RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7C Solution
EXERCISE 7C
Factorise:
FORMULA:
(i) a2 + 2ab + b2 = (a + b)2
(ii) a2 – 2ab + b2 = (a – b)2
(1) x2 + 8x + 16
= (x)2 + (2 × x × 4) + (4)2
= (x + 4)2
(2) x2 + 14x + 49
= (x)2 + (2 × x × 7) + (7)2
= (x + 7)2
(3) 1 + 2x + x2
= (1)2 + (2 × 1 × x) + (x)2
= (1 + x)2
(4) 9 + 6z + z2
= (3)2 + (2 × 3 × z) + (z)2
= (3 + z)2
(5) x2 + 6ax + 9a2
= (x)2 + (2 × x × 3a) + (3a)2
= (x + 3a)2
(6) 4y2 + 20y + 25
= (2y)2 + (2 × 2y × 5) + (5)2
= (2y + 5)2
(7) 36a2 + 36a + 9
= (6a)2 + (2 × 6a × 3) + (3)2
= (6a + 3)2
(8) 9m2 + 24m + 16
= (3m)2 + (2 × 3m × 4) + (4)2
= (3m + 4)2
(10) 49a2 + 84ab + 36b2
= (7a)2 + (2 × 7a + 6b) + (6b)2
= (7a + 6b)2
(11) p2 – 10p + 25
= (p)2 – (2 × p × 5) + (5)2
= (p – 5)2
(12) 121a2 – 88ab + 16b2
= (11a)2 – (2 × 11a × 4b) + (4b)2
= (11a – 4b)2
(13) 1 – 6x + 9x2
= (1)2 – (2 × 1 × 3x) + (3x)2
= (1 – 3x)2
(14) 9y2 – 12y + 4
= (3y)2 – (2 × 3y × 2) + (2)2
= (3y – 2)2
(15) 16x2 – 24x + 9
= (4x)2 – (2 × 4x × 3) + (3)2
= (4x – 3)2
(16) m2 – 4mn + 4n2
= (m)2 – (2 × m × 2n) + (2n)2
= (m – 2n)2
(17) a2b2 – 6abc + 9c2
= (ab)2 – (2 × ab × 3c) + (3c)2
= (ab – 3c)2
(18) m4 + 2m2n2 + n4
= (m2)2 + (2 × m2 × n2) + (n2)2
= (m2 + n2)2
(19) (l + m)2 – 4lm
= l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
= (l – m)2
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