Question

Factorize by the grouping method :

$16(a+b{)}^2-4a-4b$

• A. $4(a-3b)(4a+4b-1)$
• B. $4(2a+b)(a+4b-3)$
• C. $4(a-b)(4a-4b-1)$
• D. $4(a+b)(4a+4b-1)$

Answer 1

The correct option is D $4(a+b)(4a+4b-1)$

$16(a+b{)}^2-4a-4b$
$=16(a+b{)}^2-4(a+b)$
$=4(a+b)\left[4\right(a+b)-1]$
$=4(a+b)[4a+4b-1]$