RS Aggarwal Class 9 Solutions Chapter 2 - Polynomials Ex 2A (2.1)

RS Aggarwal Class 9 Chapter 2 - Polynomials Ex 2A (2.1) Solutions Free PDF

The RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2A help students as reference material while solving the exercise question of the RS Aggarwal textbook. These solutions prove to be an extremely important study resource for students looking to improve their performance in the exam. For clearing your doubts the RS Aggarwal solutions assist students in developing a deeper understanding of the subject. With these solutions, students are better placed in solving difficult questions easily and enhance their problem-solving skills.

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RS Aggarwal Class 9 Ex 2A Chapter 2

Question 1:

(x3 – 7x2 + 6x + 4) is divided by (x-6)

f(x) = x3 – 7x2 + 6x + 4

Now, x – 6 = 0

x = 6

By the remainder theorem, we know that when f(x) is divide by (x – 6) the remainder is f(6)

Now, f(6) = 63 – 7 x 62 + 6 x 6 + 4

= 216 – 252 + 36 + 4

= 256 – 252

= 4

The required remainder is 4.

Question 2:

(x3 – 6x2 + 13x + 60) is divided by (x+2)

f(x) = (x3 – 6x2 + 13x + 60)

Now, x + 2 = 0

x = -2

By the remainder the theorem, we know that when f(x) is divide by (x + 2) the remainder is f(-2).

Now, f(-2) = (-2)3 – 6(-2)2 + 13(-2) + 60

= -8 – 24 – 26 + 60

= -58 + 60

= 2

The required remainder is 2.

 

Question 3:

(2x4 + 6x3 + 2x2 + x – 8) is divided by (x+3)

f(x) = (2x4 + 6x3 + 2x2 + x – 8)

Now, x + 3 = 0

x = -3

By the remainder the theorem, we know that when f(x) is divide by (x + 3) the remainder is f(-3).

f(-3) = 2(-3)4 + 6(-3)3 + 2(-3)2 – 3 – 8

= 162 – 162 + 18 – 3 – 8

= 18 – 11

= 7 The required remainder is 7.

 

Question 4:

(4x3 – 12x2 + 11x – 5) is divided by (2x-1)

f(x) = (4x3 – 12x2 + 11x – 5)

Now, 2x – 1 = 0

x=1÷2

By the remainder theorem, we know that when f(x) is divided by (2x – 1) the remainder is  f(12)

Now,

f(12)= 4(12)312(12)2+11(12)5

=4×1812×14+1125
=123+1125
=16+11102
=16+122
=42

=      -2

The required remainder is -2.

 

Question 5:

(81x4 + 54x3 – 9x2 – 3x + 2) is divided by (3x+2)

f(x) = (81x4 + 54x3 – 9x2 – 3x + 2)

Now, 3x + 2 = 0

x= 23

By the remainder theorem, we know that when f(x) is divided by (3x+ 2) the remainder is f(23)

Now,

f(23)=81(23)4+54(23)39(23)23(23)+2

=81×(1681)+54(827)9(49)+2+2

=16-16-4+4

=0

The required remainder is 0.

 

Question 6:

(x3 – ax2 + 2x – a) is divided by (x-a)

f(x) = (x3 – ax2 + 2x – a)

Now,x-a=0

x =a

By the remainder theorem, we know that when f(x) is divided by (x – a) the remainder is f(a)

Now, f(a) = a3 – a a2 + 2 a – a

= a3 – a3 + 2a – a

=a

The required remainder is a.

 

Question 7:

The polynomials (ax3 + 3x2 – 3) and (2x3 – 5x + a) when divided by (x-4) leave the same remainder. Find the value of a.

Let f(x) = ax3 + 3x2 – 3

and g(x) = 2x3 – 5x + a

f(4) = a x 43 + 3 x 42 – 3

= 64a + 48 – 3

= 64a + 45

g(4) = 2 x 43 – 5 x 4 + a

= 128 – 20 + a

= 108 + a

It is given that:

f(4) = g(4)

64a + 45 = 108 + a

64a – a = 108 – 45

63a = 63

a=6363 =  1

The value of a is 1.

 

Question 8:

The polynomial f(x) = (x4 – 2x3 + 3x2 – ax + b) when divided by (x-1) and (x+1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x-2).

Let f(x) = (x4 – 2x3 + 3x2 – ax + b)

From the given information,

f(1) = 14– 2(1)3 + 3(1)2 – a (1 ) + b = 5

1 – 2 + 3 – a + b = 5

2 – a + b = 5 ….(i)

And,

f(-1) = (-1)4 – 2(4)3 + 3(4)2 – a(-1) + b = 19

1 + 2 + 3 +a+b= 19

6 + a + b = 19 ….(ii)

Adding (i) and (ii), we get

8 + 2b = 24

2b = 24 – 8 = 16

b = 162

Substituting the value of b = 8 in (i), we get

2 – a + 8 = 5

-a + 10 = 5

-a = -10 + 5

a = 5

a = 5 and b = 8

f(x) = x4 – 2x3 + 3x2 – ax + b

= x4 – 2x3 + 3x2 – 5x + 8

f(2) = (2)4 – 2(2)3 + 3(2)2 – 5(2) + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10

= 10

The required remainder is 10.

Question 9:

(x-2) is a factor of (x3 – 8)

f(x) = (x3 – 8)

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

Here, f(2) = (2)3 – 8

= 8 – 8

= 0

Therefore,  (x – 2) is a factor of (x3 – 8).

 

Question 10:

(x-3) is a factor of  (2x3 + 7x2 – 24x – 45)

f(x) = (2x3 + 7x2 – 24x – 45)

By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2 x 33 + 7 x 32 – 24 x 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117

= 0

Therefore,  (x – 3) is a factor of (2x3 + 7x2 – 24x – 45).

 

Question 11:

(x-1) is a factor of  (2x4 + 9x4 + 6x2 – 11x – 6)

f(x) = (2x4 + 9x4 + 6x2 – 11x – 6)

By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2 x 14 + 9 x 13 +6 x 12 – 11 x 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17

= 0

Therefore,  (x – 1) is factor of (2x4 + 9x3 + 6x2 – 11x – 6).

Question 12:

(x+2) is a factor of  (x4 – x2 – 12)

f(x) = (x4 – x2 – 12)

By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.

Here, f(-2) = (-2)4 – (-2)2 – 12

= 16 – 4 – 12

= 16 – 16

= 0

Therefore,  (x + 2) is a factor of (x4 – x2 – 12).

 

Question 13:

(x+5) is a factor of  2x3 + 9x2 – 11x – 30

f(x) = 2x3 + 9x2 – 11x – 30

By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.

Here, f(-5) = 2(-5)3 + 9(-5)2 – 11(-5) – 30

= -250 + 225 + 55 – 30

= -280 + 280

= 0

Therefore,  (x + 5) is a factor of (2x3 + 9x2 – 11x – 30).

Question 14:

(2x-3) is a factor of  (2x4 + x3 – 8x2 – x + 6)

f(x) = (2x4 + x3 – 8x2 – x + 6)

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

Here, 2x – 3 = 0

X = 32

f(32)=2(32)4+(32)38(32)2(32)+6
2×8116+2788×9432+6
=818+2781832+6
=81+2714412+488
=1561568

= 0

Therefore,  (2x – 3) is a factor of (2×4 + x3 – 8×2 – x + 6).

 

Question 15:

(x-2) is a factor of (7x2 – 42× – 6 = 0)

f(x) = (7x2 – 42× – 6 = 0)

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

Here,     f(2)=7(2)242×26

= 14 – 8 – 6

= 14 – 14

= 0

Therefore, (x –2 ) is a factor of (7 – 42 x – 6 = 0).

 

Question 16:

(x+2) is a factor of  (42x2+ 5x +2 = 0)

f(x) = (42x2+ 5x +2 = 0)

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

f(2)=22(2)2+5(2)+2
=22×252+2
=4252+2
=5252=0

Here,

(x + =2) is a factor of (4=2 + 5x +2= 0).

 

Question 17:

Find the value of k for which (x-1) is a factor of (2x3 + 9x2+ x + k)

f(x) = (2x3 + 9x2+ x + k)

x – 1 = 0

= x = 1

f(1) = 2 x 13+ 9 x 12 + 1 + k

= 2 + 9 + 1 + k

= 12 + k

Given that (x – 1) 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(1) = 0.

f(1) = 12 + k = 0

k = -12.

 

Question 18:

Find the value of a for which (2x3 – 3x2 – 18x + a) is divisible by (x-4).

f(x) = (2x3 – 3x2 – 18x + a)

x – 4 = 0

x = 4

f(4) = 2(4)3 – 3(4)2 – 18 x 4 + a

= 128 – 48 – 72 + a

= 128 – 120 + a

= 8 + a

Given that (x – 4) 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(4) = 0.

f(4) = 8 + a = 0

a = -8

Question 19:

Find the value of a for which (x4 – x3 – 11x2 – x + a) is divisible by (x+3).

f(x) = x4 – x3 – 11x2 – x + a

x + 3 = 0

x = -3

f(-3) = (-3)4 – (-3)3 -11 (-3)2 – (-3) + a

= 81 + 27 – 11 x 9 + 3 + a

= 81 + 27 – 99 + 3 + a

= 111 – 99 + a

= 12 + a

 

Given that f(x) is divisible by (x + 3), that is (x+3) 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(-3) = 0

. f(-3) = 12 + a =0

a = -12.

Question 20:

For what value of a is the polynomial (2x3 + ax2 + 11x + a + 3) exactly divisible by (2x-1)?

f(x) = (2x3 + ax2 + 11x + a + 3)

2x – 1 = 0

X= 12

Given that f(x) is exactly divisible by (2x – 1), that is (2x – 1) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0  therefore  f(12)0

Therefore, we have

f(12)=2(12)3+a(12)2+11×12+a+3=0
=2(18)+a(14)+112+a+3=0
14+14a+112+a+3=0
1+a+22+4a+124=0
5a+354=0

5a+35= 0

5a = -35

  a = 355  = -7

Therefore, The value of a = -7.

Question 21:

Find the values of a and b so that the polynomial (x3 – 10x2 + ax + b) is exactly divisible by (x-1) as well as (x-2).

Let f(x) = (x3 – 10x2 + ax + b), then by factor theorem

(x – 1) and (x – 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.

f(1)= 13 – 10x 12 +a x 1 +b= 0

1 – 10 + a + b = 0

a + b = 9 ….(i)

And f(2) = 23 – 10 x 22 + a x 2 + b = 0

8 – 40 + 2a + b = 0

2a + b = 32 ….(ii)

Subtracting (i) from (ii), we get

a = 23

Substituting the value of a = 23 in (i), we get

23 + b = 9

b = 9 – 23

b = -14

a= 23 and b = -14.

 

Question 22:

Find the values of a and b so that the polynomial (x4 + ax3 – 7x2 – 8x + b) is exactly divisible by (x+2) as well as (x+3).

Let f(x) = (x4 + ax3 – 7x2 – 8x + b)

Now, x + 2 = 0 x = -2 and x + 3 = 0 x = -3

By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if

f(-2) = 0 and f(-3) = 0

f(-2) = (-2)4 + a (-2)3 – 7 (-2)2 – 8 (-2) + b = 0

16 – 8a – 28 + 16 + b = 0

-8a + b = -4

8a – b = 4 …(i)

And, f(-3) = (-3)4 + a (-3)3– 7 (-3)2 – 8 (-3) + b = 0

81 – 27a – 63 + 24 + b = 0

-27a + b = -42

27a – b = 42 …. (ii)

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

19a = 38

So, a = 2 S

substituting the value of a = 2 in (i), we get

8(2) – b = 4

16 – b = 4

-b = -16 + 4

-b = -12

b = 12

a = 2 and b = 12.

Key Features of RS Aggarwal Class 9 Solutions Chapter 2– Polynomials Ex 2A (2.1)

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Motion of a motorcyclist is shown in the figure. Which of the following shows the distance-time graph for this motion?

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