Question

For what values of $m$ will the expression $y^2+2xy+2x+my-3$ can be resolved in two linear factors.

• A. $-1$
• B. $-2$
• C. $-3$
• D. $-4$

Answer 1

The correct option is B $-2$

$y^2+2xy+2x+my-3$ can be resolved into two linear factors

$\therefore △={(2x+m)}^2-4(2x-3)\ge 0$

$4x^2+4mx+m^2-8x+12\ge 0$

$4x^2+(4m-8)x+(m^2+12)\ge 0$

$△$ must be perfect square in order to resolve given expression into product of two linear factors.

$\therefore 4x^2+2(m-2)\left(2x\right)+m^2+12$ must be perfect square

$\therefore \therefore {(m-2)}^2=m^2+12$

$-4m+4=12$

$\therefore m=-2$