Remainder And Factor Theorems Class 10 Concise Exe-8A ICSE Maths Selina Solution Ch-8. In this article you would learn how tp solve questions / problems on Remainder And Factor Theorems easily. Visit official website CISCE for detail information about ICSE Board Class-10 Mathematics.
Remainder And Factor Theorems Class 10 Concise Exe-8A ICSE Maths Selina Solution Ch-8
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-8 | Remainder and Factor Theorems |
| Writer | R.K. Bansal |
| Exe-8A | Remainder and Factor Theorems |
| Edition | 2025-2026 |
Questions / Problems on Remainder and Factor Theorems with Answer / Solutions
Class 10 Concise Exe-8A ICSE Maths Selina Solution Ch-8
Que-1: Find, in each the remainder when :
(i) x4 – 3x² + 2x + 1 is divided by x-1.
(ii) x³ + 3x² – 12x + 4 is divided by x-2
(iii) x4 + 1 is divided by x+1.
Sol: (i) By remainder theorem we know that when a polynomial f(x) is divided by x – a, then the remainder is f(a).
f(x) = x4 – 3x2 + 2x + 1
Remainder = f(1)
= (1)4 – 3(1)2 + 2(1) + 1
= 1 – 3 + 2 + 1
= 1
(ii) By remainder theorem we know that when a polynomial f(x) is divided by x – a, then the remainder is f(a).
f(x) = x3 + 3x2 – 12x + 4
Remainder = f(2)
= (2)3 + 3(2)2 – 12(2) + 4
= 8 + 12 – 24 + 4
= 0
(iii) By remainder theorem we know that when a polynomial f(x) is divided by x – a, then the remainder is f(a).
f(x) = x4 + 1
Remainder = f(–1)
= (–1)4 + 1
= 1 + 1
= 2
Que-2: Show that :
(i) x-2 is a factor of 5x²+15x-50
(ii) 3x+2 is a factor of 3x²-x-2.
Sol: (i) (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.
f(x) = 5x2 + 15x – 50
f(2) = 5(2)2 + 15(2) – 50
= 20 + 30 – 50
= 0
Hence, x – 2 is a factor of 5x2 + 15x – 50
(ii) (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.
f(x) = 3x2 – x – 2
𝑓(−2/3) = 3(−2/3)² − (−2/3) − 2
= 3 × (4/9) + (2/3) − 2
= (4/3) + (2/3) −2
= (6/3) − 2
= 2 – 2
= 0
Hence, 3x + 2 is a factor of 3x2 – x – 2
Que-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
Sol: (i) 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) = 2x3 + 3x2 – 5x – 6
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).
(ii) 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) = 2x3 + 3x2 – 5x – 6
𝑓(1/2) = 2(1/2)³ + 3(1/2)² − 5(1/2) − 6
= (1/4) + (3/4) − (5/2) − 6
= (−5/2) − 5
= −15/2 ≠0
Thus, (2x – 1) is not a factor of the polynomial f(x).
(iii) 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) = 2x3 + 3x2 – 5x – 6
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).
Que-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.
Sol: (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) − 𝑎² = 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
Que-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.
Sol: Expression x3 + ax2 + bx – 12
(x – 2) is a factor i.e, at x = 2
the remainder will be xero
⇒ (2)3 + a(2)2 + b(2) – 12 = 0
⇒ 8 + 4a + 2b – 12 = 0
⇒ 4a + 2b = 4
⇒ 2a + b = 2 …(i)
when x + 3 is a factor i.e., at x = -3 the remainder will be zero.
⇒ (-3)3 + a(-3)2 + b(-3) -12 = 0
⇒ -27 + 9a – 3b – 12 = 0
⇒ 9a – 3b = 39
⇒ 3a – b = 13 …(ii)
Solving (i) and (ii) simultaneously
2a + b = 2
By adding
3a – b = 13
5a = 15
a = 3
Substituting the value of a in the equation (i)
⇒ 2 x 3 + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = -4
⇒ a = 3, b = -4.
Que-6: Find the value of k, if 2x + 1 is a factor of (3k + 2)x3 + (k – 1).
Sol: Let f(x) = (3k + 2)x3 + (k − 1)
2x + 1 = 0
⇒𝑥 = −1/2
Since, 2x + 1 is a factor of f(x), remainder is 0.
∴ (3𝑘+2) (−1/2)³ + (𝑘−1) = 0
⇒ (3𝑘+2) (−1/8) + (𝑘−1) = 0
⇒ {−(3𝑘+2)/8} + (𝑘−1) = 0
⇒ (−3𝑘−2+8𝑘−8)/8 = 0
⇒ (−3k − 2 + 8k − 8) = 0 × 8
⇒ 5k – 10 = 0
⇒ 5k = 10
⇒ k = 10/5
⇒ k = 2
Que-7: Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.
Sol: f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 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
Que-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.
Sol: Let f(x) = x3 + (3m + 1)x2 + nx – 18
x – 1 = 0 ⇒ x = 1
x – 1 is a factor of f(x).
So, remainder = 0
∴ (1)3 + (3m + 1)(1)2 + n(1) – 18 = 0
⇒ 1 + 3m + 1 + n – 18 = 0
⇒ 3m + n – 16 = 0 …(1)
x + 2 = 0 ⇒ x = –2
x + 2 is a factor of f(x).
So, remainder = 0
∴ (–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
m = 3
Putting the value of m in (1), we get,
3(3) + n – 16 = 0
9 + n – 16 = 0
n = 7
Que-9: When x3 + 2x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.
Sol: Let f(x) = x3 + 2x2 – kx + 4
x – 2 = 0 ⇒ x = 2
On dividing f(x) by x – 2, it leaves a remainder k.
∴ f(2) = k
(2)3 + 2(2)2 – k(2) + 4 = k
8 + 8 – 2k + 4 = k
20 = 3k
𝑘 = 20/3 = 6*(2/3)
Que-10: Find the value of a, if the division of ax3 + 9x2 + 4x – 10 by x + 3 leaves a remainder 5.
Sol: 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
-27a + 81 – 22 = 5
-27a + 59 = 5
-27a = 5 – 59
-27a = -54
a = −54/−27
a = 2
Que-11: If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b.
Sol: Let f(x) = x3 + ax2 + bx + 6
∴ x – 2 = 0 ⇒ x = 2
(2)3 + a(2)2 + b(2) + 6 = 0
8 + 4a + 2b + 6 = 0
4a + 2b + 14 = 0
2(2a + b + 7) = 0
2a + b + 7 = 0/2
2a + b + 7 = 0
2a + b = –7 …(i)
∴ x – 3 = 0 ⇒ x = 3
(3)3 + a(3)2 + b(3) + 6 = 3
27 + 9a + 3b + 6 = 3
9a + 3b + 33 = 3
9a + 3b = 3 – 33
9a + 3b = –30
3(3a + b) = –30
3a + b = −30/3
3a + b = –10 …(ii)
Subtracting (i) from (ii), we get,
2a + b = – 7
3a + b = – 10
– – +
– a = 3
∴ a = –3
Substituting the value of a = –3 in (i), we get,
2a + b = –7
2(–3) + b = –7
– 6 + b + 7 = 0
b = –7 + 6
∴ b = –1
Que-12: The expression 2x3 + ax2 + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
Sol: Let f(x) = 2x3 + ax2 + bx – 2
2x – 3 = 0 ⇒ 𝑥 = 3/2
On dividing f(x) by 2x – 3, it leaves a remainder 7.
∴ 2(3/2)³ + 𝑎(3/2)² + 𝑏(3/2) − 2 = 7
(27/4) + (9𝑎/4) + (3b/2) = 9
(27+9𝑎+6𝑏)/4 = 9
27 + 9a + 6b = 36
9a + 6b – 9 = 0
3a + 2b – 3 = 0 …(1)
x + 2 = 0 ⇒ x = –2
On dividing f(x) by x + 2, it leaves a remainder 0.
∴ 2(–2)3 + a(–2)2 + b(–2) – 2 = 0
–16 + 4a – 2b – 2 = 0
4a – 2b – 18 = 0 …(2)
Adding (1) and (2), we get,
7a – 21 = 0
a = 3
Substituting the value of a in (1), we get,
3(3) + 2b – 3 = 0
9 + 2b – 3 = 0
2b = –6
b = -3
Que-13: What number should be added to 3x3 – 5x2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8?
Sol: Let the number k be added and the resulting polynomial be f(x).
So, f(x) = 3x3 – 5x2 + 6x + k
It is given that when f(x) is divided by (x – 3), the remainder is 8.
∴ f(3) = 8
3(3)3 – 5(3)2 + 6(3) + k = 8
81 – 45 + 18 + k = 8
54 + k = 8
k = – 46
Thus, the required number is – 46.
Que-14: What number should be subtracted from x3 + 3x2 – 8x + 14 so that on dividing it with x – 2, the remainder is 10.
Sol: Let the number to be subtracted be k and the resulting polynomial be f(x).
So, f(x) = x3 + 3x2 – 8x + 14 – k
It is given that when f(x) is divided by (x – 2), the remainder is 10.
∴ f(2) = 10
(2)3 + 3(2)2 – 8(2) + 14 – k = 10
8 + 12 – 16 + 14 – k = 10
18 – k = 10
– k = 10 – 18 = – 8
k = 8
Thus, the required number is 8.
Que-15: The polynomials 2x3 – 7x2 + ax – 6 and x3 – 8x2 + (2a + 1)x – 16 leaves the same remainder when divided by x – 2. Find the value of ‘a’.
Sol: Let f(x) = 2x3 – 7x2 + ax – 6
x – 2 = 0 ⇒ x = 2
When f(x) is divided by (x – 2), remainder = f(2)
∴ f(2) = 2(2)3 – 7(2)2 + a(2) – 6
= 16 – 28 + 2a – 6
= 2a – 18
Let g(x) = x3 – 8x2 + (2a + 1)x – 16
When g(x) is divided by (x – 2), remainder = g(2)
∴ g(2) = (2)3 – 8(2)2 + (2a + 1)(2) – 16
= 8 – 32 + 4a + 2 – 16
= 4a – 38
By the given condition, we have:
f(2) = g(2)
2a – 18 = 4a – 38
4a – 2a = 38 – 18
2a = 20
a = 10
Thus, the value of a is 10.
Que-16: If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b
Sol: Let p(x) = 2x3 + ax2 + bx – 14
Given, (x – 2) is a factor of p(x),
⇒ Remainder = p(2) = 0
⇒ 2(2)3 + a(2)2 + b(2) – 14 = 0
⇒ 16 + 4a + 2b – 14 = 0
⇒ 4a + 2b + 2 = 0
⇒ 2a + b + 1 = 0 …(1)
Given, when p(x) is divided by (x – 3), it leaves a remainder 52
∴ p(3) = 52
∴ 2(3)3 + a(3)2 + b(3) – 14 = 52
⇒ 54 + 9a + 3b – 14 – 52 = 0
⇒ 9a + 3b – 12 = 0
⇒ 3a + b – 4 = 0 …(2)
Subtracting (1) from (2), we get,
a – 5 = 0 ⇒ a = 5
From (1),
10 + b + 1 = 0
⇒ b = –11
Que-17: Find ‘a‘ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leave the same remainder when divided by x + 3.
Sol: Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3.
The given polynomials are ax3 + 3x2 – 9
and 2x3 + 4x + a
Let p(x) = ax3 + 3x2 – 9
and q(x) = 2x3 + 4x + a
Given that p(x) and q(x) leave the same
remainder when divided by (x + 3),
Thus by Remainder Theorem, we have
p(–3) = q(–3)
⇒ a(–3)3 + 3(–3)2 – 9 = 2(–3)3 + 4(–3) + a
⇒ –27a + 27 – 9 = –54 – 12 + a
⇒ –27a + 18 = –66 + a
⇒ –27a – a = –66 – 18
⇒ -28a = –84
⇒ a = 84/28
∴ a = 3.
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