Question

Complex numbers help. *Question attached below* Will give medal and fan :)

Answer 1

So, I've been working for ages on this question, but still somehow is stuck. I've found another of the 3 roots, which is 2+i, along with 2-i which was given. But how do I go about obtaining the next root and solving for p and q? Can I have some assistance with explanation, please?

Answer 2

when the coefficients are real (as in this case) the "roots" come in complex conjugate pairs.
you found the other root 2+i

so this is what you know:

(z - (2+i)) (z - (2 - i)) (z - a) =0

if we multiply out the first two terms we get
z^2 -4z + (4+1) = z^2 -4z + 5

now divide that into the equation they gave you
we know it will go evenly, with no remainder.

Answer 3

Give me a second, let me work on it :)

Answer 4

maybe it would be easier to multiply out
(z - (2+i)) (z - (2 - i)) (z - a) =0
( z^2 -4z + 5)(z-a) = 0
and then match the cubic with what they gave you.

Answer 5

I ended up with 13/5 as my third root after matching them with the cubic equation?

Answer 6

yes

Answer 7

Oh wow! :O :D And I ended up with -7/5 for p and -27/5 for q. Are those correct?

Answer 8

yes

Answer 9

After equating coefficients of like terms, that is. (Just like I did to find the third root)

Answer 10

you can check by putting this into wolfram
http://www.wolframalpha.com

roots z^3 - (7/5)z^2 -(27/5) z + 13 =0

Answer 11

Alrighty! Thank you so much! :)

Answer 12

It said that the root was -13/5. I got 13/5 :S @phi

Answer 13

when you match up
( z^2 -4z + 5)(z-a) = 0
z^3 + (-4 -a)z^2 + (4a+5) z -5a= 0
z^3 + p z^2 + q z + 13=0

match -5a with 13: -5a= 13, a= -13/5

you found the "a" that matches
(x -a) = 0
x - -13/5 =0
x + 13/5 =0
x = -13/5 is the root.

Answer 14

Ooh, I see my mistake. I put (z+a) instead of (z-a). Thank you!