Question

if x ^3 +ax^2+ bx + 6 has (x - 2 )as a factor and leaves a remainder 3 when divided by (x - 3)find the values of a and b.

Answer 1

x³+ax²+bx+6=0
when x=2
8+4a+2b+6=0
4a+2b=-14
2a+b=-7

also
when x=3
then

3³+a3²+b3+6-3=0
27+9a+3b+3=0
9a+3b=-30
3a+b=-10

on solving

2a+b-(3a+b)=-7-(-10)
2a-3a=-7+10
-a=-3
a=3
and thus b=-19