Q2 Factorise the following expressions 

(VIII)

- 4 a ^2 + 4 ab - 4ca

Answers (1)

We have, 
            -4a^2 = -1\times 2 \times 2 \times a\times a
           4ab = 2 \times 2 \times a\times b
            4ca = 2 \times 2 \times c\times a
\therefore      on factorization we get,
-4a^2+4ab-4ca = (-1 \times 2 \times 2 \times a\times a )+( 2 \times 2 \times a\times b )- (2 \times 2 \times c\times a)                                     

 =(2 \times 2 \times a) (-1 \times a + b - c)
  = 4a(-a+b-c)

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