**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 2x^{3} + 3x^{2} – kx + 5 by x – 2, leaves a remainder 7.

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

**Ans :- (b) 13**

**Hint**

#### f(2) = (2 × 2^{3}) + (3 × 2^{2}) – (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 6x ^{3} + 5x^{2} + 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) = 6x^{3} + 5x^{2} + 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 6x^{3} + 5x^{2} + 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 2x^{3} – 5x^{2} + 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 2x^{3} – 5x^{2} + 5x

Then, f(x) = 2x^{3} – 5x^{2} + 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 x^{3} + 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) = x^{3} + 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 x^{3} + 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) = x^{3} + 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 x^{3} + 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 x^{3} – 3x^{2} + 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) = x^{3} – 3x^{2} + 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 4x^{2} – 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) = 4x^{2} – 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 2x^{3} + 6x^{2} – (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) = 2x^{3} + 6x^{2} – (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 3x^{3} + kx^{2} + 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) = 3x^{3} + kx^{2} + 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 2x^{2} – 5x + p and 2x^{2} + 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) = 2x^{2} – 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 2x^{2} – 5x + p

remainder is 0.

Then, 3 + p = 0

p = – 3

Now = 2x^{2} + 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 2x^{2} + 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 2x^{3} + 3x^{2} – 5x – 6.

(a) x + 1

(b) 2x – 1

(c) x + 2

(d) x – 2

**Ans :- (a) x + 1**

**Hint**

Let f(x) = 2x^{3} + 3x^{2} – 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 3x^{3} – 5x^{2} + 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 4x^{3} – bx^{2} + 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 x^{3} – ax^{2} + x + 2.

(a) -2 (b) -3 (c) -4 (d) -5

**Ans :- (a) -2**

**Hint**

Let f(x) = x^{3} – ax^{2} + x + 2

It is given that (x – a) is a factor of f(x).

Remainder = f(a) = 0

a^{3} – a^{3} + a + 2 = 0

a + 2 = 0

a = -2

**Question 14 **If x – 2 is a factor of x^{2} + 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) = x^{2} + 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 :-

-: also visit :-

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

ICSE Class-10 Text book Solutions, Notes , Syllabus, Paper, Notes

Please share with your ICSE friends if it is helpful

Thanks

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