At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
GIven that (3+2i) is a root of \[2x^3 + px^2 + 20x + q = 0\] and p and q are real numbers, find p and q (finding the other roots is the next part of the question which I could do much more easily, no worries :P)
put x = 3 + 2i, and simplify simplify ... and compare the real and imaginary part of both sides.
oh whoops, thank you so much :D
and for other roots, factor out (x - 3 - 2i) <-- this can be quite troublesome.
see, one of the roots has to be 3-2i, right?
oh ,, yes yes ... right, it's cubic equation.
then we multiply those two, and divide?? idk, lemme work it out and see.
best way is to guess the root by remainder theorem. else you can factor out (x^2 - 6x + 13)
what do you mean by compare the imaginary and real things on both sides??
Just set x = 3+2i for now... and simplify.
I did that. The previous part of the question said find x^2 and x^3 if x = 3+2i. I'm wondering, maybe i've got that wrong
OH MY GOD. I JUST REALIZED
What IS the cube and square of 3+2i?
so instead of multiplying normally, I went and used de moivres theorem, and got some really weird answers
de moivre? You actually used polar forms? -.-
waste of time, and energy, and brain power. UGH
So... I take it you get it now? ^_^
Hey... everything okay?