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RD Chapter 5 Factorisation of Algebraic Expressions Ex 5.1 Solutions

Question - 1 : -
Factorize

x3 + x – 3x2 – 3

Answer - 1 : -

x3 + x – 3x2 – 3
x3 – 3a2 + x – 3
⇒ x2(x – 3) + 1(x – 3)
= (x – 3) (x2 + 1)

Question - 2 : - a(a + b)3 – 3a2b(a + b)

Answer - 2 : -

a(a + b)3 – 3a2b(a + b)
=a(a + b) {(a + b)2 – 3ab}
=a(a + b) {a2 +b2 +2ab – 3ab}
=a{a + b) {a2 –ab + b2)

Question - 3 : - x(x3 – y3) + 3xy(x – y)

Answer - 3 : -

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

Question - 4 : - a2x2 + (ax2 +1)x + a

Answer - 4 : -

a2x2 + (ax2 + 1)x + a
= a2x2 + a + (ax2 + 1)x
= a(ax2 + 1) + x(ax2 + 1)
= (ax2 + 1) (a + x)
= (x + a) (ax2 + 1)

Question - 5 : - x2 + y – xy – x

Answer - 5 : -

x2 + y – xy – x
= x2-x-xy + y = x(x- l)-y(*- 1)
= (x – 1) (x – y)

Question - 6 : - X3 – 2x2y + 3xy2 – 6y3

Answer - 6 : -

x3 –2x2y + 3xy2 – 6y3
=x2(x – 2y) + 3y2(x– 2y)
=(x – 2y) (x2 + 3y2)

Question - 7 : - 6ab – b2 + 12ac – 2bc

Answer - 7 : -

6ab – b2 + 12ac – 2bc
= 6ab + 12ac – b2 – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b)

Question - 8 : - x(x – 2) (x – 4) + 4x – 8

Answer - 8 : -

x(x – 2) (x – 4) + 4x – 8
= x(x – 2) (x – 4) + 4(x – 2)
= (x – 2) [x(x – 4) + 4]
= (x – 2) (x2 – 4x + 4)
= (x – 2) [(x)2 – 2 x x x 2 + (2)2]
= (x – 2) (x – 2)2 = (x – 2)3

Question - 9 : - (a – b + c)2 + (b – c + a)2 + 2(a – b + c) (b – c + a)

Answer - 9 : -

(a– b +c)2 + ( b- c+a)2 +2(a – b +c) (b – c +a)      {∵ a2 +b2 +2ab =(a + b)2}
=[a – b + c + b- c + a]2
= (2a)2 =4a2

Question - 10 : - a2 + 2ab + b2 – c2

Answer - 10 : -

a2 + 2ab + b2 – c2
= (a2 + 2ab + b2) – c2
= (a + b)2 – (c)2         {∵  a2 –b2 = (a + b) (a – b)}
= (a + b + c) (a + b – c)

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