Question 156079
let {{{P(x)=ax^3+bx^2+cx+d}}}
The coefficient of x^2 is 3 less than the coefficient of x^3.
then {{{b=a-3}}}
The coefficient of x is 3 times the coefficient of x^2.
then  {{{c=3b}}}
The remaining coefficient is 2 more than the coefficient of x^3.
{{{d=a+2}}}
The sum of the coefficients is -4.
then
{{{a+b+c+d=-4}}}
using
{{{b=a-3}}}
{{{c=3b=3(a-3)=3a-9}}}
{{{d=a+2}}}

sum is:

{{{a+(a-3)+(3a-9)+a+2=-4}}}
{{{6a-10=-4}}}
{{{6a=6}}}
{{{a=1}}}
then
{{{b=a-3=1-3=-2}}}
{{{c=3a-9=3-9=-6}}}
{{{d=a+2=1+2=3}}}

Answer:{{{P(x)=x^3-2x^2-6x+3}}}