(x + 3)*(x - 1)*(x + 2) = x3 +4x2 + x - 6
To find this, start by assuming (x-a)*(x-b)*(x-c) = x3 +4x2 + x - 6
The left-hand side can be expanded, to get: x3 - (a+b+c)*x2 +(a*b+a*c+b*c)*x -a*b*c = x3 +4x2 + x - 6
By equating the coefficients of powers of x on both sides of this equation you get three simultaneous equations in a, b and c, which when solved, produce the result shown above.
By equating the coefficients I mean:
coeff of x2: -(a+b+c) = 4
coeff of x1: a*b + a*c +b*c = 1
coeff of x0: -a*b*c = -6