Ответ:

f(x)=(x−4)

2n+(x−3)

n−1

x 2−7x+12=(x−3)(x−4)

f(3)=(3−4)

2n+(3−3)

n−1=(−1)

2n+0

n−1=0=>f(x)⋮(x−3)

f(4)=(4−4)

2n+(4−3)

n −1=(0)

2n+1

n−1=0=>f(x)⋮(x−4)