Factorisation MCQs for ICSE Class-10 Maths

Factorisation MCQs for ICSE Class-10 Maths for Sem-1. These MCQ  / Objective Type Questions of Factorisation  is based on latest reduced syllabus according 2021-22 session on bifurcated pattern. Main motto of MCQ Type Question is cracking the next upcoming exam of council. Visit official website CISCE for detail information about ICSE Board Class-10 Maths

 MCQs of Factorisation for ICSE Class-10 Maths

Board ICSE
Class 10th ( x )
Subject Maths
Chapter Factorisation 
Syllabus  on bifurcated syllabus (after reduction)
bifurcated
pattern
Semester-1
Session 2021-22
Topic MCQ / Objective Type Question

ICSE Maths Factorisation MCQs for Class-10 of Sem-1

Question-1 Using remainder theorem, the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2, leaves a remainder 7.

(a) 12 (b) 13 (c) 16(d) 8

Ans :- (b) 13

Hint

f(2) = (2 × 23) + (3 × 22) – (k × 2) + 5

= (2 × 8) + (3 × 4) – 2k + 5

= 16 + 12 – 2k + 5

= 33 – 2k

given that, remainder = 7.

So, 7 = 33 – 2k

K = 13

Question -2 :- If (2 x + 1) is a factor of 6x3 + 5x2 + ax – 2  the value of a

(a) -2 (b) 2 (c) 3 (d) -3

Ans :- (d) -3

Hint

Let 2x + 1 = 0

Then, 2x = – 1

X = -½

Given, f(x) = 6x3 + 5x2 + ax – 2

substitute the value of x in f(x),

f (-½) = 6 (-½)3 + 5 (-½)2 + a (-½) – 2

= 6 (-1/8) + 5 (¼) – ½a – 2

= -3/4 + 5/4 – a/2 – 2

= (-3 + 4 – 2a – 8)/4

= (-6 – 2a)/4

, (2x + 1) is a factor of 6x3 + 5x2 + ax – 2

Then, remainder is 0.

So, (-6 – 2a)/4 = 0

-6 – 2a = 4 × 0

– 6 – 2a = 0

-2a = 6

a = -6/2

a = – 3

Question -3 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

Ans :- (c) 3

Hint

Let P number to be subtracted from 2x3 – 5x2 + 5x

Then, f(x) = 2x3 – 5x2 + 5x – p

Given, 2x – 3 = 0

x = 3/2

f(3/2) = 0

So, f(3/2) = 2(3/2)3 – 5(3/2)2 + 5(3/2) – p = 0

2(27/8) – 5(9/4) + 15/2 – p = 0

27/4 – 45/4 + 15/2 – p = 0

27 – 45 + 30 – 4p = 0

57 – 45 – 4p = 0

12 – 4p = 0

P = 12/4

P = 3

Question-4  If (x + 2) and (x – 3) are factors of x3 + ax + b,  the values of a and b ?

(a)  a = -7 and b = -6

(b)  a = 7 and b = 6

(c)  a = -7 and b = 6

(d)  a = 7 and b = -6

Ans :- (a)  a = -7 and b = -6

Hint

If x + 2 = 0, then x = -2

Substituting the value of x in f(x),
f(x) = x3 + ax + b

substitute the value of x in f(x),

f(-2) = (-2)3 + a(-2) + b

= -8 – 2a + b

(x + 2) is a factor of x3 + ax + b.

remainder is 0.

f(x) = 0

– 8 – 2a + b = 0

2a – b = – 8 …  (i)]

let x – 3 = 0

Then, x = 3

Given, f(x) = x3 + ax + b

, substitute the value of x in f(x),

f(3) = (3)3 + a(3) + b

= 27 + 3a + b

From the question, (x – 3) is a factor of x3 + ax + b.

Therefore, remainder is 0.

f(x) = 0

27 + 3a + b = 0

3a + b = – 27……… (ii)

adding both equation (i) and  (ii)

(2a – b) + (3a + b) = – 8 – 27

2a – b + 3a + b = -35

5a = -35

a = -35/5

a = -7

putting a = -7 in equation (i) to find  ‘b’.

2a – b = – 8

2(-7) – b = -8

-14 – b = -8

b = – 14 + 8

b = -6

value of a = -7 and b = -6

Question -5 When x3 – 3x2 + 5x – 7 is divided by x – 2, then the remainder is

(a) 0 (b) 1 (c) 2 (d) – 1

Ans :- (d) – 1

Hint

f(x) = x3 – 3x2 + 5x – 7
g(x) = x – 2, if x – 2 = 0, then x = 2
Remainder will be
∴ f(2) = (2)3 – 3(2)3 + 5 x 2 – 7
= 8 – 12 + 10 – 7
= 18 – 19
=  –1
∴ Remainder = –1.   option (d)

Question -6 If on dividing 4x2 – 3kx + 5 by x + 2, the remainder is – 3 then the value of k is

(a) 4 (b) – 4 (c) 3 (d) – 3

Ans :- (b) – 4

Hint

f(x) = 4x2 – 3kx + 5
g(x) = x + 2
Remainder = – 3
Let x + 2 = 0, then x = – 2
Now remainder will be
f(–2) = 4(–2)2 – 3k(–2) + 5
= 16 + 6k + 5
= 21 + 6k
∴ 21 + 6k = –3
⇒  6k = –3 – 21
= –24

⇒ k = -24/6 =–4
∴ k = –4.

Question -7 If on dividing 2x3 + 6x2 – (2k – 7)x + 5 by x + 3, the remainder is k – 1 then the value of k is

(a) 2 (b) – 2(c) – 3 (d) 3

Ans :- (d) 3

Hint

f(x) = 2x3 + 6x2 – (2k – 7)x + 5
g(x) = x + 3
Remainder = k – 1
If x + 3 = 0,
then x = –3
∴ Remainder will be
f(–3) = 2(–3)2 + 6(–3)2 – (2k – 7)(–3) + 5
= –54 + 54 + 3(2k – 7) + 5
= –54 + 54 + 6k – 21 + 5
= 6k – 16
∴ 6k – 16 = k – 1
6k – k = –1 + 16
⇒ 5k – 15
k = 15/5 = 3
∴ k = 3.

option (d) correct

Question-8 If x + 1 is a factor of 3x3 + kx2 + 7x + 4, then the value of k is

(a) p = 3 and q = – 2

(b) p = 3 and q = 2

(c) p = – 3 and q = 2

(d) p = – 3 and q = – 2

Ans :- (c) p = – 3 and q = 2

Hint

f(x) = 3x3 + kx2 + 7x + 4
g(x) = x + 1
Remainder = 0
Let x + 1 = 0,
then x = – 1
f(– 1) = 3(– 1)3 + k(– 1)2 + 7(– 1) + 4
= – 3 + k – 7 + 4
= k – 6
∴ Remainder = 0
∴ k – 6 = 0
⇒ k = 6.
option  (c) is correct

Question -9  If (2x + 1) is a factor of both the expressions 2x2 – 5x + p and 2x2 + 5x + q, find the value of p and q.

(a) –1 (b) 0 (c) 6 (d) 10

Ans :- (c) 6

Hint

et  2x + 1 = 0

Then, 2x = -1

x = -½

Given, p(x) = 2x2 – 5x + p

substitute the value of x in p(x),

p (-½) = 2 (-½)2 – 5(-½) + p

= 2(1/4) + 5/2 + p

= ½ + 5/2 + p

= 6/2 + p

= 3 + p

given that, (2x + 1) is a factor of both the expressions 2x2 – 5x + p

remainder is 0.

Then, 3 + p = 0

p = – 3

Now = 2x2 + 5x + q

Substitute the value of x in q(x)

q (-½) = 2 (-½)2 + 5(-½) + q

= 2(1/4) – 5/2 + q

= ½ – 5/2 + q

= (1 – 5)/2 + q

= -4/2 + q

= q – 2

given that, (2x + 1) is a factor of both the expressions 2x2 + 5x + q

, remainder is 0

q – 2 = 0

q = 2

Therefore, p = – 3 and q = 2

Question 10 Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6.

(a) x + 1

(b) 2x – 1

(c) x + 2

(d) x – 2

Ans :- (a) x + 1

Hint

Let f(x) = 2x3 + 3x2 – 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

Question 11 What number should be added to 3x3 – 5x2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8?

(a) 46  (b) -46  (c) 3  (d) -3

Ans :- (b) -46

Question 12  The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively.  then the values of b and c will be

(a) b = -8 ,c  = -3

(b) b = 8 ,c  = 3

(c)b = 8 ,c  = -3

(d) b = -8 ,c  = 3

Ans :-   (d) b = -8 , c  = 3

Question 13 Find the value of ‘a’, if (x – a) is a factor of x3 – ax2 + x + 2.

(a)  -2 (b) -3  (c) -4  (d) -5

Ans :- (a)  -2

Hint

Let f(x) = x3 – 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 14  If x – 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.

(a)  a = -5 and b  = 6

(b) a = -5 and b  = – 6

(c) a = 5 and b  = 6

(d) a = 5 and b  = -6

Ans :- (a)  a = -5 and b  = 6 

Hint

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 (i), we get,

a = -5

Substituting the value of a in (ii), we get,

b = 1 – (-5) = 6

–: End of Factorisation MCQs for ICSE Class-10 Maths  :-

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ICSE MCQs for Class-10 Subject-Wise / Chapter-Wise

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1 thought on “Factorisation MCQs for ICSE Class-10 Maths”

  1. In question no 8 the given answer is wrong whereas it is written the answer for p and q however we have yo found the answer for k

    Reply

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