Answers: 1
![\bf \textit{difference of squares} \\\\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b) \\\\\\ \textit{also recall that }i^2=-1\\\\ -------------------------------\\\\ \begin{cases} x=-1+4i\implies &x+1-4i=0\\ x=-1-4i\implies &x+1+4i=0 \end{cases} \\\\\\ (x+1-4i)(x+1+4i)=\stackrel{y}{0} \\\\\\\ [(x+1)~~-~~(4i)][(x+1)~~+~~(4i)]=y \\\\\\\ [(x+1)^2~~-~~(4i)^2]=0\implies (x^2+2x+1)~-~(4^2i^2)=y \\\\\\ (x^2+2x+1)~-~(16\cdot -1)=y\implies (x^2+2x+1)~-~(-16)=y \\\\\\ x^2+2x+1~~+16=y\implies x^2+2x+17=y](/tpl/images/0161/0836/566e0.png)
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