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Factorisation

Short Answer Type

51.
Factorise:

25a²- 4b² + 28bc - 49c²

We have -4b² + 28bc - 49c² = (-1)[4b² - 28bc + 49c²]

= -1[(2b)² - 2(2b)(7c) + (7c)²]

= - [2b - 7c]²

∴      25a²- 4b² + 28bc - 49c² = 25a² - (2b - 7c)²

Now using              a²-b² = (a + b) (a - b), we have

              [5a]² - [2b - 7c]² = (5a + 2b - 7c)(5a- 2b + 7c)

∴ 25a² -4a² + 28bc - 49c² = (5a + 2b - 7c)(5a - 2b + 7c)

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Previous Year Papers

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Short Answer Type

51. Factorise:25a2 - 4b2 + 28bc - 49c2

51.
Factorise:

25a²- 4b² + 28bc - 49c²