Question

Porfavor ayudaaaa (con resolución)

Answer 1

Respuesta:

82

Explicación paso a paso:

teorema del resto

$x^{2} +2x+2 =0$

entonces  reemplazar para hallar el resto

$R(x)=(x^{2} +2x-1)^{4} -(x+1)^{11} +(2x+2)^{2} +3\\R(x)=(x^{2} +2x+2-2-1)^{4} -(x+1)(x+1)^{10} +(2(x+1))^{2} +3\\R(x)=(-2-1)^{4} -(x+1)((x+1)^{2} )^{5} +2^{2} (x+1)^{2} +3\\R(x)=(-3)^{4} -(x+1)(x^{2} +2x+1 )^{5} +4(x^{2} +2x+1) +3\\R(x)=81 -(x+1)(x^{2} +2x+2-2+1 )^{5} +4(x^{2} +2x+2-2+1) +3\\R(x)=84 -(x+1)(-2+1 )^{5} +4(-2+1)\\R(x)=84 -(x+1)(-1 )^{5} +4(-1)\\R(x)=84 -(x+1)(-1)-4\\R(x)=80+x+1=81+x=ax +b\\entonces:a=1 ;b=81\\pide:a+b=82$