If (x3−3x2−x+3) is divided by (x-1) we get ax2−bx+c, then value of (a+b+c)
-4
-3
-2
-1
When (x3−3x2−x+3) is divided by (x-1) we get the quotient as x2−2x−3 and 0 as remainder. So, taking the coefficients of a, b, and c= 1 - 2 - 3 = - 4
x3+3x2−4x−12, when divided by (x - 2), leaves a remainder of ax2+bx+c. Find a, b, c