Precalc factoring $x^2+10x+25-9y^2$

Question

Factor $$x^2+10x+25-9y^2$$

The solution is

$$(x+5-3y)(x+5+3y)$$

I understand how to factor when there is only one variable $x$ but I am not sure how to complete this problem with the additional variable $y$.

Answer 1

Consider $x$ only: $x^2+10x+25=(x+5)^2$, then you can use it with the whole expression:

$$x^2+10x+25 - 9y^2=(x+5)^2-(3y)^2=(x+5-3y)(x+5+3y)$$

Because $a^2-b^2=(a-b)(a+b)$, you only need to substitute $a=x+5$ and $b=3y$.

Answer 2

Note that $$a^2-b^2=(a-b)(a+b)$$ Now put that $$a=x+5$$ $$b=3y$$

Answer 3

The given equation can be treated like quadratic in y assuming x to be constant.

Then y=$\frac{\pm \sqrt{36(x^2+10x+25)}}{18}$ $$3y=(x+5) \,\,or\,\,3y=-(x+5)$$