 + ab (b + 1) + b3

a2 + ab (b + 1) + b3

= a2 + ab2 + ab + b3

= a2 + ab + ab2 + b3

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

= (a + b) (a + b2)

 

Question 28:

a3+ab(1-2a)-2b2

a3+ab(1-2a)-2b2

= a3 + ab – 2a2 b – 2b2

= a (a2 + b) – 2b (a2 + b)

= (a2 + b) (a – 2b)

Question 29:

2a2 + bc – 2ab – ac

2a2 + bc – 2ab – ac

= 2a2 – 2ab – ac + bc

= 2a (a – b) – c (a – b)

= (a – b) (2a – c)

Question 30:

(ax + by)2 + (bx – ay)2

(ax + by)2 + (bx – ay)2

= a2 x2 + b2 y2 + 2abxy + b2 x2 + a2 y2 – 2abxy

= a2 x2 + b2 y2 + b2 x2 + a2 y2

= a2 x2 + b2 x2+ b2 y2 + a2 y2

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

= (a2 + b2) (x2 +y2)

 

Question 31:

a (a + b – c) – bc

a (a + b – c) – bc

= a2 + ab – ac – bc

= a(a + b) – c (a + b)

= (a – c) (a + b)

 

Question 32:

a(a – 2b – c) + 2bc

a(a – 2b – c) + 2bc

= a2 – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)

Question 33:

a2 x2 + (ax2 + 1)x + a

a2 x2 + (ax2 + 1)x + a

= a2 x2 + ax3 + x + a

= ax2 (a + x) + 1 (x + a)

= (ax2 + 1) (a + x)

Question 34:

ab (x2 + 1) + x (a2 + b2)

ab (x2 + 1) + x (a2 + b2)

= abx2 + ab + a2 x + b2 x

= abx2 + a2 x + ab + b2 x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

 

Question 35:

X2– (a + b) x + ab

X2– (a + b) x + ab

= x2 – ax – bx + ab

= x (x – a) – b(x – a)

= (x – a) (x – b)

Question 36:

x2+1x223x+3x

x2+1x223x+3x
=(x1x)23(x1x)
=(x1x)(x1x3)

 

Question 37:

25x2 – 64y2

25x2 – 64y2

= (5x)2 – (8y)2

= (5x + 8y) (5x – 8y)

[Therefore,  a2-b2 =  (a + b) (a – b)]

Question 38:

100 – 9x2

100 – 9x2

= (10)2 – (3x)2

= (10 + 3x) (10 – 3x)

[Therefore,   a2 – b2 = (a + b) (a – b)]

 

Question 39:

5x2 – 7y2

5x2 – 7y2

=(5x)2(7y)2
=(5x+7y)(5x7y)

[Therefore,   a2 – b2 = (a + b) (a – b)]

Question 40:

(3x + 5y)2 – 4z2

(3x + 5y)2 – 4z2

= (3x + 5y)2 – (2z)2

= (3x + 5y + 2z) (3x + 5y – 2z)

[Therefore,   a2 – b2 = (a + b) (a-b)]

Question 41:

150 – 6x2

150 – 6x2

= 6 (25 – x2)

= 6 ( 52 – x2)

= 6 (5 + x) (5 – x)

[Therefore,    a2 – b2 = (a + b) (a-b)]

 

Question 42:

20x2 – 45

20x2 – 45

= 5(4x2 – 9)

= 5 [(2x)2 – (3)2]

= 5 (2x + 3) (2x – 3)

[Therefore,    a2 – b2 = (a + b) (a-b)]

Question 43:

3x3 – 48x

3x3 – 48x

= 3x (x2 – 16)

= 3x [(x)2 – (4)2]

= 3x (x + 4) (x – 4)

[Therefore,    a2 – b2 = (a + b) (a-b)]

 

Question 44:

2 – 50x2

2 – 50x2

= 2 (1 – 25x2)

= 2 [(1)2 – (5x)2]

= 2 (1 + 5x) (1 – 5x)

[Therefore a2 – b2 = (a + b) (a-b)]

 

Question 45:

27a2– 48b2

27a2– 48b2

= 3 (9a2 – 16b2)

= 3 [(3a)2 – (4b)2]

= 3(3a + 4b) (3a – 4b)

[Therefore,    a2 – b2 = (a + b) (a-b)]


Practise This Question

The following figure shows the elliptical path of a planet about the sun. The two shaded parts have equal areas. If t1, t2 be the time taken by the planet to go from A to B and from C to D respectively–