Ex 12.2, 2 (viii) - Chapter 12 Class 8 Factorisation
Last updated at April 16, 2024 by Teachoo
Factorisation using identities
Factorisation using identities
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Transcript
Ex 12.2, 2 Factorise. (viii) 25𝑎^2 – 4𝑏^2 + 28bc – 49𝑐^225𝑎^2 – 4𝑏^2 + 28bc – 49𝑐^2 = 25𝑎^2 − (4𝑏^2 − 28bc +49𝑐^2) = 25𝑎^2 − (4𝒃^𝟐 + 49𝒄^𝟐 + 28bc) = 25𝑎^2 − ((2b)2 + (7c)2 − 2 × 2b × 7c) = 25𝑎^2 − "(2b − " 〖"7c)" 〗^𝟐 = (5a)2 − "(2b − " 〖"7c)" 〗^𝟐 = (5a + (2b − 7c)) (5a − (2b − 7c)) = (5a + 2b − 7c) (5a − 2b + 7c) Using 𝒙^𝟐 − 𝒚^𝟐 = (𝑥 + y) (𝑥 − y) Here x = 5a and y = 2b − 7c
