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Sagot :
keeping in mind that complex factors never come alone, their sister, its conjugate, is always along with them, so if the polynomial has a root/solution of 1 - 12i, then 1 + 12i must also be there, and the same is true for -8-i, her sister -8 + i is along, so let's check the factors by setting them to 0.
[tex]\begin{cases} x = 6\implies &x-6=0\\ x=1-12i\implies &x-1+12i=0\\ x=1+12i\implies &x-1-12i=0\\ x=-8-i\implies &x+8+i=0\\ x-8+i\implies &x+8-i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ x-6=0\implies \stackrel{factor}{(x-6)} \\\\[-0.35em] ~\dotfill\\\\ (x-1+12i)(x-1-12i)=0\implies \underset{\textit{difference of squaeres}}{[(x-1)+12i][(x-1)-12i]=0}[/tex]
[tex][(x-1)^2-(12i)^2]=0\implies [(x^2-2x+1)-(12^2i^2)]=0 \\\\\\ (x^2-2x+1)-(144(-1))=0\implies x^2-2x+1+144=0 \\\\\\ x^2-2x+145=0\implies \stackrel{factor}{(x^2-2x+145)} \\\\[-0.35em] ~\dotfill\\\\ (x+8+i)(x+8-i)=0\implies \underset{\textit{difference of squares}}{[(x+8)+i][(x+8)-i]=0} \\\\\\\ [(x+8)^2-(i)^2]=0\implies (x^2+16x+64)-(i^2)=0 \\\\\\ x^2+16x+64-(-1)=0\implies \implies x^2+16x+65=0\implies \stackrel{factor}{x^2+16x+65}[/tex]
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