ICSE Solutions for Chapter 8 Remainder and Factor Theorem Class 10 Mathematics
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).
= (x - 1)(x2 - 2x - 4x + 8)
= (x - 1)[x(x - 2) – 4(x - 2)]
= (x - 1)(x - 2)(x - 4)
Question 2: Using Remainder Theorem, factorise: x3 + 10x2 - 37x + 26 completely.
Solution 2: By Remainder Theorem,
For x = 1, the value of the given expression is the remainder.
x3 + 10x2 - 37x + 26
= (1)3 + 10(1)2 - 37(1) + 26
= 1 + 10 - 37 + 26
= 37 – 37
= 0
⇒ x - 1 is a factor of x3 + 10x2 – 37x + 26.
= (x – 1)(x2 + 13x - 2x - 26)
= (x - 1)[x(x + 13) - 2(x + 13)]
∴ x3 + 10x2 - 37x + 26 = (x - 1)(x + 13)(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 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).
= (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
⇒ x = -1, -3, 2/3
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 4: (i) 2x + 1 is a factor of f(x) = 2x2 + ax - 3.
∴ f(-1/2) = 0
⇒ 2(-1/2)2 + a(-1/2) – 3 = 0
⇒1/2 – a/2 = 3
⇒ 1 – a = 6
⇒ a = -5
(ii) 3x – 4 is a factor of g(x) = 3x2 + 2x –k.
∴ 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: 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 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).
∴ x3 + 3x2 – 6x – 8 = (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)
Question 6: If x - 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
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 (ii), we get,
a = -5
Substituting the value of a in (ii), we get,
b = 1 - (-5) = 6
Question 7: Factorise x3 + 6x2 + 11x + 6 completely using factor theorem
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).
∴ x3 + 6x2 + 11x + 6 = (x + 1)(x2 + 5x + 6)
= (x + 1)(x2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)]
= (x + 1)(x + 2)(x + 3)
Question 8: The polynomials ax3 + 3x2- 3 and 2x3 - 5x + a, when divided by x - 4, leave the same remainder in each case. Find the value of a.
Solution 8: Let f(x) = ax + 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
Question 9: Find the value of 'a', if (x - a) is a factor of x3 - ax2 + x + 2.
Solution 9: Let f(x) = x2 - 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
Question 10: Find the value of a, if the division of ax3 + 9x2 + 4x - 10 by x + 3 leaves a remainder 5.
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.
∴ f(-3) = 5
a(-3)3 + 9(-3)2 + 4(-3) - 10 = 5
-27a + 81 – 12 - 10 = 5
54 = 27a
a = 2
∴ 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: 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 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).
= (x + 1)(x2 + 4x – 2x – 8)
= (x + 1)[x(x + 4) – 2(x + 4)]
= (x + 1)(x + 4)(x – 2)
Question 6: If x - 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
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 (ii), we get,
a = -5
Substituting the value of a in (ii), we get,
b = 1 - (-5) = 6
Question 7: Factorise x3 + 6x2 + 11x + 6 completely using factor theorem
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).
= (x + 1)(x2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)]
= (x + 1)(x + 2)(x + 3)
Question 8: The polynomials ax3 + 3x2- 3 and 2x3 - 5x + a, when divided by x - 4, leave the same remainder in each case. Find the value of a.
Solution 8: Let f(x) = ax + 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
Question 9: Find the value of 'a', if (x - a) is a factor of x3 - ax2 + x + 2.
Solution 9: Let f(x) = x2 - 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
Question 10: Find the value of a, if the division of ax3 + 9x2 + 4x - 10 by x + 3 leaves a remainder 5.
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.
∴ f(-3) = 5
a(-3)3 + 9(-3)2 + 4(-3) - 10 = 5
-27a + 81 – 12 - 10 = 5
54 = 27a
a = 2