If $x^2+4xy+4y^2+4x+cy+3$ can be written as the product of two linear factors, then the value of ‘c’ is

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$\mathrm{1.4.3}+2\frac{c}{2}\mathrm{.2.2}-1.\frac{c^2}{4}-4.4-3.4$

$=0$

$12+4c-\frac{c^2}{4}-16-12=0$

$4c-\frac{c^2}{4}-16=0$

$c=8$