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RD Chapter 5 Factorisation of Algebraic Expressions Ex MCQS Solutions

Question - 11 : - The factors of x3 – 7x + 6 are
(a) x(x – 6) (x –1)                                           
(b) (x2 – 6) (x – 1)
(c) (x + 1) (x + 2) (x –3)                               
(d) (x – 1) (x + 3) (x – 2)

Answer - 11 : -

x3 -7x + 6= x3-1-7x + 7
= (x – 1) (x2 + x + 1) – 7(x – 1)
= (x – 1) (x2 + x + 1 – 7)
= (x – 1) (x2 + x – 6)
= (x – 1) [x2 + 3x – 2x – 6]
= (x – 1) [x(x + 3) – 2(x + 3)]
= (x – 1) (x+ 3) (x – 2)                          (d)

Question - 12 : - The expression x4 + 4 can befactorized as
(a) (x2 + 2x + 2) (x2 –2x +2)                       
(b) (x2 + 2x + 2) (x2 +2x – 2)
(c) (x2 – 2x – 2) (x2 –2x +2)                         
(d) (x2 + 2) (x2 –2)

Answer - 12 : -

x4 + 4 = x4 + 4 + 4x2 –4x2                (Addingand subtracting 4x2)
= (x2)2 + (2)2 + 2 x x2 x2 – (2x)2
= (x2 + 2)2 – (2x)2
= (x2 + 2 + 2x) (x2 + 2 – 2x)               {∵ a2 – b2 =(a + b) (a – b)}
= (x2 + 2x + 2) (x2 – 2x + 2)                 (a)

Question - 13 : - If 3x = a + b + c, then the value of (x – a)3 +(x –    bf + (x – cf – 3(x – a) (x – b) (x – c) is
(a) a + b +c                                                
(b) (a – b) {b – c) (c – a)
(c)0                                                                  
(d) none of these

Answer - 13 : -

3x = a + b +c                                                                     .
⇒3x-a-b-c = 0
Now, (x – a)3+ (x – b)3 + (x – c)3 –3(x – a) (x -b)  (x – c)
= {(x – a) + (x – b) + (x – c)} {(x – a)2 + (x – b)2 +(x – c)2  – (x – a) (x – b) (x – b) (x – c) – (x – c) (x – a)}
= (x – a + x – b + x – c) {(x – a)2 + (x – b)2  +(x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
= (3x – a – b -c) {(x – a)2 + (x -b)2+ (x – c)2 –(x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
But 3x-a-b-c = 0, then
= 0 x {(x – a)2 + (x – b)2 + (x – c)2 –(x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
= 0                                                       (c)

Question - 14 : - If (x + y)3 – (x – y)3 –6y(x2 – y2) = ky2, then k =
(a)1                                   
(b)2                                
(c)4                                     
(d) 8

Answer - 14 : -

(x + y)3 – (x – y)3 – 6y(x2 –y2) = ky2
LHS = (x + y)3 – (x – y)3 – 3 x (x + y)(x – y) [x + y – x + y]
= (x+y-x + y)3       {∵ a3 – b3 –3ab (a – b) = a3 – b3}
= (2y)3 = 8y3
Comparing with ky3, k = 8                    (d)

Question - 15 : - If x3 – 3x2 + 3x – 7= (x + 1) (ax2 + bx + c), then a + b + c =
(a)4                                   
(b)12                             
(c)-10                                 
(d) 3

Answer - 15 : -

x3 – 3x2 + 3x + 7 = (x + 1) (ax2 +bx + c)
= ax3 + bx2 + cx + ax2 + bx + c
x3 – 3x2 + 3x – 7 = ax3 + (b +a)2 + (c + b)x + c
Comparing the coefficient,
a = 1
b + a = -3 ⇒ b+1=-3 ⇒ b = -3-1=-4
c + b = 3 ⇒ c-4 = 3 ⇒ c= 3 + 4 = 7
a + b + c = 1- 4 + 7 = 8- 4 = 4             (a)

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