Add maths help? (remainder theorem)

When ax² + bx -6 is divided by x + 3 ,the remainder is 9. Find, in terms of 'a' only, the remainder when 2x³ - bx² + 2ax - 4 is divided by x - 2 .

Clearly a(9)+b(-3)-6=9 or 9a = 15 + 3b or b = 3a-5
And so the given remainder
2(8)-b(4)+2a(2)-4 = 12-4b+4a = 12-4(3a-5)+4a = 32 -8a.

in first part after dividing you get remainder as r1 = - 6 - 3b + 9a.............
since r1=9 given

so,

9 = - 6 - 3b + 9a
divide the whole equation by 3 we get
3 = - 2 - b +3a or 3a-b=5 or b = 3a - 5

in second division we get r2 as r2 = - 4 + 2( 2a + 4 - b )
or on simplification r2 = 4a - 2b + 4
put the value of b in r2

r2 = 4a - 2(3a - 5) + 4
so,

r2 = - 2a + 14

1.
x=-3
9a-3b-6+9=9a+15=3b
3a+5=b

16-b(2^2)+2(2)a-4=4a-4(3a+5)+12
b=-8(a+1)