Factorise $- 4 a^{2} + 4 a b - 4 c a$

- a (a - b + c)

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- 4 a (a - b + c)

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- 4 (2 a - b + c)

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- 4 a (2 a - b + 2 c)

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Solution

The correct option is D $- 4 a (a - b + c)$
We have, $4 a^{2} = 2 \times 2 \times a \times a,$
$4 a b = 2 \times 2 \times a \times b$
and, $4 c a = 2 \times 2 \times c \times a$
The three terms have 2, 2 and a as common facotrs
$- 4 a^{2} + 4 a b - 4 c a = - (2 \times 2 \times a \times a) + (2 \times 2 \times a \times b) - (2 \times 2 \times c \times a)$
$= 2 \times 2 \times a \times (- a + b - c)$
$= 4 a (- a + b - c)$
$= - 4 a (a - b + c)$

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Factorisation of - 4 a^{2} + 4 a b - 4 c a gives:

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