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

Question -1 When x³ – 3x² + 5x – 7 is divided by x – 2,then the remainder is
(a) 0
(b) 1
(c) 2
(d) – 1

Answer -1

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 -2 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

Answer-2

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.,  Option (b) 

Question -3 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

Answer -3

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-4 If x + 1 is a factor of 3x³ + kx² + 7x + 4, then the value of k is
(a) – 1
(b) 0
(c) 6
(d) 10

Answer-4

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

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

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