$\displaystyle \begin{array}{{>{\displaystyle}l}} B( 0) =a.0+b.0+c.0+d=2\Rightarrow d=2\\ B( 1) =a.1^{3} +b.1^{2} +c.1+2=8\ \Rightarrow a+b+c=6\ ( 1)\\ B( -1) =a( -1)^{3} +b( -1)^{2} +c( -1) +2=2\Rightarrow -a+b-c=0\ ( 2)\\ Mà\ a=2c\\ Lấy\ ( 1) +( 2) \ ta\ được:\ 2b=6\Rightarrow b=3\\ Thay\ b=3\ và\ a=2c\ vào\ ( 1) :\ 2c+3+c=6\Rightarrow c=1\Rightarrow a=2\\ Vậy\ a=2,\ b=3,\ c=1,\ d=2 \end{array}$
