Since our divisor is quadratic (degree 2), the remainder must be linear (degree 1) :

\mathop{R}\left(x\right)=ax+b

Note: a or b could be 0 for all we know, technically the degree is 1 or smaller.

The divisor happens to have integer roots:

x^2-5x+6=\left(x-3\right)\left(x-2\right)

If we write the result of the polynomial division as

2x^3-5x^2-4x+25=\left(x^2-5x+6\right)\cdot\mathop{Q}\left(x\right)+ax+b

then when we plug in the roots of the quadratic, \mathop{Q}\left(x\right) gets multiplied by 0, leaving only \mathop{R}\left(x\right):

\mathop{P}\left(2\right)=\mathop{R}\left(2\right),\quad\mathop{P}\left(3\right)=\mathop{R}\left(3\right)

Plugging in x=2:

2\cdot2^3-5\cdot2^2-4\cdot2+25=0\cdot\mathop{Q}\left(x\right)+a\cdot2+b

which simplifies to

13=2a+b\quad\left(1\right)

Plugging in 3 gives:

22=3a+b\quad\left(2\right)

Taking \left(2\right)-\left(1\right):

22-13=9=a

Which leads to b=-5. Thus

\boxed{\mathop{R}\left(x\right)=9x-5}

Polynomial division, factors and remainders