Question

Factorize: $3-a(4+7a)$

• A. $(a+1)(7a-3)$
• B. $(a-1)(3-7a)$
• C. $(a-1)(7a-3)$
• D. $(a+1)(3-7a)$

Answer 1

The correct option is D $(a+1)(3-7a)$

$\Rightarrow (3-a(4+7a)$
$\Rightarrow 3-4a-7a^2$
$\Rightarrow -[7a^2+4a-3]$
$\Rightarrow -[7a^2+7a-3a-3]$
$\Rightarrow -\left[7a\right(a+1)-3(a+1\left)\right]$
$\Rightarrow -(a+1)(7a-3)$
$\Rightarrow (a+1)(3-7a)$