Question

Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1?
f(x) = (x – 2i)(x – 3i)
f(x) = (x + 2i)(x + 3i)
f(x) = (x – 2)(x – 3)(x – 2i)(x – 3i)
f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)

Answer 1

I think it is A
It fulfills all the conditions

Answer 2

B and D are eliminated for sure

Answer 3

i have more for u if u wish to help

Answer 4

I can try

Answer 5

hold on a sec

Answer 6

@RowanMartinFan Do you know how MaimiGirl got the answer to your first question?

Answer 7

yeah srry
forgot to explain

Answer 8

do u know??

Answer 9

and i dont know this one too

Answer 10

yea i do i found that i have to accually read the problem too lol

Answer 11

Well you need to find the roots of each function, and that is done by setting f(x) to be 0 and solve for x, so for the first one it would be
f(x) = (x – 2i)(x – 3i) which can be broken up into:
0 = (x – 2i) => x = 2i
0 = (x – 3i) => x = 3i
and a multiplicity of one makes A the only answer.

Answer 12

oh DUA i completly forgot lol

Answer 13

OM TOO MUCH MATH

Answer 14

So for your next problem do you know where to start?

Answer 15

oh snap hold on