# SELINA Solutions for Class 10 Maths Chapter 8 - Remainder And Factor Theorems

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## Chapter 8 - Remainder And Factor Theorems Exercise Ex. 8(A)

Solution 1

By remainder theorem we know that when a polynomial f (x) is divided by x - a, then the remainder is f(a).

Solution 2

(x - a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x - a), is 0, i.e., if f(a) = 0.

Solution 3

By remainder theorem we know that when a polynomial f (x) is divided by x - a, then the remainder is f(a).

Let f(x) = 2x3 + 3x2 - 5x - 6

(i) f (-1) = 2(-1)3 + 3(-1)2 - 5(-1) - 6 = -2 + 3 + 5 - 6 = 0

Thus, (x + 1) is a factor of the polynomial f(x).

(ii)

Thus, (2x - 1) is not a factor of the polynomial f(x).

(iii) f (-2) = 2(-2)3 + 3(-2)2 - 5(-2) - 6 = -16 + 12 + 10 - 6 = 0

Thus, (x + 2) is a factor of the polynomial f(x).

Solution 4

(i) 2x + 1 is a factor of f(x) = 2x2 + ax - 3.

(ii) 3x - 4 is a factor of g(x) = 3x2 + 2x - k.

Solution 5

Let f(x) = x3 + ax2 + bx - 12

x - 2 = 0 x = 2

x - 2 is a factor of f(x). So, remainder = 0

x + 3 = 0 x = -3

x + 3 is a factor of f(x). So, remainder = 0

Adding (1) and (2), we get,

5a - 15 = 0

a = 3

Putting the value of a in (1), we get,

6 + b - 2 = 0

b = -4

Solution 6

Let f(x) = (3k + 2)x3 + (k - 1)

2x + 1 = 0

Since, 2x + 1 is a factor of f(x), remainder is 0.

Solution 7

f(x) = 2x5 - 6x4 - 2ax3 + 6ax2 + 4ax + 8

x - 2 = 0 x = 2

Since, x - 2 is a factor of f(x), remainder = 0.

2(2)5 - 6(2)4 - 2a(2)3 + 6a(2)2 + 4a(2) + 8 = 0

64 - 96 - 16a + 24a + 8a + 8 = 0

-24 + 16a = 0

16a = 24

a = 1.5

Solution 8

Let f(x) = x3 + (3m + 1) x2 + nx - 18

x - 1 = 0 x = 1

x - 1 is a factor of f(x). So, remainder = 0

x + 2 = 0 x = -2

x + 2 is a factor of f(x). So, remainder = 0

Adding (1) and (2), we get,

9m - 27 = 0

m = 3

Putting the value of m in (1), we get,

3(3) + n - 16 =0

9 + n - 16 = 0

n = 7

Solution 9

Let f(x) = x3 + 2x2 - kx + 4

x - 2 = 0 x = 2

On dividing f(x) by x - 2, it leaves a remainder k.

Solution 10

Let f(x) = ax3 + 9x2 + 4x - 10

x + 3 = 0 x = -3

On dividing f(x) by x + 3, it leaves a remainder 5.

Solution 11

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

x - 2 = 0 x = 2

Since, x - 2 is a factor, remainder = 0

x - 3 = 0 x = 3

On dividing f(x) by x - 3, it leaves a remainder 3.

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

a + 3 = 0

a = -3

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

-6 + b + 7 = 0

b = -1

Solution 12

Let f(x) = 2x3 + ax2 + bx - 2

2x - 3 = 0 x =

On dividing f(x) by 2x - 3, it leaves a remainder 7.

x + 2 = 0 x = -2

On dividing f(x) by x + 2, it leaves a remainder 0.

Adding (i) and (ii), we get,

7a - 21 = 0

a = 3

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

9 + 2b - 3 = 0

2b = -6

b = -3

Solution 13

Let the number k be added and the resulting polynomial be f(x).

So, f(x) = 3x3 - 5x2 + 6x + k

It is given that when f(x) is divided by (x - 3), the remainder is 8.

Thus, the required number is -46.

Solution 14

Let the number to be subtracted be k and the resulting polynomial be f(x).

So, f(x) = x3 + 3x2 - 8x + 14 - k

It is given that when f(x) is divided by (x - 2), the remainder is 10.

Thus, the required number is 8.

Solution 15

Let f(x) = 2x3 - 7x2 + ax - 6

x - 2 = 0 x = 2

When f(x) is divided by (x - 2), remainder = f(2)

Let g(x) = x3 - 8x2 + (2a + 1)x - 16

When g(x) is divided by (x - 2), remainder = g(2)

By the given condition, we have:

f(2) = g(2)

2a - 18 = 4a - 38

4a - 2a = 38 - 18

2a = 20

a = 10

Thus, the value of a is 10.

Solution 16

Solution 17

## Chapter 8 - Remainder And Factor Theorems Exercise Ex. 8(B)

Solution 1

(i) Let f(x) = x3 - 2x2 - 9x + 18

x - 2 = 0 x = 2

Remainder = f(2)

= (2)3 - 2(2)2 - 9(2) + 18

= 8 - 8 - 18 + 18

= 0

Hence, (x - 2) is a factor of f(x).

Now, we have:

x3 - 2x2 - 9x + 18 = (x - 2) (x2 - 9) = (x - 2) (x + 3) (x - 3)

(ii) Let f(x) = 2x3 + 5x2 - 28x - 15

x + 5 = 0 x = -5

Remainder = f(-5)

= 2(-5)3 + 5(-5)2 - 28(-5) - 15

= -250 + 125 + 140 - 15

= -265 + 265

= 0

Hence, (x + 5) is a factor of f(x).

Now, we have:

2x3 + 5x2 - 28x - 15 = (x + 5) (2x2 - 5x - 3)

= (x + 5) [2x2 - 6x + x - 3]

= (x + 5) [2x(x - 3) + 1(x - 3)]

= (x + 5) (2x + 1) (x - 3)

(iii) Let f(x) = 3x3 + 2x2 - 3x - 2

3x + 2 = 0

Hence, (3x + 2) is a factor of f(x).

Now, we have:

Solution 2

(i)

(ii) Let f(x) = 2x3 + x2 - 13x + 6

For x = 2,

f(x) = f(2) = 2(2)3 + (2)2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0

Hence, (x - 2) is a factor of f(x).

(iii) f(x) = 3x3 + 2x2 - 23x - 30

For x = -2,

f(x) = f(-2) = 3(-2)3 + 2(-2)2 - 23(-2) - 30

= -24 + 8 + 46 - 30 = -54 + 54 = 0

Hence, (x + 2) is a factor of f(x).

(iv) f(x) = 4x3 + 7x2 - 36x - 63

For x = 3,

f(x) = f(3) = 4(3)3 + 7(3)2 - 36(3) - 63

= 108 + 63 - 108 - 63 = 0

Hence, (x + 3) is a factor of f(x).

(v) f(x) = x3 + x2 - 4x - 4

For x = -1,

f(x) = f(-1) = (-1)3 + (-1)2 - 4(-1) - 4

= -1 + 1 + 4 - 4 = 0

Hence, (x + 1) is a factor of f(x).

Solution 3

Let f(x) = 3x3 + 10x2 + x - 6

For x = -1,

f(x) = f(-1) = 3(-1)3 + 10(-1)2 + (-1) - 6 = -3 + 10 - 1 - 6 = 0

Hence, (x + 1) is a factor of f(x).

Now, 3x3 + 10x2 + x - 6 = 0

Solution 4

f (x) = 2x3 - 7x2 - 3x + 18

For x = 2,

f(x) = f(2) = 2(2)3 - 7(2)2 - 3(2) + 18

= 16 - 28 - 6 + 18 = 0

Hence, (x - 2) is a factor of f(x).

Solution 5

f(x) = x3 + 3x2 + ax + b

Since, (x - 2) is a factor of f(x), f(2) = 0

(2)3 + 3(2)2 + a(2) + b = 0

8 + 12 + 2a + b = 0

2a + b + 20 = 0 ...(i)

Since, (x + 1) is a factor of f(x), f(-1) = 0

(-1)3 + 3(-1)2 + a(-1) + b = 0

-1 + 3 - a + b = 0

-a + b + 2 = 0 ...(ii)

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

3a + 18 = 0

a = -6

Substituting the value of a in (ii), we get,

b = a - 2 = -6 - 2 = -8

f(x) = x3 + 3x2 - 6x - 8

Now, for x = -1,

f(x) = f(-1) = (-1)3 + 3(-1)2 - 6(-1) - 8 = -1 + 3 + 6 - 8 = 0

Hence, (x + 1) is a factor of f(x).

Solution 6

Let f(x) = 4x3 - bx2 + x - c

It is given that when f(x) is divided by (x + 1), the remainder is 0.

f(-1) = 0

4(-1)3 - b(-1)2 + (-1) - c = 0

-4 - b - 1 - c = 0

b + c + 5 = 0 ...(i)

It is given that when f(x) is divided by (2x - 3), the remainder is 30.

Multiplying (i) by 4 and subtracting it from (ii), we get,

5b + 40 = 0

b = -8

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

c = -5 + 8 = 3

Therefore, f(x) = 4x3 + 8x2 + x - 3

Now, for x = -1, we get,

f(x) = f(-1) = 4(-1)3 + 8(-1)2 + (-1) - 3 = -4 + 8 - 1 - 3 = 0

Hence, (x + 1) is a factor of f(x).

Solution 7

f(x) = x2 + px + q

It is given that (x + a) is a factor of f(x).

g(x) = x2 + mx + n

It is given that (x + a) is a factor of g(x).

From (i) and (ii), we get,

pa - q = ma - n

n - q = a(m - p)

Hence, proved.

Solution 8

Let f(x) = ax3 + 3x2 - 3

When f(x) is divided by (x - 4), remainder = f(4)

f(4) = a(4)3 + 3(4)2 - 3 = 64a + 45

Let g(x) = 2x3 - 5x + a

When g(x) is divided by (x - 4), remainder = g(4)

g(4) = 2(4)3 - 5(4) + a = a + 108

It is given that f(4) = g(4)

64a + 45 = a + 108

63a = 63

a = 1

Solution 9

Let f(x) = x3 - ax2 + x + 2

It is given that (x - a) is a factor of f(x).

Remainder = f(a) = 0

a3 - a3 + a + 2 = 0

a + 2 = 0

a = -2

Solution 10

Let the number to be subtracted from the given polynomial be k.

Let f(y) = 3y3 + y2 - 22y + 15 - k

It is given that f(y) is divisible by (y + 3).

Remainder = f(-3) = 0

3(-3)3 + (-3)2 - 22(-3) + 15 - k = 0

-81 + 9 + 66 + 15 - k = 0

9 - k = 0

k = 9

## Chapter 8 - Remainder And Factor Theorems Exercise Ex. 8(C)

Solution 1

Let f(x) = x3 - 7x2 + 14x - 8

f(1) = (1)3 - 7(1)2 + 14(1) - 8 = 1 - 7 + 14 - 8 = 0

Hence, (x - 1) is a factor of f(x).

Solution 2

Solution 3

Let f(x) = x3 + 3x2 - mx + 4

According to the given information,

f(2) = m + 3

(2)3 + 3(2)2 - m(2) + 4 = m + 3

8 + 12 - 2m + 4 = m + 3

24 - 3 = m + 2m

3m = 21

m = 7

Solution 4

Let the required number be k.

Let f(x) = 3x3 - 8x2 + 4x - 3 - k

According to the given information,

f (-2) = 0

3(-2)3 - 8(-2)2 + 4(-2) - 3 - k = 0

-24 - 32 - 8 - 3 - k = 0

-67 - k = 0

k = -67

Thus, the required number is -67.

Solution 5

Let f(x) = x3 + (a + 1)x2 - (b - 2)x - 6

Since, (x + 1) is a factor of f(x).

Remainder = f(-1) = 0

(-1)3 + (a + 1)(-1)2 - (b - 2) (-1) - 6 = 0

-1 + (a + 1) + (b - 2) - 6 = 0

a + b - 8 = 0 ...(i)

Since, (x - 2) is a factor of f(x).

Remainder = f(2) = 0

(2)3 + (a + 1) (2)2 - (b - 2) (2) - 6 = 0

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

4a - 2b + 10 = 0

2a - b + 5 = 0 ...(ii)

Adding (i) and (ii), we get,

3a - 3 = 0

a = 1

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

1 + b - 8 = 0

b = 7

f(x) = x3 + 2x2 - 5x - 6

Now, (x + 1) and (x - 2) are factors of f(x). Hence, (x + 1) (x - 2) = x2 - x - 2 is a factor of f(x).

f(x) = x3 + 2x2 - 5x - 6 = (x + 1) (x - 2) (x + 3)

Solution 6

Let f(x) = x2 + ax + b

Since, (x - 2) is a factor of f(x).

Remainder = f(2) = 0

(2)2 + a(2) + b = 0

4 + 2a + b = 0

2a + b = -4 ...(i)

It is given that:

a + b = 1 ...(ii)

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

a = -5

Substituting the value of a in (ii), we get,

b = 1 - (-5) = 6

Solution 7

Let f(x) = x3 + 6x2 + 11x + 6

For x = -1

f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6

= -1 + 6 - 11 + 6 = 12 - 12 = 0

Hence, (x + 1) is a factor of f(x).

Solution 8

Let f(x) = mx3 + 2x2 - 3

g(x) = x2 - mx + 4

It is given that f(x) and g(x) leave the same remainder when divided by (x - 2). Therefore, we have:

f (2) = g (2)

m(2)3 + 2(2)2 - 3 = (2)2 - m(2) + 4

8m + 8 - 3 = 4 - 2m + 4

10m = 3

m =

Solution 9

Let f(x) = px3 + 4x2 - 3x + q

It is given that f(x) is completely divisible by (x2 - 1) = (x + 1)(x - 1).

Therefore, f(1) = 0 and f(-1) = 0

f(1) = p(1)3 + 4(1)2 - 3(1) + q = 0

p + q + 1 = 0 ...(i)

f(-1) = p(-1)3 + 4(-1)2 - 3(-1) + q = 0

-p + q + 7 = 0 ...(ii)

Adding (i) and (ii), we get,

2q + 8 = 0

q = -4

Substituting the value of q in (i), we get,

p = -q - 1 = 4 - 1 = 3

f(x) = 3x3 + 4x2 - 3x - 4

Given that f(x) is completely divisible by (x2 - 1).

Solution 10

Let the required number be k.

Let f(x) = x2 + x + 3 + k

It is given that f(x) is divisible by (x + 3).

Remainder = 0

f (-3) = 0

(-3)2 + (-3) + 3 + k = 0

9 - 3 + 3 + k = 0

9 + k = 0

k = -9

Thus, the required number is -9.

Solution 11

It is given that when the polynomial x3 + 2x2 - 5ax - 7 is divided by (x - 1), the remainder is A.

(1)3 + 2(1)2 - 5a(1) - 7 = A

1 + 2 - 5a - 7 = A

- 5a - 4 = A ...(i)

It is also given that when the polynomial x3 + ax2 - 12x + 16 is divided by (x + 2), the remainder is B.

x3 + ax2 - 12x + 16 = B

(-2)3 + a(-2)2 - 12(-2) + 16 = B

-8 + 4a + 24 + 16 = B

4a + 32 = B ...(ii)

It is also given that 2A + B = 0

Using (i) and (ii), we get,

2(-5a - 4) + 4a + 32 = 0

-10a - 8 + 4a + 32 = 0

-6a + 24 = 0

6a = 24

a = 4

Solution 12

Let f(x) = (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15

It is given that (3x + 5) is a factor of f(x).

f(x) = (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15

= 3x3 + 5x2 - 9x - 15

Solution 13

If (x - 3) divides f(x) = x3 - px2 + x + 6, then,

Remainder = f(3) = 33 - p(3)2 + 3 + 6 = 36 - 9p

If (x - 3) divides g(x) = 2x3 - x2 - (p + 3) x - 6, then

Remainder = g(3) = 2(3)3 - (3)2 - (p + 3) (3) - 6 = 30 - 3p

Now, f(3) = g(3)

36 - 9p = 30 - 3p

-6p = -6

p = 1

Solution 14

f(x) = 2x3 + x2 - 13x + 6

Factors of constant term 6 are 1, 2, 3, 6.

Putting x = 2, we have:

f(2) = 2(2)3 + 22 - 13 (2) + 6 = 16 + 4 - 26 + 6 = 0

Hence (x - 2) is a factor of f(x).

Solution 15

Let f(x) = 2x3 + 3x2 - kx + 5

Using Remainder Theorem, we have

f(2) = 7

2(2)3 + 3(2)2 - k(2) + 5 = 7

16 + 12 - 2k + 5 = 7

33 - 2k = 7

2k = 26

k = 13

Solution 16

Here, f(x) = 16x3 - 8x2 + 4x + 7

Let the number subtracted be k from the given polynomial f(x).

Given that 2x + 1 is a factor of f(x).

Therefore 1 must be subtracted from 16x3 - 8x2 + 4x + 7 so that the resulting expression has 2x + 1 as a factor.

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