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Ex 14.1, 2 (viii) - Chapter 14 Class 8 Factorisation
Last updated at March 31, 2023 by Teachoo
Ex 14.1
Ex 14.1, 1 (i)
Ex 14.1, 1 (ii)
Ex 14.1, 1 (iii) Important
Ex 14.1, 1 (iv) Important
Ex 14.1, 1 (v)
Ex 14.1, 1 (vi) Important
Ex 14.1, 1 (vii)
Ex 14.1, 1 (viii) Important
Ex 14.1, 2 (i)
Ex 14.1, 2 (ii) Important
Ex 14.1, 2 (iii)
Ex 14.1, 2 (iv) Important
Ex 14.1, 2 (v)
Ex 14.1, 2 (vi)
Ex 14.1, 2 (vii)
Ex 14.1, 2 (viii) Important You are here
Ex 14.1, 2 (ix)
Ex 14.1, 2 (x) Important
Ex 14.1, 3 (i)
Ex 14.1, 3 (ii) Important
Ex 14.1, 3 (iii)
Ex 14.1, 3 (iv) Important
Ex 14.1, 3 (v) Important
Transcript
Ex 14.1, 2 (Method 1) Factorise the following expressions. (viii) – 4𝑎^2 + 4 ab – 4 ca Now, – 4𝑎^2 = −4 × 𝑎^2 = −1 × 4 × 𝑎^2 = −1 × 2 × 2 × 𝑎^2 = −1 × 2 × 2 × a × a 4ab = 4 × a × b = 2 × 2 × a × b 4ca = 4 × c × a = 2 × 2 × c × a So, 2, 2 and a are the common factors. – 4𝑎^2 + 4ab −4ca = (−1 × 2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a) Taking 2 × 2 × a common, = 2 × 2 × a × ((−1 × a) + b − c) = 4a (−a + b − c) Ex 14.1, 2 (Method 2) Factorise the following expressions. (viii) – 4𝑎^2 + 4 ab – 4 ca – 4𝑎^2 + 4 ab – 4 ca Taking 4 common, = 4 (–𝑎^2 + ab − ca) = 4 ((−a × a) + (a × b) − (c × a)) Taking a common, = 4a (−a + b − c)
