Remainder and Factor Theorem MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions

Remainder and Factor Theorem MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-8. We Provide Step by Step Answer of MCQs  on Remainder and Factor Theorem as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

Remainder and Factor Theorem MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions

Remainder and Factor Theorem MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-8

Board ICSE
Subject Maths
Class 10th
Chapter-8 Remainder and Factor
Writer/Book RS Aggarwal
Topics Solution of MCQs
Academic Session 2024-2025

Multiple Choice Questions 

Que-1: When a polynomial f(x) is divided by (x+a), then the remainder is
(a) a   (b) -a   (c) f(a)    (d) f(-a) 

Sol: (d) f(-a)

Reason:  Polynomial is f(x)
And, g(x) = x+a
x+a = 0
x = -a
So, f(-a).

Que-2: If p(x) = x+4, then p(x) + p(-x) = ?
(a) 0        (b) 2x         (c) 8            (d) -8

Sol:  (c) 8

Reason:  Let: p(x) = x+4
∴p(−x) = −x+4
We have, p(x) + p(-x)
= x+4 + (-x+4)
= x+4-x+4
= 8.

Que-3: If p(x) = x²-2√2x+1, p(2√2) = ?
(a) 0        (b) 1         (c) -1          (d) -8

Sol: (b) 1

Reason:   p(x) = x²-2√2x+1
Then, p(2√2) is
= (2√2)² – 2√2(2√2) + 1
= 8 – 8 + 1
= 1.

Que-4: If f(x) 3x-5x²-1, then f(-1) = ?
(a) 1        (b) -1         (c) 7         (d) -9

Sol: (d) -9

Reason:   f(x) 3x-5x²-1
Then, f(-1) is
= 3(-1) – 5(-1)² – 1
= -3 -5(1) – 1
= -3-5-1
= -9.

Que-5: If (x^101 + 101) is divided by (x+1), then the remainder is :
(a) 102       (b) 100         (c) 0        (d) -101

Sol: (b) 100

Reason:  p(x) = (x^101 + 101)
When p(x) is divided by (x+1), it leaves remainder p(-1).
Now p(-1) = (-1)^101 + 101
= -1+101
= 100.

Que-6: If (3x³-5x²+3x-7) is divided by (x-2), then the remainder is :
(a) -5          (b) 6        (c) 3         (d) -8

Sol:  (c) 3

Reason:  P(x) = 3x³-5x²+3x-7
When p(x) is divided by (x-2), it leaves remainder p(2).
Now, p(2) = 3(2)³ – 5(2)² + 3(2) – 7
= 3(8) – 5(4) + 6 – 7
= 24-20-1
= 3.

Que-7: When f(x) = x⁴+2x³-3x²+x-1 is divided by (x-2) the remainder is :
(a) 0           (b) 1            (c) -13            (d) 21

Sol:  (d) 21

Reason:  f(x) = x⁴+2x³-3x²+x-1
When f(x) is divided by (x-2), it leaves remainder f(2).
Now, f(2) = (2)⁴ + 2(2)³ – 3(2)² + (2) – 1
= 16 + 2(8) – 3(4) +1
=17+16-12
= 21.

Que-8: If p(x) = x³-3x²-4x+10 is divided by (x+2), then the remainder is :
(a) 0         (b) -1         (c) 2          (d) -2

Sol:  (d) -2

Reason:   p(x) = x³-3x²-4x+10
When p(x) is divided by (x+2), it leaves remainder p(-2).
Now, p(-2) = (-2)³ – 3(-2)² – 4(-2) + 10
= -8 – 3(4) + 8 + 10
= -12+10
= -2.

Que-9: If f(x) = x⁴-ax³+x²-ax is divided by (x-a), then the remainder is :
(a) 0          (b) a            (c) -a             (d) 2a²

Sol:  (a) 0

Reason:  f(x) = x⁴-ax³+x²-ax
When f(x) is divided by (x-a), it leaves remainder f(a).
Now, p(a) = (a)⁴ – a(a)³ + (a)² – a(a)
= a⁴ – a⁴ + a² – a²
= 0.

Que-10: When f(x) = x³+ax²+2x+a is divided by (x+a), then the remainder is :
(a) 0              (b) -a             (c) a            (d) 3a

Sol:  (b) -a

Reason:   f(x) = x³+ax²+2x+a
When f(x) is divided by (x+a), it leaves remainder f(-a).
Now, f(-a) = (-a)³ + a(-a)² + 2(-a) + a
= -a³ + a³ -2a + a
= -a.

Que-11: If (x²-7x+a) leaves a remainder 1 when divided by (x+1), then the value of a is :
(a) 1                 (b) -1            (c) -7           (d) -5

Sol:  (c) -7

Reason:  f(x) = x²-7x+a
Remainder = 1
x+1 = 0
x = -1
f(-1)
(-1)² – 7(-1) + a = 1
1+7+a = 1
8+a = 1
a = 1-8
a = -7

Que-12: The remainder when, (x³+1) is divided by (x+1) is
(a) 0           (b) 1            (c) 2            (d) 4

Sol:  (a) 0

Reason:  f(x) = x³+1
x+1 = 0
x = -1
f(-1) = (-1)³ + 1
= -1+1
= 0.

Que-13: In the division of a cubic polynomial f(x) by a linear polynomial, the remainder is f(-2), then the divisor must be :
(a) (x-2)          (b) (x+2)           (c) (2x+1)            (d) (2x-1)

Sol:  (b) x+2

Reason:  The remainder theorem tells us that if a polynomial p(x) is divided by x-a then the remainder is equal to p(a).Here since p(-2) is the remainder, so x-(-2) = x+2 must be the divisor.
x+2.

Que-14: If (2x³-7x²+4x-7) is divided by x, then the remainder is :
(a) 0            (b) -1             (c) -2            (d) -7

Sol:  (d) -7

Reason:  f(x) = 2x³-7x²+4x-7
x = 0
p(0) = 2(0)³ – 7(0)² + 4(0) – 7
= -7.

Que-15: If p(t) = t²-t-2, then the value of p(-1/3) is :
(a) -2              (b) -10/9           (c) -14/9            (d) 0

Sol:  (c) -14/9

Reason:  p(t) = t²-t-2
p(-1/3)
= (-1/3)² – (-1/3) – 2
= (1/9) + (1/3) – 2
= (1+3-18)/9
= -14/9.

Que-16: If f(x) = 4x³-3x²+5 is divided by (2x+1), then the remainder is :
(a) -15/4          (b) -9/4          (c) -5/4           (d) 15/4

Sol:  (d) 15/4

Reason:  f(x) = 4x³-3x²+5
2x+1 = 0
x = -1/2
f(-1/2) = 4(-1/2)³ – 3(-1/2)² + 5
= 4(-1/8) – 3(1/4) + 5
= (-1/2) – (3/4) + 5
= (-2-3+20)/4
= 15/4.

Que-17: If (x^50 – 1) is divided by (x-1), then the remainder is :
(a) 0          (b) -2           (c) 49            (d) 51

Sol:  (a) 0

Reason:  f(x) = x^50 – 1
x-1 = 0
x = 1
f(1) = 1^50 – 1
= 1-1
= 0.

Que-18: When 2x²-3kx+k is divided by (x+2), then the remainder obtained is 5. The value of k is :
(a) -2/5          (b) -3/5         (c) -3/7            (d) -5/
7

Sol:  (c) -3/7

Reason: f(x) = 2x²-3kx+k
remainder = 5
x+2 = 0
x = -2
Now, f(-2) is
5 = 2(-2)² – 3k(-2) + k
5 = 2(4) + 6k +k
5 = 8+7k
5-8 = 7k
-3/7 = k.

Que-19: If the polynomial p(x) = x³-4kx+3 is divided by (2x+1), then the remainder obtained is -3. The value of k is :
(a) -17/16         (b) -47/8         (c) -27/16        (d) -47/16

Sol:  (d) -47/16

Reason:  p(x) = x³-4kx+3
remainder = -3
2x+1 = 0
x = -1/2
Now, f(-1/2) is
(-1/2)³ – 4k(-1/2) + 3 = -3
(-1/8) + 2k + 3 = -3
2k = -3-3+(1/8)
2k = (-24-24+1)/8
2k = -47/8
k = -47/16.

Que-20: f(x) is a polynomial in x and a is a real number. If (x-a) is a factor of f(x), then f(a) must be
(a) zero        (b) a          (c) negative        (d) positive

Sol:  (a) zero

Reason: The Factor Theorem states that if f(x) is a polynomial and (x−a) is a factor of f(x), then f(a)=0.
Conversely, if f(a)=0, then (x−a) is a factor of f(x).
Thus, for the given question,
f(a) = 0

Que-21: If (x+5) is a factor of f(x) = x³-20x+5k, then k = ?
(a) -2           (b) -3           (c) 5           (d) -5

Sol:  (c) 5

Reason:  (x+5) is a factor of p(x) = x³-20x+5k
p(x) = x³-20x+5k
p(-5) = (-5)³-20(-5)+5k
p(-5) = -125+100+5k
p(-5) = 5k-25
Equating it with 0,
5k-25 = 0
5k = 25
k = 25/5
k = 5

Que-22: For what value of k is the polynomial f(x) = 2x³-kx²+3x+10 exactly divisible by (x+2)?
(a) 3           (b) -3           (c) -1/3        (d) -4

Sol:  (b) -3

Reason:  p(x) = 2x3 – kx2 + 3x + 10
When it is exactly divided by x – 2,
p(-2) = 0
2(-2)3 – k(-2)2 + 3(-2) + 10 = 0
2(-8) – k(4) – 6 + 10 = 0
-16 – k(4) – 6 + 10 = 0
-16 – 4k – 6 + 10 = 0
-12 – 4k = 0
-12 = 4k
∴ k = -12/4
= -3
The value of k = -3

Que-23: If (x^50 + 2x^49 + k) is divisible by (x+1), then the value of k is :
(a) 0         (b) -1          (c) 1          (d) -2

Sol:  (c) 1

Reason:  f(x) = x^50 + 2x^49 + k
x+1 = 0
x = -1
f(-1) = (-1)^50 + 2(-1)^49 + k
= 1 + 2(-1) + k
= 1-2+k
k = 1.

Que-24: If x-1 is a factor of x³-kx²+11x-6, then the value of k should be
(a) 0          (b) 3            (c) 4            (d) 6

Sol:  (d) 6

Reason:  f(x) = x³-kx²+11x-6
x-1 = 0
x = 1
f(1) = (1)³ – k(1)² + 11(1) – 6
= 1-k+11-6
= 6-k
k = 6.

Que-25: If a specific number a is substituted for the variable x in a polynomial f(x) so that the value of the polynomial becomes zero, then x=a is said to be :
(a) zero of the polynomial      (b) zero coefficient     (c) rational number     (d) factor of f(x)

Sol:  (a) zero of the polynomial

Reason:  If a specific number a is substituted for the variable xxx in a polynomial f(x) such that the value of the polynomial becomes zero, then x=a is said to be a root or zero of the polynomial f(x).

Que-26: (x+1) is a factor of the polynomial :
(a) x³+x²-x+1      (b) x³+2x²-x-2       (c) x³+4x²-x+2         (d) x³+x²+1

Sol:  (b) x³+2x²-x-2

Reason:  If (x+1) is a factor of a polynomial f(x)f(x)f(x), then substituting x = −1 in the polynomial should make the polynomial equal to zero. In other words, f(−1) = 0.
So, x+1 is a factor of
x³+2x²-x-2

Que-27: If (x-p) is a factor of (x³-px²+2x+p-1), then the value of p is :
(a) 1/2         (b) -1/2          (c) 1/3           (d) -1/3

Sol: (c) 1/3

Reason:  p(x) = x³-px²+2x+p-1
x-p = 0
x = p
f(p) = (p)³ – p(p)² + 2(p) + p – 1
= p³ – p³ + 2p + p -1
= 3p-1
3p = 1
p = 1/3.

Que-28: What number should be subtracted from 2x³-5x²+5x, so that the resulting polynomial has (2x-3) as a factor ?
(a) 2           (b) -2           (c) -3            (d) 3

Sol:  (d) 3

Reason:  Let the number to be subtracted be k and the resulting polynomial be f(x), then
f(x) = 2x³-5x²+5x-k
2x-3 = 0
x = 3/2
f(3/2) = 2(3/2)³ – 5(3/2)² + 5(3/2) – k = 0
= 2(27/8) – 5(9/4) + (15/2) – k = 0
= (27/4) – (45/4) + (15/2) – k = 0
= (27-45+30-4k)/4 = 0
= 12-4k = 0
= 12 = 4k
= k = 3.

Que-29: If (x²+ax+b) is divided by (x+c), then the remainder is :
(a) -c²+ac+b      (b) c²+ac+b        (c) c²-ac-b        (d) c²-ac+b

Sol:  (d) c²-ac+b

Reason:  f(x) = x²+ax+b
x+c = 0
x = -c
f(-c) = (-c)² + a(-c) + b
= c²-ac+b.

Que-30: If (x-a) is a factor of  f(x) = ax²+bx+c, then which of the following is true ?
(a) f(a) = 2         (b) f(-a) = 0         (c) f(a) = a          (d) f(a) = 0

Sol:  (d) f(a) = 0

Reason:  f(x) = ax²+bx+c
x-a = 0
x = a
f(a) = a(a)² + b(a) +c
= a³+ab+c
Since (x – a) is a factor of p(x), it implies that when x = a, p(x) = 0. Therefore, p(a) = 0.
Thus, f(a) = 0.

Que-31: For two polynomials f(x) and g(x), (x-a) and (x-b) are their respective factors. Which of the following is true ?
(a) f(a)+g(b)=1    (b) f(a)+g(b)=a-b    (c) f(a)+g(b)=0    (d) f(a)+g(b)=a+b

Sol:  (c) f(a)+g(b) = 0

Reason:  If (x−a) is a factor of the polynomial f(x), it means that f(a) = 0.
Similarly, if (x−b) is a factor of the polynomial g(x), it means that g(b) = 0.
Therefore, since f(a) = 0 and g(b) = 0, we have:
f(a) + g(b) = 0+0 = 0.

Que-32: If (x-2) is a factor of x³-kx-12, then the value of k is :
(a) 3          (b) 2            (c) -2           (d) -3

Sol:  (c) -2

Reason:  f(x) = x³-kx-12
x-2 = 0
x = 2
f(2) = (2)³ – (2)k – 12
= 8-2k-12
2k = -4
k = -2.

Que-33: For a polynomial f(x), f(-1) and f(2) are both equal to zero. Which of the following is a factor of f(x) ?
(a) x²+x-2         (b) x²-x-2         (c) x²+x+2         (d) x²-2x+1

Sol:  (b) x²-x-2

Reason:  If f(−1) = 0 and f(2) = 0, it means that (x+1) and (x−2) are factors of the polynomial f(x).
To find which option is a factor, we can multiply these linear factors:
(x+1)(x−2) = x²−2x+x−2
= x²-x-2

Que-34: If (x+2) is a factor of the polynomial x³-kx²-5x+6, then the value of k is :
(a) 1          (b) 2          (c) 3           (d) -2

Sol:  (b) 2

Reason:  f(x) = x³-kx²-5x+6
x+2 = 0
x = -2
f(-2) = (-2)³ – k(-2)² – 5(-2) + 6
= -8 – k(4) + 10 + 6
= -8-4k+16
4k = 8
k = 2.

–: Remainder and Factor Theorem MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions :–

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