Question

The function \(f(x)=x^4−10x^3+40x^2−80x+64\) has four complex roots, one of which is \(2−2i\). What is the sum of all real and imaginary coefficients of these roots?

Answer 1

if one i 2-2i then one more will be 2+2i

Answer 2

Should I proceed like this? :

\((2-2i)\times z = f(x) \) ?

Answer 3

And then find the value of z by dividing? ... Ok! @shubham.bagrecha

Answer 4

So I can do : \((2-2i)(2+2i)(z) = f(x)\)

Answer 5

other two will also be conjugates

Answer 6

How can we say?

Answer 7

you said that all four are complex and complex roots occur in conjugate forms

Answer 8

Ok.

Answer 9

I am getting another factor as : \(\cfrac{1}{8}\) . (real factor) .

Answer 10

hmn I am wrong somewhere .

Answer 11

let the four roots be :
2-2i ,2+2i ,a+bi ,a-bi
from product of roots taken one at a time , a^2 + b^2 = 8

Answer 12

Yep.

Answer 13

4+2a=10
a=3

(sum of roots)

Answer 14

Ok so I have : 3 + bi and 3 -bi yet as two another factors.

Answer 15

I can calculate \(3^2 + b^2 = 8\)
\(b^2 =1\)
\(b = \pm 1\)

Answer 16

b^2 = -1

Answer 17

Oh yes , sorry : \(b = i\)

Answer 18

so something is wrong , b can't be imaginery

Answer 19

a^2 + b^2 = 8

There is wrong in ^ above.

Answer 20

Well how did you get 8 there?

Answer 21

no it is correct

Answer 22

which means our roots are :
2+2i,2-2i,2,4

Answer 23

That is we don't have more complex roots than 2+2i and 2-2i

Answer 24

yes

Answer 25

So 10 is what I haev

Answer 26

and the other roots are satisfying f(x)

Answer 27

^ have.

Answer 28

Got it. Thanks @shubham.bagrecha . Good work :)

Answer 29

thanks