Question

因式分解t³+t²+t+3

要详细过程，给好评哦

Answer 1

因式分解的过程，同等于求解t^3+t^2+t+3=0的根
令x=t+1/3 t=x-1/3，代入方程
(x-1/3)^3+(x-1/3)^2+(x-1/3)+3=0
x^3-x^2+x/3-1/27+x^2-2x/3+1/9+x-1/3+3=0
x^3+2x/3+74/27=0
△=(37/27)^2+(2/9)^3=17/9>0
所以方程有且仅有一个实根
x=[-37/27+√△]^(1/3)+[-37/27-√△]^(1/3)
=[-37/27+√17/3]^(1/3)+[-37/27-√17/3]^(1/3)
=(1/3)*[-37+9√17]^(1/3)+(1/3)*[-37-9√17]^(1/3)
所以t=x-1/3=(1/3)*[(9√17-37)^(1/3)-(9√17+37)^(1/3)-1]
所以t^3+t^2+t+3
={t-(1/3)*[(9√17-37)^(1/3)-(9√17+37)^(1/3)-1]}{t^2+(1/3)*[(9√17-37)^(1/3)-(9√17+37)^(1/3)-1]*t-74/9[(9√17-37)^(1/3)-(9√17+37)^(1/3)-1]}