The expression $$2x^{3}+ax^{2}+bx+3$$ where a and b are constants, has a factor of x-1 and leaves a remainder of 15 when divided by x + 2. Then, (a, b) =

Let f(x) = $$2x^{3}+ax^{2}+bx+3$$ 
It is given that (x-1) is a factor of f(x) , so it means x = 1 is a root of f(x)
    f(1) = a + b + 5 = 0
     this implies , a + b = -5 .... Equation 1

It is said that f(x) leaves remainder of 15 when it is divided by (x+2) , so it implies : 
     f(x) = $$\left[2\left[(x+\alpha\right)\left(x\ +\ \beta\right)(x+2)\right]$$ + 15
     $$f(x)-15$$ = $$\left[2\left[(x+\alpha\right)\left(x\ +\ \beta\right)(x+2)\right]$$ 

So, (x+2) is a factor of "f(x)-15" .
  at x = -2, the value of f(x) - 15 = 0
Hence, 15 = -16 + 4a -2b + 3
this implies,  4a - 2b = 28 .... Equation 2

Therefore, by solving both equations 1 & 2 , we get a = 3 & b = -8