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Let f(x) = x^3+bx+c  where b and c are integers.

If f(5+\sqrt 3)=0 determine b+c

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waffles  Jan 4, 2018
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If   5 + √3   is a root then so is  5 - √3

 

There is no x^2 term......and the sum of the roots  = 0/a  =  0

 

So

 

 (5 + √3)  + (5 - √3)  + R  = 0   where R is the remaining root

 

10  + R  =  0    ⇒  R   = -10

 

And  ....let  R1, R2  and R3   represent the roots...and we have that

 

R1*R2   +  R1*R3  + R2*R3   =   b  / a   =  b   .....so.....

 

-10(5 + √3) + -10(  5 - √3)  +   (5 + √3) ( 5 - √3)  =  b

 

-50 + -50  + 22   =  b   ⇒   -78

 

And the product of the roots  =  -c   ....so....

 

-10(22)  =  -c

 

220  =  c

 

So   ...b +  c  =  -78   +  220   =  142

 

 

cool cool cool

CPhill  Jan 4, 2018

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