Factoring $x^2 + 8x + 16 - y^2$

Question

Factor$$x^2 + 8x + 16 - y^2 .$$

First approach: \begin{gather*} (x^2 + 8x + 16) – y^2\\ (x + 4)^2 – y^2\\ [(x + 4) + y][(x + 4) – y] \end{gather*}

Second approach where I mess up: \begin{gather*} (x^2 + 8x) + (16 - y^2)\\ x(x + 8) + (4 + y)(4 - y)\\ x + 1(x + 8)(4 + y)(4 - y) \end{gather*}

I think I am breaking one of algebra's golden rules, but I cannot find it.

Answer 1

Let us fix your second approach. $$ \begin{align} &x^2+8x+16-y^2\\ =&x^2+8x+(4-y)(4+y)\\ =&x^2+x\big[(4-y)+(4+y)\big]+(4-y)(4+y)\\ =&(x+4-y)(x+4+y). \end{align} $$