ML Aggarwal Factorisation MCQs Class 10 ICSE Maths Solutions Ch-6. We Provide Step by Step Answer of MCQs Questions for Factorisation as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

### ML Aggarwal Factorisation MCQs Class 10 ICSE Maths Solutions Ch-6

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-6 | Factorisation |

Writer / Book | Understanding |

Topics | Solutions of MCQs |

Academic Session | 2024-2025 |

### ML Aggarwal Ch-6 **Factorisation MCQs** Class 10 ICSE Maths Solutions

Choose the correct answer from the given four options (1 to 4) :

**Question -1 ****When x**^{3} – 3x^{2} + 5x – 7 is divided by x – 2,then the remainder is

**(a) 0**

**(b) 1**

**(c) 2**

**(d) – 1**

^{3}– 3x

^{2}+ 5x – 7 is divided by x – 2,then the remainder is

**Answer -1**

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 -2 ****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**

^{2}– 3kx + 5 by x + 2, the remainder is – 3 then the value of k is

**Answer-2**

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., **Option (b) **

**Question -3 ****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**

^{3}+ 6x

^{2}– (2k – 7)x + 5 by x + 3, the remainder is k – 1 then the value of k is

**Answer -3**

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-4 ****If x + 1 is a factor of 3x**^{3} + kx^{2} + 7x + 4, then the value of k is

**(a) – 1**

**(b) 0**

**(c) 6**

**(d) 10**

^{3}+ kx

^{2}+ 7x + 4, then the value of k is

**Answer-4**

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

— : End of ML Aggarwal Factorisation MCQs Class 10 ICSE Maths Solutions Ch-6 : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

Thanks

Please Share with Your Friends