Let px+q be the quotient when the polynomial is divided by x²-5x leaving
a remainder of 6x-15, and
Let rx+s be the quotient when the polynomial is divided by x²-5x+8 leaving
a remainder of 2x-7,

We use the fact that

(divisor)(quotient) + remainder = polynomial of degree 3

   (x²-5x)(px+q) + (6x-15) = polynomial 
  (x²-5x+8)(rx+s) + (2x-7) = polynomial 

Thus we have the identity

   (x²-5x)(px+q) + (6x-15) = (x²-5x+8)(rx+s) + (2x-7) 

which must be true for all values of x. Lots of terms will become 0
if we substitute x=0 and x=5

If we substitute x=0 and solve for s, we get s = -1

   (x²-5x)(px+q) + (6x-15) = (x²-5x+8)(rx-1) + (2x-7)

If we substitute x=5, we get r = 1/2 = 0.5

Thus the polynomial is equal to

  (x²-5x+8)(0.5x-1) + (2x-7)

When we multiply that out, we get

  0.5x³-3.5x²+11x-15   <-- answer

--------------------------------------------

Checking:
                   0.5x- 1                            0.5x- 1 
x²-5x+0)0.5x³-3.5x²+11x-15         x²-5x+8)0.5x³-3.5x²+11x-15
        0.5x³-2.5x²+ 0x                    0.5x³-2.5x²+ 4x 
                -x²+11x-15                         -x²+ 7x-15
                -x²+ 5x+ 0                         -x²+ 5x- 8
                     6x-15                              2x- 7
                  
Edwin