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# RD Chapter 5- Factorisation of Algebraic Expressions Ex-MCQS Interview Questions Answers

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Question 1 : The factors of x3 – x2y-xy2 + y3 are
(a) (x + y) (x2 -xy + y2)
(b) (x+y)(x2 + xy + y2)
(c) (x + y)2 (x – y)
(d) (x – y)2 (x + y)

x3 – x2y – xy2 + y3
= x3 + y3 –x2y – xy2
= (x + y) (x2 -xy + y2)- xy(x + y)
= (x + y) (x2 – xy + y2 – xy)
= (x + y) (x2 – 2xy + y2)
= (x + y) (x – y)2         (d)

Question 2 : The factors of x3 – 1 +y3 +3xy are
(a) (x – 1 + y)  (x2 + 1 + y2 +x + y – xy)
(b) (x + y + 1)  (x2 + y2 +1- xy – x – y)
(c) (x – 1 + y)   (x2 – 1– y+ x + y + xy)
(d) 3(x + y – 1) (x2 + y2 –1)

x3 – 1 + y3 + 3xy
= (x)3 + (-1)3 + (y)3 – 3x  x  x (-1) x y
= (x – 1 + y) (x2 + 1 + y2 + x + y – xy)
= (x- 1 + y) (x2+ 1 + y2 + x + y – xy)     (a)

Question 3 : The factors of 8a3 + b3 –6ab + 1 are
(a) (2a + b – 1) (4a2 + b2 +1 – 3ab – 2a)
(b) (2a – b + 1) (4a2 + b2 –4ab + 1 – 2a + b)
(c) (2a + b+1) (4a2 + b2 +1 – 2ab – b – 2a)
(d) (2a – 1 + b)(4a2 + 1 – 4a – b –2ab)

8a3 + b3 – 6ab + 1
= (2a)3 + (b)3 + (1)3 – 3 x 2ax b x 1
= (2a + b + 1) [(2a)2 + b2+1-2a x b- b x 1 – 1 x 2a]
= (2a + b + 1) (4a2 + b2+1-2ab-b-2a)            (c)

Question 4 : (x + y)3 – (x – v)3 canbe factorized as
(a) 2y (3x2 + y2)
(b) 2x (3x2 + y2)
(c) 2y (3y2 + x2)
(d) 2x (x2 + 3y2)

(x + y)3 – (x – y)3
= (x + y -x + y) [(x + y)2 + (x +y) (x -y) + (x – y)2]
= 2y(x2 + y2 + 2xy + x2-y2 +x2+y2 – 2xy)
= 2y(3x2 + y2)          (a)

Question 5 : The expression (a – b)3 + (b – c)3 +(c – a)3 can be factorized as
(a) (a -b) (b- c) (c – a)
(b) 3(a – b) (b – c) (c – a)
(c) -3(a – b) (b – c) (a – a)
(d) (a + b + c) (a2 + b2 +c2 – ab – bc – ca)

(a – b)3 + (b – c)3 + (c – a)3
Let a – b = x, b – a = y, c – a = z
x3 +y3 + z3
x+y + z = a- b + b- c + c – a = 0
x3 +y3 +z3 = 3xyz
(a – b)3 + (b – a)3 + (c – a)3
= 3 (a – b) (b – c) (c – a)        (b)

Question 6 :

Question 7 :

Question 8 : The factors of a2 – 1 – 2x – x2 are
(a) (a – x + 1) (a – x –1)
(b) (a + x – 1) (a – x + 1)
(c) (a + x + 1) (a – x –1)
(d) none of these

a2 – 1- 2x – x2
a2 –(1 + 2x + x2)
= (a)2 – (1 + x)2
= (a + 1 + x) (a – 1 – x)                        (c)

Question 9 : The factors of x4 + x2 +25 are
(a) (x2 + 3x + 5) (x2 –3x +5)
(b) (x2 + 3x + 5) (x2 +3x – 5)
(c) (x2 + x + 5) (x2 –x +5)
(d) none of these

x4 + x2 + 25 = x4 +25 +x2
= (x2)2 + (5)2 + 2 x x2 x5- 9x2
= (x2 + 5)2 – (3x)2
= (x2 + 5 + 3x) (x2 + 5 – 3x)
= (x2 + 3x + 5) (x2 – 3x + 5)                (a)

Question 10 : The factors of x2 + 4y2 +4y – 4xy – 2x – 8 are
(a) (x – 2y – 4) (x – 2y +2)
(b)  (x – y  +   2) (x – 4y – 4)
(c) (x + 2y – 4) (x + 2y +2)
(d)    none of these

x2 + 4y2 + 4y – 4xy – 2x – 8
x2 + 4y + 4y – 4xy – 2x – 8
= (x)2 + (2y)2– 2 x x x 2y + 4y-2x-8
= (x – 2y)2 – (2x – 4y) – 8
= (x – 2y)2 – 2 (x – 2y) – 8
Let x – 2y = a, then
a2– 2a – 8 = a2– 4a + 2a – 8
= a(a – 4) + 2(a – 4)
= (a-4) (a + 2)
= (x2 -2y-4) (x2 -2y +2)                       (a)

krishan