How do I solve $x^2+10x-y^2+25=A×(x+y+5) \Rightarrow A=?$

Question

$x^2+10x-y^2+25=A×(x+y+5) \Rightarrow A=?$

Answer 1

You can write $x^2+10x+25=(x+5)^2$, so $(x+5)^2-y^2=(x+5+y)(x+5-y)$.

Hence $A=x+5-y$, assuming $x+y \neq -5$.

Answer 2

Assume $A=(ax+by+c)$ then

$$(ax+by+c)(x+y+5)=ax^2+(a+b)xy+by^2+(5a+c)x+(5b+c)y+5c$$

that is

• $a=1$
• $a+b=0\implies b=-1$
• $5a+c=10\implies c=5$

that is $A=x-y+5$.