Factorise x^2 - 4x - y^2 - 6y - 5.

Step-by-step solution

1. 1

   Group the x terms and the y terms separately. The mixed term $xy$ is absent, which is the clue that the expression can be turned into a difference of two perfect squares:

   $\left(x^{2}-4x\right)-\left(y^{2}+6y\right)-5$

   Note the minus sign in front of the $y$ group changes the sign of both $y^2$ and $6y$.

2. 2

   Complete the square in x. $x^2-4x=(x-2)^2-4$.

3. 3

   Complete the square in y. $y^2+6y=(y+3)^2-9$, so $-\left(y^{2}+6y\right)=-(y+3)^{2}+9$.

4. 4

   Recombine the constants.

   $(x-2)^{2}-4-(y+3)^{2}+9-5=(x-2)^{2}-(y+3)^{2}$

   The three loose constants $-4+9-5$ cancel exactly, which is the sign that the grouping was the intended one.

5. 5

   Apply the difference of squares. With $A=x-2$ and $B=y+3$:

   $A^{2}-B^{2}=(A-B)(A+B)=\big[(x-2)-(y+3)\big]\big[(x-2)+(y+3)\big]=(x-y-5)(x+y+1)$

   $\boxed{(x-y-5)(x+y+1)}$

6. 6

   Verify by expanding. $(x-y-5)(x+y+1)=x^2+xy+x-xy-y^2-y-5x-5y-5=x^2-4x-y^2-6y-5$. ✓