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RD Chapter 5 Factorisation of Algebraic Expressions Ex 5.3 Solutions

Question - 1 : - Factorize:
64a3 + 125b3 + 240a2b + 300ab2

Answer - 1 : -

64a3┬а+ 125b3┬а+ 240a2b+ 300ab2
= (4a)3┬а+ (5b)3┬а+ 3 x (4a)2┬аx5b + 3(4a) + (5b)2
= (4a + 5b)3
= (4a + 5b) (4a + 5b) (4a + 5b)

Question - 2 : - 125x3 тАУ 27y3 тАУ 225x2y + 135xy2

Answer - 2 : -

125x3┬атАУ 27y3┬атАУ 225x2y+ 135xy2
= (5x)3┬атАУ (3y)3┬атАУ 3 x (5x)2┬аx(3y) + 3- x 5x x (3y)2
= (5x тАУ┬а3y)3
=┬а
(5x┬атАУ 3y) (5x тАУ┬а3y) (5x тАУ┬а3y)

Question - 3 : -

Answer - 3 : -


Question - 4 : - 8x3 + 27y3 + 36x2y + 54xy2

Answer - 4 : -

8x3┬а+ 27y3┬а+ 16x2y +54xy2
= (2x)3┬а+ (3y)3┬а+ 3 x (2x)2┬аx3y┬а +┬а 3 x 2x x (3y)2
= (2x + 3y)3
= (2x + 3y) (2x + 3y) (2x + 3y)

Question - 5 : - a3 тАУ 3a2b + 3ab2 тАУ b3 + 8

Answer - 5 : -

a3┬атАУ 3a2b + 3ab2┬атАУ b3┬а+8
= (a тАУ b)3┬а+ (2)3
= (a тАУ b + 2) [(a -b)2тАУ (a тАУ b) x 2 + (2)2]
= (a- b + 2) (a2┬а+ b2┬а-2ab тАУ 2a + 2b + 4)

Question - 6 : - x3 + 8y3 + 6x2y + 12xy2

Answer - 6 : -

x3┬а+ 8y3┬а+ 6x2y +12xy2
= (x)3┬а+ (2y)3┬а+ 3 x x2x 2y + 3 x xx (2y)2
= (x + 2y)3
= (x + 2y) (x + 2y) (x + 2y)

Question - 7 : - 8x3┬а+ y3┬а+ 12x2y+ 6xy2

Answer - 7 : -

8x3┬а+ y3┬а+ 12x2y +6xy2
= (2x)3┬а+ (y)3┬а+ 3 x (2x)2┬аxy + 3 x 2x x y2
= (2x + y)3
= (2x + y) (2x + y) (2x + y)

Question - 8 : - 8a3┬а+ 27b3┬а+ 36a2b+ 54ab2

Answer - 8 : -

8a3┬а+ 27b3┬а+ 16a2b +54ab2
= (2a)3┬а+ (3b)3┬а+ 3 x (2a)2┬аx3b + 3 x 2a x (3b)2
= (2a + 3b)3
= (2a + 3b) (2a + 3b) (2a + 3b)

Question - 9 : - 8a3┬атАУ 27b3┬атАУ 36a2b+ 54ab2

Answer - 9 : -

8a3┬атАУ 27b3┬атАУ 36a2b +54ab2
= (2a)3┬атАУ (3b)3┬атАУ 3 x (2a)2┬аx3b + 3 x 2a x (3b)2
= (2a тАУ 3b))3
= (2a тАУ 3b) (2a тАУ 3b) (2a тАУ 3b)

Question - 10 : - x3 тАУ 12x(x тАУ 4) тАУ 64

Answer - 10 : -

x3┬атАУ 12x(x тАУ 4) тАУ 64
= x3┬атАУ 12x2┬а+ 48x тАУ 64
= (x)3┬атАУ 3 x x2┬аx 4 + 3 x x┬аx (4)2тАУ┬а(4)3
= (x тАУ 4)3
= (x тАУ 4) (x тАУ 4) (x тАУ 4)

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