Question

$\text{Factorisation of}-4a^2+4ab-4ca\text{gives:}$

• A. $4a(a+b-c)$
• B. $4a(-a-b+c)$
• C. $4a(-a+b-c)$
• D. $4a(a-b-c)$

Answer 1

The correct option is C $4a(-a+b-c)$

The irreducible factor form of each of the terms is,
$4a^2=2\times 2\times a\times a4ab=2\times 2\times a\times b4ca=2\times 2\times c\times a\text{The common factors are 2, 2 and}a.\therefore -4a^2+4ab-4ca=-(2\times 2\times a\times a)+(2\times 2\times a\times b)-(2\times 2\times c\times a)=2\times 2\times a[-(a)+b-c]=4a(-a+b-c)$