Selina Textbook Solutions: Remainder and Factor Theorems | Mathematics Class 10 ICSE PDF Download

 Page 1


Remainder and Factor Theorems 
 
 
Question 1. 
 
 
Solution: 
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the 
remainder is f(a). 
 
 
Question 2. 
 
 
Solution: 
 
(x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is 
Page 2


Remainder and Factor Theorems 
 
 
Question 1. 
 
 
Solution: 
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the 
remainder is f(a). 
 
 
Question 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. 
 
Question 3. 
Use the Remainder Theorem to find which of the following is a factor of 2x
3
 + 3x
2
 – 5x – 
6. 
(i) x + 1 
(ii) 2x – 1 
(iii) x + 2 
 
Solution: 
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) = 2x
3
 + 3x
2
 – 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). 
 
 
 
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). 
Page 3


Remainder and Factor Theorems 
 
 
Question 1. 
 
 
Solution: 
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the 
remainder is f(a). 
 
 
Question 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. 
 
Question 3. 
Use the Remainder Theorem to find which of the following is a factor of 2x
3
 + 3x
2
 – 5x – 
6. 
(i) x + 1 
(ii) 2x – 1 
(iii) x + 2 
 
Solution: 
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) = 2x
3
 + 3x
2
 – 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). 
 
 
 
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). 
Question 4. 
(i) If 2x + 1 is a factor of 2x
2
 + ax – 3, find the value of a. 
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x
2
 + 2x – k. 
 
Solution: 
 
 
Question 5. 
Find the values of constants a and b when x – 2 and x + 3 both are the factors of 
expression x
3
 + ax
2
 + bx – 12. 
 
Solution: 
Page 4


Remainder and Factor Theorems 
 
 
Question 1. 
 
 
Solution: 
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the 
remainder is f(a). 
 
 
Question 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. 
 
Question 3. 
Use the Remainder Theorem to find which of the following is a factor of 2x
3
 + 3x
2
 – 5x – 
6. 
(i) x + 1 
(ii) 2x – 1 
(iii) x + 2 
 
Solution: 
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) = 2x
3
 + 3x
2
 – 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). 
 
 
 
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). 
Question 4. 
(i) If 2x + 1 is a factor of 2x
2
 + ax – 3, find the value of a. 
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x
2
 + 2x – k. 
 
Solution: 
 
 
Question 5. 
Find the values of constants a and b when x – 2 and x + 3 both are the factors of 
expression x
3
 + ax
2
 + bx – 12. 
 
Solution: 
 
Question 6. 
find the value of k, if 2x + 1 is a factor of (3k + 2)x
3
 + (k – 1). 
 
Solution: 
 
 
Page 5


Remainder and Factor Theorems 
 
 
Question 1. 
 
 
Solution: 
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the 
remainder is f(a). 
 
 
Question 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. 
 
Question 3. 
Use the Remainder Theorem to find which of the following is a factor of 2x
3
 + 3x
2
 – 5x – 
6. 
(i) x + 1 
(ii) 2x – 1 
(iii) x + 2 
 
Solution: 
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) = 2x
3
 + 3x
2
 – 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). 
 
 
 
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). 
Question 4. 
(i) If 2x + 1 is a factor of 2x
2
 + ax – 3, find the value of a. 
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x
2
 + 2x – k. 
 
Solution: 
 
 
Question 5. 
Find the values of constants a and b when x – 2 and x + 3 both are the factors of 
expression x
3
 + ax
2
 + bx – 12. 
 
Solution: 
 
Question 6. 
find the value of k, if 2x + 1 is a factor of (3k + 2)x
3
 + (k – 1). 
 
Solution: 
 
 
Question 7. 
Find the value of a, if x – 2 is a factor of 2x
5
 – 6x
4
 – 2ax
3
 + 6ax
2
 + 4ax + 8. 
 
Solution: 
 
f(x) = 2x
5
 – 6x
4
 – 2ax
3
 + 6ax
2
 + 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 
Question 8. 
Find the values of m and n so that x – 1 and x + 2 both are factors of x
3
 + (3m + 1) x
2
 + 
nx – 18. 
 
Solution: