RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6

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RD Sharma Solutions for Class 8 Chapter 7 Factorization Ex 7.6 Download PDF

Factorize each of the following algebraic expressions :
Question 1.
4x² + 12xy + 9y²
Solution:
4x² + 12xy + 9y² = (2x)² + 2 x 2x x 3y + (3y)² {∵ a² + 2ab + b² = (a +b)²}= (2x + 3y)²    

Question 2.
9a² – 24ab + 16b²
Solution:
9a² – 24ab + 16b²
= (3a)² – 2 x 3a x 4b + (4b)²     {∵ a² – 2ab + b² = (a – b)²}
= (3a – 4b)²

Question 3.
36a² – 6pqr + 9r²
Solution:
p²q² – 6pqr + 9r²
= (pq)² – 2 x pq x3r + (3r)² {∵ a² – 2ab + b² = (a -b)²}
= (pq-3r)²

Question 4.
36a² + 36a + 9
Solution:
36a² + 36a + 9
= (6a)² + 2 x 6a x 3 + (3)²   {∵ a² + 2ab + b² = (a + b)²
= (6a + 3)²

Question 5.
a² + 2ab + b² – 16
Solution:
a² + 2ab + b² – 16
= (a + b)² – (4)²     {∵ a² + 2ab + b² = (a + b)² and a² – b² = (a + b) (a – b)}
= (a + b + 4) (a + b – 4)

Question 6.
9z² – x² + 4xy – 4y²
Solution:
9z² – x² + 4xy – 4y²    {∵ a² – b² = (a + b) (a – b) and a² – 2ab + b² (a – b)²}
= 9z² – (x² – 4xy + 4y²)
= (3z)² – [(x)² – 2 x x x 2y + (2y)²]
= (3z)²-(x-2y)²
= (3z + x – 2y) (3z – x + 2y)

Question 7.
9a⁴ – 24a²b² + 16b⁴ – 256
Solution:

Question 8.
16 – a⁶ + 4a³b³ – 4b⁶
Solution:

Question 9.
a² – 2ab + b² – c²
Solution:

Question 10.
x² + 2x + 1 – 9y²
Solution:

Question 11.
a² + 4ab + 3b²
Solution:
a² + 4ab + 3b²
= a² + 4ab+ 4b² – b²
= (a)² + 2 x a x 2b + (2b)² – b² (∵ 3b² = 4b² – b²)
= (a + 2b)² – (b)²   {∵ a² – b² = (a +b) (a – b)}
= (a + 2b + b) (a + 2b- b)

Question 12.
96 – 4x-x²
Solution:
96 – 4x – x² = 96 – (4x + x²)
= 96 – [(x)² + 2 x x x 2 + (2)²] + (2)²   (on completing the square)
= 96 + 4 – (x + 2)² = 100 – (x + 2)²
= (10)² – (x + 2)²
= (10 + x + 2) (10 – x- 2)
= (x + 12) (-x + 8)

Question 13.
a⁴ + 3a² + 4
Solution:
a⁴ + 3a² + 4
= (a²)² + (2)² + 2 x a² x 2 – a²   (on completing the square)
= (a² + 2)² – (a)²
= (a² + 2 + a) (a² + 2 – a)
= (a¹ + a + 2) (a² – a + 2)

Question 14.
4a⁴ + 1
Solution:
4x⁴ + 1 = (2a²)² + (1)² + 2 x 2x² x 1 – 2 x 2x² x 1  (completing the square)
= (2x² + 1)² – 4a²
= (2x² + 1)² – (2a)²   {a² – b² = (a + b) (a – b)}
= (2x² + 1 + 2a) (2a² + 1 – 2a)
= (2a² + 2a + 1) (2a² – 2a + 1)

Question 15.
4x⁴+y⁴
Solution:
4a⁴ + y⁴ = (2x²)² + (y²)² + 2 x 2x²y² – 2 x 2x²y²
= (2x² + y²)² – 4x²y²
= (2x² + y²)² – (2xy)²
= (2x² + y² + 2xy) (2x² + y² – 2xy)
= (2x² + 2xy + y²) (2x² – 2xy + y²)

Question 16.
(x+ 2)² – 6 (a + 2) + 9
Solution:
(x + 2)² – 6 (x + 2) + 9
=  (x + 2)² – 2 x (x + 2) x 3 + (3)²
= (x + 2 – 3)²
= (x-1)² = (x-1)(x-1)

Question 17.
25 – p² – q² – 2pq
Solution:

Question 18.
a² + 9y² – 6xy – 25a²
Solution:
a² + 9y² – 6xy – 25a²

Question 19.
49 – a² + 8ab – 16b²
Solution:

Question 20.
a² – 8ab + 16b² – 25c²
Solution:

Question 21.
x² -y²+ 6y- 9
Solution:

Question 22.
25x² – 10x + 1 – 36y²
Solution:

Question 23.
a²-b² + 2bc – c²
Solution:

Question 24.
a² + 2ab + b² -c²
Solution:

Question 25.
49 -x² – y² + 2xy
Solution:

Question 26.
a² + 4b² – 4ab – 4c²
Solution:

Question 27.
x² -y² – 4xz + 4z²
Solution:

All Chapter RD Sharma Solutions For Class 8 Maths

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All Subject NCERT Solutions For Class 8

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