Polynomial Functions again... Imaginary Numbers...

[Guest]

15b) Find the values of the function. Remember: i= sqrt(-1). Simplify.

h(2i)=2(2i)^2-5(2i)+6


I got two answers for this question, and neither are right. Answer 1) -2i+12. Answer 2) 3i+2. Correct Answer) -2-10i.

ANSWER 1)

h(2i)=2(2i)^2-5(2i)+6
h(2i)=2(2i)(2i)-5(2i)+6
h(2i)=2(4+2i+2i+i^2)-5(2i)+6
h(2i)=8+8i+2i^2-10i+6
h(2i)=2i^2+2i+14
h(2i)=-2+2i+14
h(2i)=-2i+12



ANSWER 2)

h(2i)=2(2i)^2-5(2i)+6
h(2i)=2(2i)(2i)-5(2i)+6
h(2i)=2(i^2+4i+4)-5i-10+6
h(2i)=2i^2+8i-5i+4
h(2i)=2i^2+3i+4
h(2i)=-2+3i+4
h(2i)=3i+2


HELP ME PLEASE!!! Imaginary numbers really kill me. Any advice in general too???

 

[stapel]

You seem to be converting "two times i" into "two plus i". Don't.

. . . . .(2i)² = (2i)(2i) = (2×2)(i×i) = 4i² = -4

See if that helps.

Eliz.

 

[Guest]

Hmm... I found the work I did on Thurdsay with this one, in which I have to right answer. I must be over-thinking the situation... That's not uncommon for me...



h(2i)=2(2i)^2-5(2i)+6
h(2i)=4i^2-10i+6
h(2i)=-4-10i+6
h(2i)=-10i-2



But I don't know how I got the answer... Hmm... back to the binder...

 

[stapel]

You dropped a factor:

. . . . .2(2i)² = 2(-4) = -8

Eliz.