# Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems

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## Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Download PDF

### Exercises of Concise Selina Solutions Class 10 Maths Chapter 8 Remainder and Factor Theorems

Exercise 8(A) Solutions

Exercise 8(B) Solutions

Exercise 8(C) Solutions

## Access Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems

Exercise 8(A) Page No: 108

1. Find, in each case, the remainder when:

(i) x4 â€“ 3x2 + 2x + 1 is divided by x â€“ 1.

(ii) x3 + 3x2 â€“ 12x + 4 is divided by x â€“ 2.

(ii) x4 + 1 is divided by x + 1.

Solution:

From remainder theorem, we know that when a polynomial f (x) is divided by (x â€“ a), then the remainder is f(a).

(i) Given, f(x) = x4 â€“ 3x2 + 2x + 1 is divided by x – 1

So, remainder = f(1) = (1)4 â€“ 3(1)2 + 2(1) + 1 = 1 â€“ 3 + 2 + 1 = 1

(ii) Given, f(x) = x3 + 3x2 â€“ 12x + 4 is divided by x â€“ 2

So, remainder = f(2) = (2)3 + 3(2)2 â€“ 12(2) + 4 = 8 + 12 â€“ 24 + 4 = 0

(iii) Given, f(x) = x4 + 1 is divided by x + 1

So, remainder = f(-1) = (-1)4 + 1 = 2

2. Show that:

(i) x â€“ 2 is a factor of 5x2 + 15x â€“ 50

(ii) 3x + 2 is a factor of 3x2 â€“ x â€“ 2

Solution:

(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.

(i) f(x) = 5x2 + 15x â€“ 50

f(2) = 5(2)2 + 15(2) â€“ 50 = 20 + 30 â€“ 50 = 0

As the remainder is zero for x = 2

Thus, we can conclude that (x â€“ 2) is a factor of 5x2 + 15x â€“ 50

(ii) f(x) = 3x2 â€“ x â€“ 2

f(-2/3) = 3(-2/3)2 â€“ (-2/3) â€“ 2 = 4/3 + 2/3 â€“ 2 = 2 â€“ 2 = 0

As the remainder is zero for x = -2/3

Thus, we can conclude that (3x + 2) is a factor of 3x2 â€“ x â€“ 2

3. Use the Remainder Theorem to find which of the following is a factor of 2x3Â + 3x2Â – 5x – 6.

(i) x + 1 (ii) 2x – 1

(iii) x + 2

Solution:

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

Here, f(x) = 2x3Â + 3x2Â – 5x – 6

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

â‡’ Remainder is zero for x = -1

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

(ii) f(1/2) = 2(1/2)3Â + 3(1/2)2Â â€“ 5(1/2) â€“ 6

= Â¼ + Â¾ – 5/2 â€“ 6

= -5/2 â€“ 5 = -15/2

â‡’ Remainder is not equals to zero for x = 1/2

Therefore, (2x – 1) is not a factor of the polynomial f(x).

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

â‡’ Remainder is zero for x = -2

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

4. (i) If 2x + 1 is a factor of 2x2Â + ax – 3, find the value of a.

(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2Â + 2x – k.

Solution:

(i) Given, 2x + 1 is a factor of f(x) = 2x2Â + ax – 3.

So, f(-1/2) = 0

2(-1/2)2Â + a(-1/2) – 3 = 0

Â½ – a/2 â€“ 3 = 0

1 â€“ a â€“ 6 = 0

a = -5

(ii) Given, 3x – 4 is a factor of g(x) = 3x2Â + 2x – k.

So, f(4/3) = 0

3(4/3)2Â + 2(4/3) â€“ k = 0

16/3 + 8/3 â€“ k = 0

24/3 = k

k = 8

5. Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3Â + ax2Â + bx – 12.

Solution:

Here, f(x) = x3Â + ax2Â + bx â€“ 12

Given, x â€“ 2 and x + 3 both are the factors of f(x)

So,

f(2) and f(-3) both should be equal to zero.

f(2) = (2)3Â + a(2)2Â + b(2) â€“ 12

0 = 8 + 4a + 2b â€“ 12

0 = 4a + 2b â€“ 4

2a + b = 2 â€¦. (1)

Now,

f(-3) = (-3)3Â + a(-3)2Â + b(-3) â€“ 12

0 = -27 + 9a â€“ 3b â€“ 12

9a â€“ 3b â€“ 39 = 0

3a â€“ b = 13 â€¦. (2)

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

5a = 15

Thus, a = 3

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

6 + b = 2

Thus, b = -4

6. Find the value of k, if 2x + 1 is a factor of (3k + 2)x3Â + (k – 1).Â

Solution:

Let take f(x) = (3k + 2)x3Â + (k – 1)

Now, 2x + 1 = 0

x = -1/2

As, 2x + 1 is a factor of f(x) then the remainder should be 0.

f(-1/2) = (3k + 2)(-1/2)3Â + (k – 1) = 0

5k = 10 = 0

k = 2

7. Find the value of a, if x – 2 is a factor of 2x5Â – 6x4Â – 2ax3Â + 6ax2Â + 4ax + 8.

Solution:

Given, f(x) = 2x5Â – 6x4Â – 2ax3Â + 6ax2Â + 4ax + 8 and x â€“ 2 is a factor of f(x).

So, x – 2 = 0;Â x = 2

Hence, f(2) = 0

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

64 – 96 – 16a + 24a + 8a + 8 = 0

-24 + 16a = 0

16a = 24

Thus, a = 1.5

8. Find the values of m and n so that x – 1 and x + 2 both are factors of x3Â + (3m + 1) x2Â + nx – 18.

Solution:

Let f(x) = x3Â + (3m + 1) x2Â + nx â€“ 18

Given, (x â€“ 1) and (x + 2) are the factors of f(x).

So,

x – 1 = 0;Â x = 1 and x + 2 = 0;Â x = -2

f(1) and f(-2) both should be equal to zero.

(1)3Â + (3m + 1)(1)2Â + n(1) â€“ 18 = 0

1 + 3m + 1 + n â€“ 18 = 0

3m + n â€“ 16 = 0 â€¦.. (1)

And,

(-2)3Â + (3m + 1)(-2)2Â + n(-2) â€“ 18 = 0

8 + 12m + 4 – 2n â€“ 18 = 0

12m â€“ 2n â€“ 22 = 0

6m â€“ n â€“ 11 = 0 â€¦.. (2)

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

9m – 27 = 0

Thus, m = 3

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

3(3) + n – 16 =0

9 + n – 16 = 0

Therefore, n = 7

Exercise 8(B) Page No: 111

1. Using the Factor Theorem, show that:

(i) (x – 2) is a factor of x3Â – 2x2Â – 9x + 18. Hence, factorise the expression x3Â – 2x2Â – 9x + 18 completely.

(ii) (x + 5) is a factor of 2x3Â + 5x2Â – 28x – 15. Hence, factorise the expression 2x3Â + 5x2Â – 28x – 15 completely.

(iii) (3x + 2) is a factor of 3x3Â + 2x2Â – 3x – 2. Hence, factorise the expression 3x3Â + 2x2Â – 3x – 2 completely.

Solution:

(i) Here, f(x) = x3Â – 2x2Â – 9x + 18

So, x – 2 = 0Â â‡’ x = 2

Thus, remainder = f(2)

= (2)3Â – 2(2)2Â – 9(2) + 18

= 8 – 8 – 18 + 18

= 0

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

Now, performing division of polynomial f(x) by (x â€“ 2) we have

Thus, x3Â – 2x2Â – 9x + 18 = (x – 2) (x2Â – 9) = (x – 2) (x + 3) (x – 3)

(ii) Here, f(x) = 2x3Â + 5x2Â – 28x – 15

So, x + 5 = 0Â â‡’ x = -5

Thus, remainder = f(-5)

= 2(-5)3Â + 5(-5)2Â – 28(-5) – 15

= -250 + 125 + 140 – 15

= -265 + 265

= 0

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

Now, performing division of polynomial f(x) by (x + 5) we get

So, 2x3Â + 5x2Â – 28x – 15 = (x + 5) (2x2Â – 5x – 3)

Further, on factorisation

= (x + 5) [2x2Â – 6x + x – 3]

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

Thus, f(x) is factorised as (x + 5) (2x + 1) (x – 3)

(iii) Here, f(x) = 3x3Â + 2x2Â – 3x â€“ 2

So, 3x + 2 = 0Â â‡’ x = -2/3

Thus, remainder = f(-2/3)

= 3(-2/3)3Â + 2(-2/3)2Â â€“ 3(-2/3) â€“ 2

= -8/9 + 8/9 + 2 â€“ 2

= 0

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

Now, performing division of polynomial f(x) by (3x + 2) we get

Thus, 3x3Â + 2x2Â – 3x â€“ 2 = (3x + 2) (x2 – 1) = (3x + 2) (x â€“ 1) (x + 1)

2. Using the Remainder Theorem, factorise each of the following completely.

(i)Â 3x3Â + 2x2Â âˆ’ 19x + 6

(ii) 2x3Â + x2Â – 13x + 6

(iii) 3x3Â + 2x2Â – 23x – 30

(iv) 4x3Â + 7x2Â – 36x – 63

(v) x3Â + x2Â – 4x – 4

Solution:

(i) Let f(x) = 3x3Â + 2x2Â âˆ’ 19x + 6

For x = 2, the value of f(x) will be

= 3(2)3Â + 2(2)2Â â€“ 19(2) + 6

= 24 + 8 â€“ 38 + 6 = 0

As f(2) = 0, so (x – 2) is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x -2) (3x2 + 8x – 3)

= (x – 2) (3x2 + 9x – x – 3)

= (x – 2) [3x(x + 3) -1(x + 3)]

= (x – 2) (x + 3) (3x – 1)

(ii) Let f(x) = 2x3Â + x2Â – 13x + 6

For x = 2, the value of f(x) will be

f(2) = 2(2)3Â + (2)2Â – 13(2) + 6 = 16 + 4 – 26 + 6 = 0

As f(2) = 0, so (x – 2) is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x -2) (2x2 + 5x – 3)

= (x – 2) [2x2 + 6x – x – 3]

= (x – 2) [2x(x + 3) -1(x + 3)]

= (x – 2) [2x(x + 3) -1(x + 3)]

= (x – 2) (2x â€“ 1) (x + 3)

(iii) Let f(x) = 3x3Â + 2x2Â – 23x – 30

For x = -2, the value of f(x) will be

f(-2) = 3(-2)3Â + 2(-2)2Â – 23(-2) – 30

= -24 + 8 + 46 – 30 = -54 + 54 = 0

As f(-2) = 0, so (x + 2) is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x + 2) (3x2 â€“ 4x – 15)

= (x + 2) (3x2 â€“ 9x + 5x – 15)

= (x + 2) [3x(x â€“ 3) + 5(x â€“ 3)]

= (x + 2) (3x + 5) (x â€“ 3)

(iv) Let f(x) = 4x3Â + 7x2Â – 36x – 63

For x = 3, the value of f(x) will be

f(3) = 4(3)3Â + 7(3)2Â – 36(3) – 63

= 108 + 63 – 108 – 63 = 0

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

Now, performing long division we have

Thus, f(x) = (x + 3) (4x2 â€“ 5x – 21)

= (x + 3) (4x2 â€“ 12x + 7x – 21)

= (x + 3) [4x(x â€“ 3) + 7(x – 3)]

= (x + 3) (4x + 7) (x – 3)

(v) Let f(x) = x3Â + x2Â – 4x – 4

For x = -1, the value of f(x) will be

f(-1) = (-1)3Â + (-1)2Â – 4(-1) – 4

= -1 + 1 + 4 – 4 = 0

As, f(-1) = 0 so (x + 1) is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x + 1) (x2 – 4)

= (x + 1) (x – 2) (x + 2)

3. Using the Remainder Theorem, factorise the expression 3x3Â + 10x2Â + x – 6. Hence, solve the equation 3x3Â + 10x2Â + x – 6 = 0.

Solution:

Letâ€™s take f(x) = 3x3Â + 10x2Â + x – 6

For x = -1, the value of f(x) will be

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

As, f(-1) = 0 so (x + 1) is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x + 1) (3x2 + 7x – 6)

= (x + 1) (3x2 + 9x â€“ 2x – 6)

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

= (x + 1) (x + 3) (3x – 2)

Now, 3x3Â + 10x2Â + x – 6 = 0

(x + 1) (x + 3) (3x – 2) = 0

Therefore,

x = -1, -3 or 2/3

4. Factorise the expression f (x) = 2x3Â – 7x2Â – 3x + 18. Hence, find all possible values of x for which f(x) = 0.

Solution:

Let f(x) = 2x3Â – 7x2Â – 3x + 18

For x = 2, the value of f(x) will be

f(2) = 2(2)3Â – 7(2)2Â – 3(2) + 18

= 16 – 28 – 6 + 18 = 0

As f(2) = 0, (x – 2) is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x – 2) (2x2 â€“ 3x – 9)

= (x – 2) (2x2 â€“ 6x + 3x – 9)

= (x – 2) [2x(x â€“ 3) + 3(x – 3)]

= (x – 2) (x – 3) (2x + 3)

Now, for f(x) = 0

(x – 2) (x – 3) (2x + 3) = 0

Hence x = 2, 3 or -3/2

5. Given that x – 2 and x + 1 are factors of f(x) = x3Â + 3x2Â + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

Solution:

Let f(x) = x3Â + 3x2Â + ax + b

As, (x – 2) is a factor of f(x), so f(2) = 0

(2)3Â + 3(2)2Â + a(2) + b = 0

8 + 12 + 2a + b = 0

2a + b + 20 = 0 … (1)

And as, (x + 1) is a factor of f(x), so f(-1) = 0

(-1)3Â + 3(-1)2Â + a(-1) + b = 0

-1 + 3 – a + b = 0

-a + b + 2 = 0 … (2)

Subtracting (2) from (1), we have

3a + 18 = 0

a = -6

On substituting the value of a in (ii), we have

b = a – 2 = -6 â€“ 2 = -8

Thus, f(x) = x3Â + 3x2Â – 6x – 8

Now, for x = -1

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

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

Now, performing long division we have

Hence, f(x) = (x + 1) (x2 + 2x – 8)

= (x + 1) (x2 + 4x â€“ 2x – 8)

= (x + 1) [x(x + 4) â€“ 2(x + 4)]

= (x + 1) (x + 4) (x – 2)

Exercise 8(C) Page No: 112

1. Show that (x – 1) is a factor of x3Â – 7x2Â + 14x – 8. Hence, completely factorise the given expression.

Solution:

Let f(x) = x3Â – 7x2Â + 14x – 8

Then, for x = 1

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

Thus, (x – 1) is a factor of f(x).

Now, performing long division we have

Hence, f(x) = (x – 1) (x2 â€“ 6x + 8)

= (x – 1) (x2 â€“ 4x â€“ 2x + 8)

= (x – 1) [x(x – 4) -2(x – 4)]

= (x – 1) (x – 4) (x – 2)

2. Using Remainder Theorem,Â factorise:

Â x3Â + 10x2Â – 37x + 26 completely.

Solution:

Let f(x) = x3Â + 10x2Â – 37x + 26

From remainder theorem, we know that

For x = 1, the value of f(x) is the remainder

f(1) = (1)3Â + 10(1)2Â â€“ 37(1) + 26 = 1 + 10 â€“ 37 + 26 = 0

As f(1) = 0, x â€“ 1 is a factor of f(x).

Now, performing long division we have

Thus, f(x) = (x – 1) (x2 + 11x – 26)

= (x – 1) (x2 + 13x – 2x – 26)

= (x – 1) [x(x + 13) â€“ 2(x + 13)]

= (x – 1) (x + 13) (x – 2)

3. When x3Â + 3x2Â – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.

Solution:

Let f(x) = x3Â + 3x2Â – mx + 4

From the question, we have

f(2) = m + 3

(2)3Â + 3(2)2Â – m(2) + 4 = m + 3

8 + 12 – 2m + 4 = m + 3

24 – 3 = m + 2m

3m = 21

Thus, m = 7

4. What should be subtracted from 3x3Â – 8x2Â + 4x – 3, so that the resulting expression has x + 2 as a factor?

Solution:

Letâ€™s assume the required number to be k.

And let f(x) = 3x3Â – 8x2Â + 4x – 3 – k

From the question, we have

f(-2) = 0

3(-2)3Â – 8(-2)2Â + 4(-2) – 3 – k = 0

-24 – 32 – 8 – 3 – k = 0

-67 – k = 0

k = -67

Therefore, the required number is -67.

5. If (x + 1) and (x – 2) are factors of x3Â + (a + 1)x2Â – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

Solution:

Letâ€™s take f(x) = x3Â + (a + 1)x2Â – (b – 2)x – 6

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

Then, remainder = f(-1) = 0

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

-1 + (a + 1) + (b – 2) – 6 = 0

a + b – 8 = 0 … (1)

And as, (x – 2) is a factor of f(x).

Then, remainder = f(2) = 0

(2)3Â + (a + 1) (2)2Â – (b – 2) (2) – 6 = 0

8 + 4a + 4 – 2b + 4 – 6 = 0

4a – 2b + 10 = 0

2a – b + 5 = 0 … (2)

Adding (1) and (2), we get

3a – 3 = 0

Thus, a = 1

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

1 + b – 8 = 0

Thus, b = 7

Hence, f(x) = x3Â + 2x2Â – 5x – 6

Now as (x + 1) and (x – 2) are factors of f(x).

So, (x + 1) (x – 2) = x2Â – x – 2 is also a factor of f(x).

Therefore, f(x) = x3Â + 2x2Â – 5x – 6 = (x + 1) (x – 2) (x + 3)

6. If x – 2 is a factor of x2Â + ax + b and a + b = 1, find the values of a and b.

Solution:

Let f(x) = x2Â + ax + b

Given, (x – 2) is a factor of f(x).

Then, remainder = f(2) = 0

(2)2Â + a(2) + b = 0

4 + 2a + b = 0

2a + b = -4 … (1)

And also given that,

a + b = 1 … (2)

Subtracting (2) from (1), we have

a = -5

On substituting the value of a in (2), we have

b = 1 – (-5) = 6

7. Factorise x3Â + 6x2Â + 11x + 6 completely using factor theorem.

Solution:

Let f(x) = x3Â + 6x2Â + 11x + 6

For x = -1, the value of f(x) is

f(-1) = (-1)3Â + 6(-1)2Â + 11(-1) + 6

= -1 + 6 – 11 + 6 = 12 – 12 = 0

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

Therefore, f(x) = (x + 1) (x2Â + 5x + 6)

= (x + 1) (x2Â + 3x + 2x + 6)

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

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

8. Find the value of ‘m’, if mx3Â + 2x2Â – 3 and x2Â – mx + 4 leave the same remainder when each is divided by x – 2.

Solution:

Let f(x) = mx3Â + 2x2Â â€“ 3 and g(x) = x2Â – mx + 4

From the question, itâ€™s given that f(x) and g(x) leave the same remainder when divided by (x – 2). So, we have:

f(2) = g(2)

m(2)3Â + 2(2)2Â – 3 = (2)2Â – m(2) + 4

8m + 8 – 3 = 4 – 2m + 4

10m = 3

Thus, m =Â 3/10

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.