Question

The factors of $x^2+4y^2+4y-4xy-2x-8$ are

• A. $(x-2y-4)(x-2y+2)$
• B. $(x-y+2)(x-4y-4)$
• C. $(x+2y-4)(x+2y+2)$
• D. none of these

Answer 1

The correct option is A $(x-2y-4)(x-2y+2)$

$x^2+4y^2+4y-4xy-2x-8$

This can be arrange in the form,

$x^2+4y^2+4y-4xy-2x-8$ $=[x^2-4xy+4y^2]-2(x-2y)-8$

$=\left[\right(x{)}^2-2\times x\times 2y+\left(2y{)}^2\right]-2(x-2y)-2(x-2y)-8$

$=(x-2y{)}^2-2(x-2y)-8$

Let $a=(x-2y)$, then expression becomes,

$x^2+4y^2+4y-4xy-2x-8$ $=a^2-2a-8$

$=a^2-(4-2)a-8$

$=a^2-4a+2a-8$

$=a(a-4)+2(a-4)$

$=(a-4)(a+2)$

Now, re-substitute $a=(x-2y)$

$x^2+4y^2+4y-4xy-2x-8$ $=\left[\right(x-2y)-4]\left[\right(x-2y)+2]$

$\therefore$ $x^2+4y^2+4y-4xy-2x-8$ $=(x-2y-4)(x-2y+2)$