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## RS Aggarwal Class 9 Ex 2A Chapter 2

**Question 1:**

**(x**^{3}** – 7x**^{2}** + 6x + 4) is divided by (x-6)**

f(x) = x^{3} – 7x^{2} + 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) = 6^{3} – 7 x 6^{2} + 6 x 6 + 4

= 216 – 252 + 36 + 4

= 256 – 252

= 4

The required remainder is 4.

**Question 2:**

**(x**^{3}** – 6x**^{2}** + 13x + 60) is divided by (x+2)**

f(x) = (x^{3} – 6x^{2} + 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**:

**(2x**^{4}** + 6x**^{3}** + 2x**^{2}** + x – 8) is divided by (x+3)**

f(x) = (2x^{4} + 6x^{3} + 2x^{2} + 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**:

**(4x**^{3}** – 12x**^{2}** + 11x – 5) is divided by (2x-1)**

f(x) = (4x^{3} – 12x^{2} + 11x – 5)

Now, 2x – 1 = 0

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

Now,

= -2

The required remainder is -2.

**Question 5**:

**(81x**^{4}** + 54x**^{3}** – 9x**^{2}** – 3x + 2) is divided by (3x+2)**

f(x) = (81x^{4} + 54x^{3} – 9x^{2} – 3x + 2)

Now, 3x + 2 = 0

x=

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

Now,

=16-16-4+4

=0

The required remainder is 0.

**Question 6**:

**(x**^{3}** – ax**^{2}** + 2x – a) is divided by (x-a)**

f(x) = (x^{3} – ax^{2} + 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) = a^{3} – a a^{2} + 2 a – a

= a^{3} – a^{3} + 2a – a

=a

The required remainder is a.

**Question 7**:

**The polynomials (ax**^{3}** + 3x**^{2}** – 3) and (2x**^{3}** – 5x + a) when divided by (x-4) leave the same remainder. Find the value of a.**

Let f(x) = ax^{3} + 3x^{2} – 3

and g(x) = 2x^{3} – 5x + a

f(4) = a x 4^{3} + 3 x 4^{2} – 3

= 64a + 48 – 3

= 64a + 45

g(4) = 2 x 4^{3} – 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

The value of a is 1.

**Question 8**:

**The polynomial f(x) = (x**^{4}** – 2x**^{3}** + 3x**^{2}** – 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) = (x^{4} – 2x^{3} + 3x^{2} – ax + b)

From the given information,

f(1) = 1^{4}– 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 =

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) = x^{4} – 2x^{3} + 3x^{2} – ax + b

= x^{4} – 2x^{3} + 3x^{2} – 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 (x**^{3}** – 8)**

f(x) = (x^{3} – 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 (x^{3} – 8).

**Question 10**:

**(x-3) is a factor of (2x**^{3}** + 7x**^{2}** – 24x – 45)**

f(x) = (2x^{3} + 7x^{2} – 24x – 45)

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

Here, f(3) = 2 x 3^{3} + 7 x 3^{2} – 24 x 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117

= 0

Therefore, (x – 3) is a factor of (2x^{3} + 7x^{2} – 24x – 45).

**Question 11**:

**(x-1) is a factor of (2x**^{4}** + 9x**^{4}** + 6x**^{2}** – 11x – 6)**

f(x) = (2x^{4} + 9x^{4} + 6x^{2} – 11x – 6)

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

Here, f(1) = 2 x 1^{4} + 9 x 1^{3} +6 x 1^{2} – 11 x 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17

= 0

Therefore, (x – 1) is factor of (2x^{4} + 9x^{3} + 6x^{2} – 11x – 6).

**Question 12**:

**(x+2) is a factor of (x**^{4}** – x**^{2}** – 12)**

f(x) = (x^{4} – x^{2} – 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 (x^{4} – x^{2} – 12).

**Question 13**:

**(x+5) is a factor of 2x**^{3}** + 9x**^{2}** – 11x – 30**

f(x) = 2x^{3} + 9x^{2} – 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 (2x^{3} + 9x^{2} – 11x – 30).

**Question 14**:

**(2x-3) is a factor of (2x**^{4}** + x**^{3}** – 8x**^{2}** – x + 6)**

f(x) = (2x^{4} + x^{3} – 8x^{2} – x + 6)

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

Here, 2x – 3 = 0

X =

= 0

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

**Question 15**:

**(x- 2–√) is a factor of (7x**

^{2}**– 4**2–√× – 6 = 0)

f(x) = (7x^{2} – 4

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

Here,

= 14 – 8 – 6

= 14 – 14

= 0

Therefore, (x –

**Question 16: **

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

f(x) = (

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

Here,

(x +

**Question 17**:

**Find the value of k for which (x-1) is a factor of (2x**^{3}** + 9x**^{2}**+ x + k)**

f(x) = (2x^{3} + 9x^{2}+ x + k)

x – 1 = 0

= x = 1

f(1) = 2 x 1^{3}+ 9 x 1^{2} + 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 (2x**^{3}** – 3x**^{2}** – 18x + a) is divisible by (x-4).**

f(x) = (2x^{3} – 3x^{2} – 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 (x**^{4}** – x**^{3}** – 11x**^{2}** – x + a) is divisible by (x+3).**

f(x) = x^{4} – x^{3} – 11x^{2} – 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 (2x**^{3}** + ax**^{2}** + 11x + a + 3) exactly divisible by (2x-1)?**

f(x) = (2x^{3} + ax^{2} + 11x + a + 3)

2x – 1 = 0

X=

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

Therefore, we have

Therefore, The value of a = -7.

**Question 21**:

**Find the values of a and b so that the polynomial (x**^{3}** – 10x**^{2}** + ax + b) is exactly divisible by (x-1) as well as (x-2).**

Let f(x) = (x^{3} – 10x^{2} + 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)= 1^{3} – 10x 1^{2} +a x 1 +b= 0

1 – 10 + a + b = 0

a + b = 9 ….(i)

And f(2) = 2^{3} – 10 x 2^{2} + 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 (x**^{4}** + ax**^{3}** – 7x**^{2}** – 8x + b) is exactly divisible by (x+2) as well as (x+3).**

Let f(x) = (x^{4} + ax^{3} – 7x^{2} – 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.

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