Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8A Goyal Brothers Solutions Ch-8. Step by step solutions of questions on Remainder and Factor Theorems. Visit official Website CISCE for detail information about ICSE Board Class-10.

Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8A Goyal Brothers Solutions 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

Practice Questions / Problems on Remainder and Factor Theorems with Answer / Solutions

Class 10 RS Aggarwal Exe-8A Goyal Brothers Solutions 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.
∴ (3⁢𝑘+2)⁢ (−1/2)³ + (𝑘−1) = 0
⇒ (3⁢𝑘+2) ⁢(−1/8) + (𝑘−1) = 0
⇒ {−(3⁢𝑘+2)/8} + (𝑘−1) = 0
⇒ (−3⁢𝑘−2+8⁢𝑘−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) + (9⁢𝑎/4) + (3⁢b/2) = 9
(27+9⁢𝑎+6⁢𝑏)/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.

— : End of Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8A Goyal Brothers :-

Return to:-  ICSE Class 10 Maths RS Aggarwal Solutions
Please share with your friend
Thanks