Respuesta :

Tucon

 

[tex]\displaystyle\bf\\x^2-4x-21+10y-y^2=\\=x^2-y^2 -7x+3x+7y+3y -21=\\\\=(x-y)(x+y) -7x+7y +3x+3y-3\cdot7=\\\\=\boxed{\bf(x+y-7) (x-y+3)}[/tex]

 

Answer:

(x - y + 3)(x + y - 7)

Step-by-step explanation:

x^2 - 4x - 21 + 10y - y^2

= x^2 - y^2 +10y - 4x - 21

= (x - y)(x + y) + 10y - 4x - 21

This is a stab in the dark - I think this might factor to (x - y + 3)(x + y - 7)  or

(x - y - 7)(x + y + 3) - I think I've come across one like this before.

Lets try the first one:

(x - y + 3)(x + y - 7) = x^2 + xy - 7x - y^2 - xy +3y + 3x + 7y - 21

= x^2 -7x + 3x -21 + 10y  - y^2.

This is correct but isn't a very logical way to do this. There must be a better way!

The logical way is  by reducing the expression to the difference of 2 squares: (x - 2)^2 - (y - 5)^2

This factors to ( x + y -7)(x - y + 3).