Question

Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Answer 1

$\bf \begin{cases} x=1\implies &x-1=0\\ x=1\implies &x-1=0\\ x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\ x=2+i\implies &x-2-i=0\\ x=2-i\implies &x-2+i=0 \end{cases} \\\\\\ (x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0} \\\\\\ (x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)+1] \\\\\\ (x^2-2x+1)(2x+1)~[x^2-4x+5] \\\\\\ (x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use  (x-1)(x-1)(2x+1)(x²-4x+5)  as well.