Question

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

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

For $a{x}^2+2hxy+b{y}^2+29x+2fy+c=0$ to be written as linear factors

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

$\Rightarrow 1\left(4\right)\left(3\right)+2\left(2\right)\left(2\right)\left(\frac{c}{2}\right)-1{\left(\frac{c}{2}\right)}^2-4(2{)}^2-3(2{)}^2=0.$

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

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

$\Rightarrow c^2-16c+64=0$

$\Rightarrow (c-8{)}^2=0$

$\Rightarrow c=8$