Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(C)

Problems based on both factor theorem and remainder theorem are included in this exercise. If students wish to get a strong grip on this chapter, the Selina Solutions for Class 10 Maths is the best place to start with. All the solutions are prepared in a simple language to make it easier for students to understand. For solutions to the Concise Selina Solutions for Class 10 Maths Chapter 8 Remainder and Factor Theorems Exercise 8(C), the PDF is available in the links given below.

Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(C) Download PDF

 

selina solutions concise maths class 10 chapter 8c
selina solutions concise maths class 10 chapter 8c
selina solutions concise maths class 10 chapter 8c
selina solutions concise maths class 10 chapter 8c

 

Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems

Exercise 8(A) Solutions

Exercise 8(B) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(C)

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

Selina Solutions Concise Class 10 Maths Chapter 8 ex. 8(C) - 1

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)

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

Selina Solutions Concise Class 10 Maths Chapter 8 ex. 8(C) - 2

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)

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

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.

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).

Selina Solutions Concise Class 10 Maths Chapter 8 ex. 8(C) - 3

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

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

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).

Selina Solutions Concise Class 10 Maths Chapter 8 ex. 8(C) - 4

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)

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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