+0  
 
+1
611
2
avatar+466 

Which has \(1+2\sqrt{3}, 3-\sqrt{2}\) as roots, and such that \(f(0) = -154\)

 Dec 10, 2020
 #1
avatar+127748 
+1

The conjugates will also be  roots

 

So.....the  polynomial is

 

P(x)  =  a [x - ( 1+ 2sqrt (3))]  [ x - (1 - 2sqrt (3)) ]  [ x - (3-sqrt (2))] [ x - ( 3 + sqrt (2)) ]  =

 

(I used WolframAlpha to expand this  )

 

a (x^4 - 8 x^3 + 8 x^2 + 52 x - 77)

 

Since  f(0)  = -154

 

Then

 

a (-77)  = -154

a =  -154/-77  =   2

 

So....the polynomial  is

 

2 [ x^4 - 8 x^3 + 8 x^2 + 52 x - 77 ]  =

 

2x^4  - 16x^3 + 16x^2  + 104x - 154

 

And  we  just need to  add all the  coefficients plus the  constant term to get f(1)   =

 

2 - 16 + 16  + 104  - 154   = 

 

-48

 

 

cool cool cool

 Dec 10, 2020
 #2
avatar+466 
+2

You are correct!!!!!! Thanks for answering my question

 Dec 10, 2020

4 Online Users

avatar
avatar
avatar