Let Q(x) = x^2+bx+c where b and c are integers. If Q\left((1+\sqrt 2)^8\right)=0 determine b+c .

(1  +   √2)^8  evaluates to  577 + 408√2

So.....this is a root....and so is the conjugate, 577 - 408√2

And the sum of the roots  = -b

So

577  +  577  =  - b  ⇒  1154  ⇒  b  =  -1154

And the product of the roots  = c

So

(  577 + 408√2)  (  577 - 408√2)   =  c   =   ( 332929  - 332928)  =   1

So

b  +  c  =   -1154  +   1   =   -1153