We want the numerical coefficient of the term in x27
of the complete expansion of (x²+5x+2)5(x²-7x+3)9.

(x²+5x+2)5(x²-7x+3)9 equals this product of 14 trinomials:

(x²+5x+2)(x²+5x+2)(x²+5x+2)(x²+5x+2)(x²+5x+2)*(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)

The complete expansion can be considered as the sum of all products of 14
factors where one factor is taken from each of the 14 trinomials.

since 27 = 13·2+1 The only way to get a term in x27 is to take the
x² term from 13 of the trinomials and the x-term from the remaining one. 

There will be 14 terms in x27 to combine, of which 

5 will be of this form  

(5x)*(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²) = 5x27

with the (5x) in the 5 positions 1-5

and the other 9 will be of this form:

(x²)(x²)(x²)(x²)(x²)*(-7x)*(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²) = -7x27

with the (-7x) in the 9 positions 6-14

Therefore the term in x27 is

5(5x27) + 9(-7x27) = 25x27 - 63x27 = -38x27

The coefficient of that term is -38.
 
The answer is -38.

Edwin