Polynomials

Let f(x) = Q(x) * (x+2)(x+3) + ax + b
By remainder theorem :
f(-2) = 0 + a(-2) + b = 11
b - 2a = 11......(1)
f(-3) = 0 + a(-3) + b = 6
b - 3a = 6......(2)
(1) - (2) :
a = 5
sub it to (1):
b - 2*5= 11
b = 21
f(x) becomes Q(x)*(x+2)(x+3) + 5x + 21
The remainder when f(x) is divided by (x+2)(x+3) is 5x + 21