Answer:
another three roots are 1-2i, 3i ,-3i
Step-by-step explanation:
[tex]2x^3+14x^2-18x+45 = 0[/tex]
has one root = 1+2i
then another root will be = 1-2i [ since complex roots always occurs in pair]
therefore together both roots makes function as
[tex](x-1+2i) (x-1-2i) =0 \\(x-1)^2=-4 \\(x^2-2x+1=-4\\(x^2-2x+5)=0[/tex]
since this polynomial is formed by its roots so it must divide the parent polynomial therefore on dividing the parent polynomial by obtained polynomial, we get
[tex]x^2+9 = 0 \\x^2 = -9 \\x = +3i \\x =-3i[/tex]
therefore we have three other roots are
1-2i
-3i
3i
