Question

A polynomial p(z)= z^5+az^4+bz^3+cz^2+dz+e
with real coefficients a,b,c,d,e has -1,2, and 1-i as roots. find it's fifth root and the coefficient e.

Answer 1

it's just to confirm if i got the right answer which is =
x^4-3x^3+2x^2+6x-4

Answer 2

given polynomial is of degree 5 right ?

Answer 3

yes

Answer 4

can you take a screenshot of original question and attach
i feel some info is missing..

Answer 5

sure hold on

Answer 6

question 8

Answer 7

Okay do you notice you have missed one root during copy pasting!

Answer 8

You're given four roots right ?

Answer 9

\[\color{red}{1},~-1,~~2,~~1-i\]

Answer 10

ohh yes

Answer 11

thanks i missed that one

Answer 12

1+i is also a root as the complex roots of a polynomial with real coefficients always come in conjugate pairs

Answer 13

so your five roots are
\[\color{red}{1},~-1,~~2,~~1-i,~~1+i\]

Answer 14

so it would beasically look like this (x-1)(x+1)(x-2)[x-(1-i)] [x-(1+i)]

Answer 15

basically*

Answer 16

yes but why do you want to know how it looks ?

Answer 17

just confirmation I was doing the right step lol

Answer 18

thats right! but you dont need to construct the polynomial to find the value of e

Answer 19

http://en.wikipedia.org/wiki/Vieta's_formulas

Answer 20

e = -(product of roots)

Answer 21

\[e = - (1(-1)(2)(1-i)(1+i)) = - (-4) = 4\]

Answer 22

oh wow, that's much faster, i'll take a look into that

Answer 23

you could also get that by constructing the polynomial

Answer 24

maybe expand out and see if you really get e = 4 or not :)

Answer 25

you should get the constant term = 4 by expanding below
\[ (x-1)(x+1)(x-2)[x-(1-i)] [x-(1+i)]\]

Answer 26

awesome , you were right i got e=4

Answer 27

thanks, i'll use that other method