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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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