[tex]\bf \textit{difference of squares}
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(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)
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\textit{also recall that }i^2=-1\\\\
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\begin{cases}
x=-1+4i\implies &x+1-4i=0\\
x=-1-4i\implies &x+1+4i=0
\end{cases}
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(x+1-4i)(x+1+4i)=\stackrel{y}{0}
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[(x+1)~~-~~(4i)][(x+1)~~+~~(4i)]=y
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[(x+1)^2~~-~~(4i)^2]=0\implies (x^2+2x+1)~-~(4^2i^2)=y
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(x^2+2x+1)~-~(16\cdot -1)=y\implies (x^2+2x+1)~-~(-16)=y
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x^2+2x+1~~+16=y\implies x^2+2x+17=y[/tex]
