Question

The polynomial $ax^{3} + bx^{2} + x - 6$ has x+2 as a factor and leaves a remainder 4 when divided by x-2. Find the values of a and b..

Answer 1

on dividing from (x + 2) in (ax³ + bx² + x -6) remainder is 4(b-2a) - 8
for (x + 2) be a factor 4(b-2a) - 8 = 0 
b-2a = 2-------------------------(1)
on dividing from (x - 2) in (ax³ + bx² + x - 6) remainder is 4(2a + b) - 4
remainder = 4                       (given)
4(2a + b) - 4 = 4
2a + b = 2 ---------------------------(2)
add equation (1) & (2)
2b = 4
b = 2
put value of b in equation (2)
2a + 2 = 2
2a = 0 
a = 0
hence value of a = 0 ,& value of b = 2