Question

Solve for a polynomial with roots -3,6+i

Answer 1

\[If 6+\iota is aroot then 6-\iota is also aroot.\]
required polynomial is \[=\left( x+3 \right)\left( x-6-\iota \right)\left( x-6+\iota \right)\]
solve it.

Answer 2

How do I solve for it?

Answer 3

\[\left( x-6-\iota \right)\left( x-6+\iota \right)=\left\{ \left( x-6 \right)-\iota \right\}\left\{ \left( x-6 \right)+\iota \right\}\]
\[=\left( x-6 \right)^{2}-\iota ^{2}=x ^{2}-12x+36+1=x ^{2}-12x+37\]
\[\left( x+3 \right)\left( x-6-\iota \right)\left( x-6+\iota \right)=\left(x+3 \right)\left( x ^{2}-12x+37 \right)\]
\[=x ^{3}-12x ^{2}+37x+3x ^{2}-36x+111=x ^{3}-9x ^{2}+x+111\]