Question

Factorizing the given expression gives:

$-4a^2+4ab-4ca$

• 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 A $4a(-a+b-c)$

$(4a^2=2\times 2\times a\times a$

$4ab=2\times 2\times a\times b$

$4ca=2\times 2\times c\times a$

$\therefore -4a^2+4ab-4ca=-(2\times 2\times a\times a)+(2\times 2\times a\times b)-(2\times 2\times c\times a)$

$=2\times 2\times a[-(a)+b-c]$

$=4a(-a+b-c)$