Practicing Success
The factors of $x^2+4 y^2+4 y-4 x y-2 x-8$ are: |
$(x-2 y-4)(x-2 y+2)$ $(x^2-2 y-4)(x^2-2 y+2)$ $(x+2 y-4)(x+2 y+2)$ $(x^2-2 y-4)(x^2+2 y+2)$ |
$(x-2 y-4)(x-2 y+2)$ |
= x2+ 4y2 − 4xy + 4y − 2x − 8 Thus A2 − 2A − 8 Re-substitute the value of A |