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RD Chapter 6- Factorisation of Polynomials Ex-MCQS Interview Questions Answers

Question 1 : If x – 2 is a factor of x2 +3 ax – 2a, then a =
(a) 2
(b) -2
(c)1                          
(d) -1

Answer 1 :

  x – 2 is a factor of
f(x) = x2 + 3 ax – 2a
Remainder = 0
Let x – 2 = 0, then x = 2
Now f(2) = (2)2 + 3a x 2 – 2a
= 4 + 6a – 2a = 4 + 4a
Remainder = 0
 4 + 4a = 0 4a =-4
a= 44 = -1
 a=-1                                            (d)

Question 2 : If x3 + 6x2 +4x + k is exactly divisible by x + 2, then k =
(a)-6                        
(b) -7
(c)-8                        
(d) -10

Answer 2 :

f(x) – x3 + 6x2 + 4x + k isdivisible by x + 2
Remainder = 0
Let x + 2 = 0, then x = -2
 f(-2) = (-2)3 + 6(-2)2 + 4(-2) + k
= -8 + 24-8 + k = 8 + k
 x + 2 is a factor
Remainder =0                                .
 8 + k= 0 k = -8
k =-8                                                (c)

Question 3 : If x – a is a factor of x3 –3x2 a + 2a2x + b, then the value of b is
(a)0                          
(b) 2
(c)1                          
(d) 3

Answer 3 :

 x – a is a factor of x3 – 3x2 a + 2a2x+ b
Let f(x) = x3 – 3x2 a + 2a2x+ b
and x – a = 0, then x = a
f(a) = a3 – 3a2.a + 2a2.a + b
= a3 – 3a3 + 2a3 + b = b
 x – a is a factor of f(x)
  b = 0                                         (a)

Question 4 : If x140 +2x151 + k is divisible by x + 1, then the value of k is
(a) 1                        
(b) -3
(c)2                          
(d) -2

Answer 4 :

x + 1is a factor of f(x) = x140 + 2x151 + k
Remainder will be zero
Let x + 1 = 0, then x = -1
 f(-1) = (-1)140 + 2(-1)151 + k
= 1 + 2 x (-1) + k          {
  140 is even and 151 is odd}
=1-2+k=k-1
Remainder = 0
k –1=0  k=1                                  (a)

Question 5 : If x + 2 is a factor of x2 +mx + 14, then m =
(a)7                          
(b) 2      
(c)9                          
(d) 14

Answer 5 :

x + 2is a factor of(x) = x2 + mx + 14
Let x + 2 = 0, then x = -2
f(-2) = (-2)2 + m{-2) + 14
= 4 – 2m + 14 = 18 – 2m
 x + 2 is a factor of f(x)
 Remainder = 0
 18 – 2m = 0
2m = 18 
 m = 182 =9                       (c)

Question 6 : If x – 3 is a factor of x2 –ax – 15, then a =
(a) -2                         
(b) 5
(c)-5                         
(d) 3

Answer 6 :

x – 3is a factor of(x) = x2 – ax – 15
Let x – 3 = 0, then x = 3
f(3) =(3)2 – a(3) – 15
= 9 -3a- 15
= -6 -3a
x – 3is a factor
 Remainder = 0
-6 – 3a = 0
 3a = -6
  a = 63 = -2            (a)

Question 7 : If x51 + 51is divided by x + 1, the remainder is
(a)0                           
(b) 1
(c)49                         
(d) 50

Answer 7 :

Letf(x) = x51 + 51 is divisible by x + 1
Let x+1=0, then x = -1
 f(-1)= (-1)51 + 51 =-1+51   ( power 51 is an odd integer)
= 50                                               (d)

Question 8 : If x+ 1 is a factor of thepolynomial 2x2 + kx, then k =
(a)-2                           
(b) -3
(c) 4                           
(d)  2

Answer 8 :

x + 1is a factor of the polynomial 2x2 + kx
Let x+1=0, then x = -1
Now f(x) = 2x2 + kx
Remainder =f(-1) =  0
= 2(-1)2 + k(-1)
= 2 x 1 + k x (-1) = 2 – k
x + 1is a factor of f(x)
Remainder = 0
2 – k= 0  k = 2                                 (d)

Question 9 : If x + a is a factor of x4 –a2x2 + 3x – 6a, then a =
(a)0                           
(b) -1
(c)1                           
(d) 2

Answer 9 :

x + a is a factor o f(x) = x4– a2x2 +3x – 6a
Let x + a = 0, then x = -a
Now, f(-a) – (-a)4 -a2(-a)2 + 3 (-a)– 6a
= a4-a4-3a-6a = -9a
x + ais a factor of f(x)
Remainder= 0
-9a =0 a =0                               (a)

Question 10 : The value of k for which x– 1 is a factor of 4x3 + 3x2 – 4x + k, is
(a) 3
(b) 1
(c)-2                          
(d) -3

Answer 10 :

x- 1 is a factor of f(x) = 4x3 + 3x2 –4x + k
Let x – 1 = 0, then x = 1
f(1) = 4(1 )3 + 3(1)2 – 4 x 1 + k
= 4+3-4+k=3+k
x- 1is a factor of f(x)
Remainder = 0
3 + k= 0 k =-3                              (d)


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RD Chapter 6- Factorisation of Polynomials Ex-MCQS Contributors

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