[tex]\bf \textit{difference of squares}
\\ \quad \\
(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=5+i\implies &x-5-i=0\\
x=5-i\implies &x-5+i=0
\end{cases}
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(x-5-i)(x-5+i)=\stackrel{\textit{original polynomial}}{0}
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[(x-5)~~-~~i]~[(x-5)~~+~~i]=0\implies [(x-5)^2~~-~~i^2]=0
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(x^2-10x+25)~~-~~(-1)=0\implies x^2-10x+25+1=y
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x^2-10x+26=y[/tex]
