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

 

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

Selina Class 10 ICSE Solution Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems will give strength to you and the board will also clear the exam, Mathematics Subject demands time and practice, then keep practicing and you will definitely get success.

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Remainder and Factor Theorems Chapter 8  Exercise 8(A)

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

(i) x4 – 3x2 + 2x + 1 is divided by x – 1.

(ii) x3 + 3x2 – 12x + 4 is divided by x – 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 – a), then the remainder is f(a).

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

So, remainder = f(1) = (1)4 – 3(1)2 + 2(1) + 1 = 1 – 3 + 2 + 1 = 1

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

So, remainder = f(2) = (2)3 + 3(2)2 – 12(2) + 4 = 8 + 12 – 24 + 4 = 0

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

So, remainder = f(-1) = (-1)4 + 1 = 2

Question 2. Show that:

(i) x – 2 is a factor of 5x2 + 15x – 50

(ii) 3x + 2 is a factor of 3x2 – x – 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 – 50

f(2) = 5(2)2 + 15(2) – 50 = 20 + 30 – 50 = 0

As the remainder is zero for x = 2

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

(ii) f(x) = 3x2 – x – 2

f(-2/3) = 3(-2/3)2 – (-2/3) – 2 = 4/3 + 2/3 – 2 = 2 – 2 = 0

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

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

Question 3. Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 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) = 2x3 + 3x2 – 5x – 6

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

⇒ Remainder is zero for x = -1

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

(ii) f(1/2) = 2(1/2)3 + 3(1/2)2 – 5(1/2) – 6

= ¼ + ¾ – 5/2 – 6

= -5/2 – 5 = -15/2

⇒ 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)3 + 3(-2)2 – 5(-2) – 6 = -16 + 12 + 10 – 6 = 0

⇒ Remainder is zero for x = -2

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

Question 4. (i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.

(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2 + 2x – k.

Solution:

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

So, f(-1/2) = 0

2(-1/2)2 + a(-1/2) – 3 = 0

½ – a/2 – 3 = 0

1 – a – 6 = 0

a = -5

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

So, f(4/3) = 0

3(4/3)2 + 2(4/3) – k = 0

16/3 + 8/3 – k = 0

24/3 = k

k = 8

Question 5. Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3 + ax2 + bx – 12.

Solution:

Here, f(x) = x3 + ax2 + bx – 12

Given, x – 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)3 + a(2)2 + b(2) – 12

0 = 8 + 4a + 2b – 12

0 = 4a + 2b – 4

2a + b = 2 …. (1)

Now,

f(-3) = (-3)3 + a(-3)2 + b(-3) – 12

0 = -27 + 9a – 3b – 12

9a – 3b – 39 = 0

3a – b = 13 …. (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

Question 6. Find the value of k, if 2x + 1 is a factor of (3k + 2)x3 + (k – 1). 

Solution:

Let take f(x) = (3k + 2)x3 + (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)3 + (k – 1) = 0

5k = 10 = 0

k = 2

Question 7. Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.

Solution:

Given, f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8 and x – 2 is a factor of f(x).

So, x – 2 = 0; x = 2

Hence, f(2) = 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

Thus, a = 1.5

Question 8. Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18.

Solution:

Let f(x) = x3 + (3m + 1) x2 + nx – 18

Given, (x – 1) and (x + 2) are the factors of f(x).

So,

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

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

(1)3 + (3m + 1)(1)2 + n(1) – 18 = 0

1 + 3m + 1 + n – 18 = 0

3m + n – 16 = 0 ….. (1)

And,

(-2)3 + (3m + 1)(-2)2 + n(-2) – 18 = 0

8 + 12m + 4 – 2n – 18 = 0

12m – 2n – 22 = 0

6m – n – 11 = 0 ….. (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

      

Remainder and Factor Theorems Chapter 8  Exercise 8(B)

Question 1. Using the Factor Theorem, show that:

(i) (x – 2) is a factor of x3 – 2x2 – 9x + 18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.

(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.

(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2 completely.

Solution:

(i) Here, f(x) = x3 – 2x2 – 9x + 18

So, x – 2 = 0 ⇒ x = 2

Thus, remainder = f(2)

= (2)3 – 2(2)2 – 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 – 2) we have

Thus, x3 – 2x2 – 9x + 18 = (x – 2) (x2 – 9) = (x – 2) (x + 3) (x – 3)

(ii) Here, f(x) = 2x3 + 5x2 – 28x – 15

So, x + 5 = 0 ⇒ x = -5

Thus, remainder = f(-5)

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

Further, on factorisation

= (x + 5) [2x2 – 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) = 3x3 + 2x2 – 3x – 2

So, 3x + 2 = 0 ⇒ x = -2/3

Thus, remainder = f(-2/3)

= 3(-2/3)3 + 2(-2/3)2 – 3(-2/3) – 2

= -8/9 + 8/9 + 2 – 2

= 0

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

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

Thus, 3x3 + 2x2 – 3x – 2 = (3x + 2) (x2 – 1) = (3x + 2) (x – 1) (x + 1)

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

(i) 3x3 + 2x2 − 19x + 6

(ii) 2x3 + x2 – 13x + 6

(iii) 3x3 + 2x2 – 23x – 30

(iv) 4x3 + 7x2 – 36x – 63

(v) x3 + x2 – 4x – 4

Solution:

(i) Let f(x) = 3x3 + 2x2 − 19x + 6

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

= 3(2)3 + 2(2)2 – 19(2) + 6

= 24 + 8 – 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) = 2x3 + x2 – 13x + 6

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

f(2) = 2(2)3 + (2)2 – 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 – 1) (x + 3)

(iii) Let f(x) = 3x3 + 2x2 – 23x – 30

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

f(-2) = 3(-2)3 + 2(-2)2 – 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 – 4x – 15)

= (x + 2) (3x2 – 9x + 5x – 15)

= (x + 2) [3x(x – 3) + 5(x – 3)]

= (x + 2) (3x + 5) (x – 3)

(iv) Let f(x) = 4x3 + 7x2 – 36x – 63

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

f(3) = 4(3)3 + 7(3)2 – 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 – 5x – 21)

= (x + 3) (4x2 – 12x + 7x – 21)

= (x + 3) [4x(x – 3) + 7(x – 3)]

= (x + 3) (4x + 7) (x – 3)

(v) Let f(x) = x3 + x2 – 4x – 4

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

f(-1) = (-1)3 + (-1)2 – 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)

Question 3. Using the Remainder Theorem, factorise the expression 3x3 + 10x2 + x – 6. Hence, solve the equation 3x3 + 10x2 + x – 6 = 0.

Solution:

Let’s take f(x) = 3x3 + 10x2 + x – 6

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

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

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

= (x + 1) (x + 3) (3x – 2)

Now, 3x3 + 10x2 + x – 6 = 0

(x + 1) (x + 3) (3x – 2) = 0

Therefore,

x = -1, -3 or 2/3

Question 4. Factorise the expression f (x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = 0.

Solution:

Let f(x) = 2x3 – 7x2 – 3x + 18

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

f(2) = 2(2)3 – 7(2)2 – 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 – 3x – 9)

= (x – 2) (2x2 – 6x + 3x – 9)

= (x – 2) [2x(x – 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

Question 5. Given that x – 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

Solution:

Let f(x) = x3 + 3x2 + ax + b

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

(2)3 + 3(2)2 + 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)3 + 3(-1)2 + 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 – 2 = -8

Thus, f(x) = x3 + 3x2 – 6x – 8

Now, for x = -1

f(-1) = (-1)3 + 3(-1)2 – 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 – 2x – 8)

= (x + 1) [x(x + 4) – 2(x + 4)]

= (x + 1) (x + 4) (x – 2)

Remainder and Factor Theorems Chapter 8  Exercise 8(C)

Question 1. Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.

Solution:

Let f(x) = x3 – 7x2 + 14x – 8

Then, for x = 1

f(1) = (1)3 – 7(1)2 + 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 – 6x + 8)

= (x – 1) (x2 – 4x – 2x + 8)

= (x – 1) [x(x – 4) -2(x – 4)]

= (x – 1) (x – 4) (x – 2)

Question 2. Using Remainder Theorem, factorise:

 x3 + 10x2 – 37x + 26 completely.

Solution:

Let f(x) = x3 + 10x2 – 37x + 26

From remainder theorem, we know that

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

f(1) = (1)3 + 10(1)2 – 37(1) + 26 = 1 + 10 – 37 + 26 = 0

As f(1) = 0, x – 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) – 2(x + 13)]

= (x – 1) (x + 13) (x – 2)

Question 3. When x3 + 3x2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.

Solution:

Let f(x) = x3 + 3x2 – mx + 4

From the question, we have

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

Thus, m = 7

Question 4. What should be subtracted from 3x3 – 8x2 + 4x – 3, so that the resulting expression has x + 2 as a factor?

Solution:

Let’s assume the required number to be k.

And let f(x) = 3x3 – 8x2 + 4x – 3 – k

From the question, we have

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

Therefore, the required number is -67.

Question 5. If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

Solution:

Let’s take f(x) = x3 + (a + 1)x2 – (b – 2)x – 6

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

Then, 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 … (1)

And as, (x – 2) is a factor of f(x).

Then, 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 … (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) = x3 + 2x2 – 5x – 6

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

So, (x + 1) (x – 2) = x2 – x – 2 is also a factor of f(x).

Therefore, f(x) = x3 + 2x2 – 5x – 6 = (x + 1) (x – 2) (x + 3)

Question 6. If x – 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.

Solution:

Let f(x) = x2 + ax + b

Given, (x – 2) is a factor of f(x).

Then, remainder = f(2) = 0

(2)2 + 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

Question 7. Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.

Solution:

Let f(x) = x3 + 6x2 + 11x + 6

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

f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6

= -1 + 6 – 11 + 6 = 12 – 12 = 0

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

Therefore, f(x) = (x + 1) (x2 + 5x + 6)

= (x + 1) (x2 + 3x + 2x + 6)

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

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

Question 8. Find the value of ‘m’, if mx3 + 2x2 – 3 and x2 – mx + 4 leave the same remainder when each is divided by x – 2.

Solution:

Let f(x) = mx3 + 2x2 – 3 and g(x) = x2 – mx + 4

From the question, it’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)3 + 2(2)2 – 3 = (2)2 – m(2) + 4

8m + 8 – 3 = 4 – 2m + 4

10m = 3

Thus, m = 3/10

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