6. Find the polynomial function with roots 11 and 2i

Question

6.	Find the polynomial function with roots 11 and 2i

anonymous · Answer

one more root exists which is the conjugate of 2i. what is the conjugate of 2i?

marissalovescats · Answer

Well 2i would have 2 pairs because it is imaginary. And we know one of the binomials will be (x-11)

marissalovescats · Answer

So basically (x-11)(x-2i)(x+2i) And what you need to know to solve this is that i^2=-1.

marissalovescats · Answer

So first FOIL (x-2i)(x+2i) and then when you get that distribute (x-11) to that. And remember i^2=-1. Does that make sense @Smartzman?

anonymous · Answer

yes thank you very much :)

marissalovescats · Answer

I have the answer I can check yours when you get it!

anonymous · Answer

ok one moment

anonymous · Answer

x^3-11x^2+4x-44

anonymous · Answer

is what I got

anonymous · Answer

:) or :
(

whpalmer4 · Answer

:-)

Now, I feel obligated to point out that there is not just one polynomial with those roots.  There's an infinite variety of them, obtained by putting a constant multiplier in the mix with your (x-root) factors.  So, either you should multiply by that constant, or harass the teacher about sloppy question-writing :-)