Question

The linear factor(s) of the equation $x^2+4xy+4y^2+3x+6y-4=0$ is/are

• A. $x+2y+4=0$
• B. $x+2y-4=0$
• C. $x+2y-1=0$
• D. $x+2y+1=0$

Answer 1

Answer: (A), (C)

$x+2y-1=0$

$x^2+4xy+4y^2+3x+6y-4=0$
Writing equation in terms of $y,$ we get
$4y^2+(4x+6)y+(x^2+3x-4)=0$
Now, using quadratic formula, we have
$y=\frac{-(4x+6)\pm \sqrt{(4x+6{)}^2-4\cdot 4\cdot (x^2+3x-4)}}{2\cdot 4}$
$\Rightarrow y=\frac{-(4x+6)\pm 10}{2\cdot 4}$
$\Rightarrow y=\frac{-(2x+3)\pm 5}{4}\Rightarrow y=\frac{-x+1}{2}\text{or}y=\frac{-x-4}{2}\Rightarrow x+2y-1=0\text{or}x+2y+4=0$



Alternate Solution:
$x^2+4xy+4y^2+3x+6y-4=0\Rightarrow (x^2+4xy+4y^2)+(3x+6y)-4=0\Rightarrow (x+2y{)}^2+3(x+2y)-4=0\Rightarrow ((x+2y)+4)((x+2y)-1)=0\Rightarrow x+2y+4=0\text{or}x+2y-1=0$