Question

Help Please! Determine all other roots, picture of problem below.

Answer 1

@mathhelp9 @mathstudent55 @Fruitbasket
anyone know how to do this?

Answer 2

If any polynomial with real roots has a complex root, then it also has a root that is the complex conjugate of the other root. See if that helps.

Answer 3

*real coefficients, whoops.

Answer 4

so, +4i?
I really have no idea on this one. :/

Answer 5

Yeah. Since you're given the complex root -4i, then you know 4i is also a root. From this you can tell that (x + 4i) and (x - 4i) are factors of f(x).

Answer 6

okay, that's not all the roots though is there? there's got to be more.

Answer 7

There is one more; do you know how to divide a polynomial by another?

Answer 8

With synthetic division?
I kind of know, I've never done one with nonreal numbers like 4i

Answer 9

Actually, I've just seen a cleaner way than division... if you multiply all the roots together, you'll get the constant in f(x), -48. This gives us:
\[-4i \times 4i \times c = -48\]
Where c is the third root.
I hope this makes sense!

Answer 10

Sorry, I meant the negatives of the roots.

Answer 11

c=3?
meaning the root is -3?

Answer 12

It's actually c = -3, I made a mistake when I wrote that last post. It should have been
\[-4i \times 4i \times -c = -48\]
Which makes the root 3. Sorry for the confusion, I'm a bit tired. :P

Answer 13

that's okay.
so the roots are :
-4i 4i and 3?

Answer 14

Yes indeedy.

Answer 15

thank you so much!