Respuesta :
[tex]\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}[/tex]
[tex]\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)+1] \\\\\\ (x^2-2x+1)(2x+1)~[x^2-4x+5] \\\\\\ (x^2-2x+1)(2x+1)(x^2-4x+5)[/tex]
of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.
[tex]\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)+1] \\\\\\ (x^2-2x+1)(2x+1)~[x^2-4x+5] \\\\\\ (x^2-2x+1)(2x+1)(x^2-4x+5)[/tex]
of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.